241.  10 playing cards and a dark room
   In a dark room there is a deck of cards with 10 face up and the rest face down. Your task is to divide the total deck into two piles so that the same number of face up cards is in each pile. Remember, it is dark so you can't see the cards. How do you do it?
Answer
Problem 241 Answer
Problem 241 Answer
Place ten cards in one pile with the other 42 cards in the other. Then, flip over the tencard pile.
Michael Shackleford
Solution
Problem 241 Solution
Problem 241 Solution
Place ten cards in one pile with the other 42 cards in the other. Let's say the 10card pile has x face up cards and 10x face down cards. The 42card pile will have 10x face up cards.
Then flip over the 10card pile. It will now have 10x face up cards, the same as the 42card pile.
Michael Shackleford
  113.  10,000 digits of e problem
   The following is a distribution of the first 10,000 digits of e: 0 974 1 989 2 1004 3 1008 4 982 5 992 6 1079 7 1008 8 996 9 968 It is speculated that the number 6 appears a disproportionately high number of times and thus the digits are not distributed randomly. Test the hypothesis that these digits form a random sample such that the outcome of 10,000 truly random digits would pass the test 95% of the time.
Answer
Problem 113 Answer
Problem 113 Answer
The digits pass the test of randomness.
Michael Shackleford, A.S.A.
Solution
Problem 113 Answer
Problem 113 Solution
Let Q be the chisquared goodness of fit statistic.
Q = The sum for i=1 to 10 of ( (N_{i}  np_{i})^{2}/np_{i} )
= The sum for i=1 to 10 of ( (N_{i}  1000)^{2}/1000 )
= (676+121+16+64+324+64+6241+64+16+1024)/1000 = 8.61
According the the chisquared table with 9 degrees of freedom a Q statistic of greater
than 16.92 shold fail the test. 8.61 is well under 16.92 so it should pass the test. In
other words the digits do appear to be random.
At a level of significance such that 10,000 random digits passed 60% of the time the
10,000 digits of e would still pass, but at a 50% level they would not.
Michael Shackleford, A.S.A.
  190.  100 coin problem
   There is a table in a dark room with 100 coins. You can't see anything nor feel which sides are up. 80 are heads up and 20 are tails up. Another table in the room has nothing on it. How can you get the same number of coins tails up on both tables with at least one penny on each table?
Answer
Problem 190 Answer
Problem 190 Answer
First, move any 20 pennies to the other table, keeping the same side on each penny up.
Second, flip all 20 pennies on the other table.
Explanation
Initially table 1 will have 80 heads and 20 tails, and table 2 will have 0 heads and 0 tails.
Let t be the number tails moved, and h be the number of heads.
After the move table 1 will have 80h heads and 20t tails, and table 2 will have h heads and t tails.
After the flip table 1 will have 80h heads and 20t tails, and table 2 will have t heads and h tails.
So, at this point table 1 has 20t tails and table 2 has h tails. We moved 20 coins: h heads, and t tails, so h+t=20, or h=20t. 20t and h must be the same amount, thus both tables have the same number of tails, although we don't know what that number is.
Thanks to Dan Florentin for this problem.
Michael Shackleford, A.S.A.
  26.  100 single women problem
   The king has 100 young ladies in his court each with an individual dowry. No two dowrys are the same. The king says you may marry the one with the highest dowry if you correctly choose her. The king says that he will parade the ladies one at a time before you and each will tell you her dowry. Only at the time a particular lady is in front of you may you select her. The question is what is the strategy that maximizes your chances to choose the lady with the largest dowry?
Hint: You should let x ladies go by and choose the first one with a dowry greater than the maximum of the first x. Of course you will pass the highest one if she is among the first x, but that is part of the game.
Answer
Problem 26 Answer
Problem 26 Answer
You should let 37 ladies go by and select the first
one with a dowry greater than the maximum of the
first 37 dowrys.
Michael Shackleford, A.S.A.
Solution
Problem 26 solution
Problem 26 Solution
Let x be the number of ladies you pass before choosing
the next lady with a dowry greater than the first
x ladies. Let max _{x} be the maximum dowry
in the first x ladies.
The probability of winning is sum for i=x+1 to 100
of the probability that the highest dowry is in the
ith position multiplied by the probability that you
choose it if it is in that position.
This is based on the condition probability formula:
Pr(A)=Sum for i=1 to n of Pr( A  B_{i} ) * Pr( B_{i} ),
where the sum for i=1 to n of Pr(B_{i})=1.
For all i the probability that the
highest dowry is in that position is 1/100.
The probability that you will choose the dowry if
it is in the ith position is equal to the probability
that the highest of the first i1 dowries belongs
to one of the first x ladies. This equals x/(i1). If
the highest dowry of the first i1 were not in the first x ladies
you would choose it before getting to the largest dowry,
thus losing the game.
For example if you let 30 ladies pass the probability that the
highest dowry is in the 75th postion and that you will choose it
equals 1/100 * 30/74.
Let A denote the overall probability of winning.
The proability of winning given x is the sum for i=x+1 to 100
of x/(i1) * 1/n. This equals:
Pr(1/n * [ 1 + x/(x+1) + x/(x+2) + ... + x/99 ] ).
Some value of x must be optimal to maximize this formula. This value
of x must the first such that Pr(Ax+1)  Pr(Ax) < 0.
Pr(Ax+1)  Pr(Ax) = 1/n * [ 1/(x+2) + 1/(x+3) + 1/(x+4) + ... + 1/99  x/(x+1) ].
let y=x+1.
Pr(Ax+1)  Pr(Ax) = 1/n * [ 1/(y+1) + 1/(y+2) + 1/(y+3) + ... + 1/99  (y1)/y ]. =
Pr(Ax+1)  Pr(Ax) = 1/n * [ 1/(y+1) + 1/(y+2) + 1/(y+3) + ... + 1/99  1 + 1/y ]. =
Pr(Ax+1)  Pr(Ax) = 1/n * [ 1/y + 1/(y+1) + 1/(y+2) + 1/(y+3) + ... + 1/99  1 ].
If Pr(Ax+1)  Pr(Ax) = 0, then 1/y + 1/(y+1) + 1/(y+2) + 1/(y+3) + ... + 1/99 = 1.
Theorem: The integral from r to n of 1/x = log(n)  log(r) = log(n/r).
The equation above the theorm is an approximation of the integral in the theorem. By
applying the theorem log(100/y) = 1.
Taking e to the power of both sides: 100/y = e.
y=100/e.
Since e=~2.7182818 y=~36.79, since x=y1 x=~35.79.
Since the integral is an underestimate of the series look at some specific
values of Pr(A) for values of x close to 35.79:
x A
 
35 37.0709%
36 37.1015%
37 37.1043%
38 37.0801%
So the optimal strategy is to let 37 ladies pass and then
choose the first lady with a dowry larger than the greatest of the first 37.
In general the answer is going to be n/e, where n is the total
number of ladies. Because the answer must be an
exact integer and error in applying the theorem mentioned above you should
check the few integers just above the optimal value. As n increases the
error decreases. The probability of winning turns out to approach 1/e =~ 36.79%
as n approaches infinity (I'll leave this proof up to you).
Here are some optimal values of x for various values of n and
the probability of winning given x:
n x Pr(winning)
  
3 2 50.00%
4 2 43.83%
5 3 43.33%
6 3 42.78%
7 3 41.43%
8 4 40.98%
9 4 40.60%
10 4 39.87%
15 6 38.94%
20 8 38.42%
30 12 37.87%
50 19 37.42%
100 38 37.10%
1000 369 36.82%
For more information I suggest section 2.5 of Probability and Statistics,
second edition by Morris H. DeGroot.
Here another version of my solution to this problem
Let x be the number of ladies you pass before choosing
the next lady with a dowry greater than the first
x ladies. Let max_{x} be the maximum dowry
in the first x ladies.
The probability you will select the highest dowry is:
Sum for i=1 to 100x of: ( Probability that i of the
remaining ladies will have a dowry greater than max_{x} )
* (1/i).
This probability for any i is:
(100x)*(99x)*(98x)*...*(101xi) * x

100*99*98*...*(100i)
For example suppose you let 30 ladies go my first and planned to
choose the next one with a dowry greater than the maximum of
the first 30. max _{30} is the largest dowry of the
first 30 ladies. Let Pr(n _{i}) equal the probability
that there are i ladies left with a dowry greater than max _{x}.
The probability of you winning would be:
Pr(n_{1}) * 1 +
Pr(n_{2}) * 1/2 +
Pr(n_{3}) * 1/3 +
Pr(n_{4}) * 1/4 +
Pr(n_{5}) * 1/5 +
.
.
.
Pr(n_{70}) * 1/70.
To derive Pr(n_{i}) consider the probability that the i highest
dowrys fall in the last 100x ladies, multiplied the the probability that
the (i+1)th highest dowry falls in the first x ladies. For example:
Pr(n_{3}) = 70/100 * 69/99 * 68/98 * 30/97.
70/100 = Probability that the highest dowry is not in first 30.
69/99 = Probability that the second highest dowry is not in first 30 (99 positions left, 69
not if first 30).
68/98 = Probability that the third highest dowry is not in first 30.
30/97 = Probability that fourth highest dowry is in first 30.
Personally I had to use a spreadsheet to find that the maximum
probability was for x=37, which was 37.1%.
If you change the number of total ladies from 100 to any other number the strategy
seems to still be to let 37% of them go by first.
Michael Shackleford, A.S.A.
  127.  100 story building and a billiard ball problem
   It is your task to determine how high you can drop a billiard ball without it breaking. There is a 100 story building and you must determine which is the highest floor you can drop a ball from without it breaking. You have only two billiard balls to use as test objects, if both of them break before you determine the answer then you have failed at your task. How can you determine the breaking point in which the maximum necessary dropping is at a minimum. Click on 'answer' for just the minimum maximum and 'solution' for how to determine the breaking point under this minimum.
Answer
Problem 127 Answer
The minimum maximum number of droppings is 14.
Michael Shackleford, A.S.A., 10/31/1998
Solution
Problem 127 Solution
First drop the first ball from the 14th floor. If it breaks you can
determine the exact breaking point with the other ball in at most
13 more droppings, starting at the bottom and going up one floor
at a time.
If the first ball survives the 14 floor drop then drop it again
from the 27th (14+13) floor. If it breaks you can determine the exact
breaking point with at most 12 more droppings.
If the first ball survives the 27 floor drop then drop it again
from the 39th (14+13+12) floor. If it breaks you can determine the exact
breaking point with at most 11 more droppings.
Keep repeating this process always going up one less floor than the
last dropping until the first ball breaks. If it breaks on the x^{th} dropping you will only need at most 14x more droppings
with the second ball to find the breaking point. By the 11th dropping
of the first ball, if you get that far, you will have reached the
99th floor.
Thanks to Alon Amit for this problem.
Michael Shackleford, A.S.A., 10/31/1998
  84.  12 pearls and a scale problem
   In front of you are 12 pearls, 11 being real and one fake. The real ones all weigh the same and the fake one differs in weight from the real ones (may weigh more or less). With a balance scale and three weighings how can you weed out the fake one and determine whether it is too heavy or too light?
Answer
Problem 84 Answer
Problem 84 Answer
Note: The following solution may not be the only one possible. At least one other solution has been submitted to me.
First weigh any 4 pearls against any other 4.
If one side weighs heavier than the other then:
Label the pearls on the heavy side H, the pearls on the light side L, and the
4 other pearls G. Next, weigh two H pearls and two L pearls against 2 G pearls, 1 H pearl, and 1 L pearl
(HHLL vs GGHL).
If the HHLL side goes down then either one of the two H pearls on the left
side from step 2 is heavy or the L pearl on the right side of step 2 is light. Finally
weigh ( HL vs GG ). If the HL side goes down the H pearl is heavy, if the HL side goes up
the L pearl is light, if they stay the same the other H pearl is heavy.
If the HHLL side (from step 2) goes up then either one of the two L pearls on the left
side from step 2 is light or the H pearl on the right side of step 2 is heavy. Finally
weigh ( HL vs GG ). If the HL side goes down the H pearl is heavy, if the HL side goes up
the L pearl is light, if they stay the same the other L pearl is light.
If the two sides from step 2 stay the same you have left one H pearl and one L pearl.
Weigh ( HL vs GG ). If the left side goes down the H pearl is heavy, if it goes up the
L pearl is light.
If the two sides from step one stay the same then label the 8 pearls used in step 1 as
G and the 4 others pearls as B. For the second weighing weigh 3 B pearls against
3 G pearls ( BBB vs GGG).
If the left side goes down one of the B pearls is heavy. Weigh them 2 of
them against each other
the heavy side will have the heavy pearl, if the sides stay the same the
other pearl is heavy.
If the left side goes up one of the B pearls is light. Weigh them 2 of
them against each other
the light side will have the light pearl, if the sides stay the same the
other pearl is light.
If the two sides from the second weighing stay the same then the last B
pearl is heavy or light. Use the third weighing against a good pearl
to determine if it is heavy or light.
Michael Shackleford, A.S.A.
  80.  128 pennies and a blindfold problem
   You are blindfolded before a table. On the table are a very large number of pennies. You are told 128 of the pennies are heads up and the rest are tails up. How can you create two subgroups of pennies, each with the same number of heads facing up?
Answer
Problem 80 Answer
Problem 80 Answer
Create a subgroup of any 128 pennies. Then flip over all 128. That group of
128 and the group of all the remaining pennies will have the same number of heads
facing up.
Michael Shackleford, A.S.A., February 4, 1999
  118.  13 pirates and a safe problem
   Thirteen pirates put their treasure in a safe. They decide that the safe should be able to be opened if any majority of pirates agree but not be able to be opened if any minority agree. The pirates don't trust each other so they consult a locksmith. The locksmith puts a specific number of locks on the safe such that every lock must be opened to open the safe. Then he distributes keys to the pirates such that every pirate has some but not all of the keys. Any given lock can have multiple keys but any given key can only open one lock. What is the least number of locks required?
Answer
Problem 118 Answer
Problem 118 Answer
1716 locks are required.
Michael Shackleford, A.S.A.
Solution
Problem 118 Solution
Problem 118 Solution
The number of ways you can choose 7 out of 13 pirates is 13!/(7!*6!) = 1716, where x! = 1*2*...*x.
Next put 1716 locks on the safe, one for each way to group 7 pirates. For each lock give 7 keys to
a unique group of 7 pirates. This way any given lock
will have a keyholder in any group of 7 or more. For any group of 6 there will be exactly one lock
in which the other 7 pirates have the key. Obviously any group of less than 6 would also be missing
at least one key to at least one lock.
Here are the number of keys required for other numbers of pirates:
Number
of
Pirates Number of Locks
 
3 3
5 10
7 35
9 126
11 462
13 1,716
15 6,435
17 24,310
19 92,378
21 352,716
23 1,352,078
25 5,200,300
27 20,058,300
29 77,558,760
31 300,540,195
33 1,166,803,110
35 4,537,567,650
37 17,672,631,900
39 68,923,264,410
41 269,128,937,220
43 1,052,049,481,860
45 4,116,715,363,800
47 16,123,801,841,550
49 63,205,303,218,876
51 247,959,266,474,050
53 973,469,712,824,060
Michael Shackleford, A.S.A.
  204.  154 Rolls of the Dice in Craps
   On May 23, 2009, a craps player held the dice for 154 rolls at the Borgata casino in Atlantic City (source). What is the probability of going 154 rolls or longer in craps? Please give an expression of the answer and/or a numeric answer.
Answer
Problem 204 Answer
Problem 204 Answer
Question
On May 23, 2009, a craps player held the dice for 154 rolls at the Borgata casino in Atlantic City (source). What is the probability of going 154 rolls or longer in craps? Please give an expression of the answer and/or a numeric answer.
Answer
The answer is P1 + P2 + P3 + P4, where
[ P1 ] [12 3 4 5]^153 [1]
[ P2 ] [ 6 27 0 0] [0]
[ P3 ] = (1/36)^153 × [ 8 0 26 0] × [0]
[ P4 ] [10 0 0 25] [0]
The exact answer is:
P1 = 3.12763 × 10^{11}
P2 = 4.63460 × 10^{11}
P3 = 4.95558 × 10^{11}
P4 = 5.17044 × 10^{11}
The sum of these probabilities is 1.78882 × 10^{10} = 1 in 5,590,264,072
Acknowledgement: My thanks to BruceZ for this help with this problem.
Michael Shackleford, ASA — June 2, 2009
Solution
Problem 204 Solution
Problem 204 Solution
Question
On May 23, 2009, a craps player held the dice for 154 rolls at the Borgata casino in Atlantic City (source). What is the probability of going 154 rolls or longer in craps? Please give an expression of the answer and/or a numeric answer.
Answer
Solution
There are four possible states the shooter can be in. Let's define them as follows.
State 1 = Come out roll
State 2 = Point of 4 or 10
State 3 = Point of 5 or 9
State 4 = Point of 6 or 8
The probabilities are recursive. Let p(x,r) represent the probability of being in state x before roll r. Based on simple dice probabilities, p(x,r) can be expressed as follows:
(1) P(1,r) = (12/36)×p(1,r1) + (3/36)×p(2,r1) + (4/36)×p(3,r1) + (5/36)× p(4,r1)
(2) P(2,r) = (6/36)×p(1,r1) + (27/36)×p(2,r1)
(3) P(3,r) = (8/36)×p(1,r1) + (26/36)×p(3,r1)
(4) P(4,r) = (10/36)×p(1,r1) + (25/36)×p(4,r1)
In more plain simple English, equation (1) is saying that the probability of being in a come out roll before roll r is the sum of the following:
 Product of the probability of being in a come out roll the previous turn, and the probability of staying in a come out roll (12/36).
 Product of the probability of rolling for a 4 or 10 the previous roll, and the probability of making the point (3/36), resulting in a come out roll.
 Product of the probability of rolling for a 5 or 9 the previous roll, and the probability of making the point (4/36), resulting in a come out roll.
 Product of the probability of rolling for a 6 or 8 the previous roll, and the probability of making the point (5/36), resulting in a come out roll.
Equation (2) is saying the probability of rolling for a 4 or 10 before roll r is the sum of:
 Product of the probability of being in a come out roll the previous turn, and the probability of rolling a 4 or 10 (6/36).
 Product of the probability of rolling for a point of 4 or 10 the previous turn, and the probability of not rolling the desired point or a 7 (27/36).
Equation (3) is saying the probability of rolling for a 5 or 9 before roll r is the sum of:
 Product of the probability of being in a come out roll the previous turn, and the probability of rolling a 5 or 9 (8/36).
 Product of the probability of rolling for a point of 5 or 9 the previous turn, and the probability of not rolling the desired point or a 7 (26/36).
Equation (4) is saying the probability of rolling for a 6 or 8 before roll r is the sum of:
 Product of the probability of being in a come out roll the previous turn, and the probability of rolling a 6 or 8 (10/36).
 Product of the probability of rolling for a point of 6 or 8 the previous turn, and the probability of not rolling the desired point or a 7 (25/36).
Next, let's express the these four equations in the form of a matrix.
[ P(1,r) ] [12 3 4 5] [ P(1,r1) ]
[ P(2,r) ] [ 6 27 0 0] [ P(2,r1) ]
[ P(3,r) ] = (1/36) × [ 8 0 26 0] × [ P(3,r1) ]
[ P(4,r) ] [10 0 0 25] [ P(4,r1) ]
The intial roll is always a come out roll, so the initial state is:
[ 1 ]
[ 0 ]
[ 0 ]
[ 0 ]
The probability of each state before the 154th roll is expressed as follows. The reason we take the matrix to the 153rd power, and not the 154th, is there are 153 transformations since the initial state.
[ P(1,154) ] [12 3 4 5]^153 [1]
[ P(2,154) ] [ 6 27 0 0] [0]
[ P(3,154) ] = (1/36)^153 × [ 8 0 26 0] × [0]
[ P(4,154) ] [10 0 0 25] [0]
The exact answer is:
P(1,154) = 3.12763 × 10^{11}
P(2,154) = 4.63460 × 10^{11}
P(3,154) = 4.95558 × 10^{11}
P(4,154) = 5.17044 × 10^{11}
The sum of these probabilities is 1.78882 × 10^{10} = 1 in 5,590,264,072
Acknowledgement: My thanks to BruceZ for this help with this problem.
Michael Shackleford, ASA — June 2, 2009
  132.  21 squares in a big square problem
   Find a way to fit 21 squares, each with a unique integer side length, that can be arranged into the shape of a square (this problem is very challenging).
Answer
Problem 132 Answer
Problem 132 Answer
The following are the lengths of the various sides, according to
the lables in the image:
a=50
b=35
c=27
d=8
e=19
f=15
g=17
h=11
i=6
j=24
k=29
l=25
m=9
n=2
o=7
p=18
q=16
r=42
s=4
t=37
u=33
Thanks to Terry W. Ryder of San Leandro, CA and
New Scientist magazine for the problem and Terry for the graphic.
This problem inspired my friend Dick Tucker to not only solve the
problem but create a piece of art with it using acrylic plexiglas.
Michael Shackleford, A.S.A., February 23, 1999
Solution
Problem 132 Solution
Problem 132 Solution
Cosinder all the ways you can define the length of an edge of the outer
square, where each variable represents the length of the side or the
corresponding square in the image:
 a+b+c
 a+b+d+e
 a+f+g+h+e
 a+f+g+i+j
 k+l+m+n+g+h+e
 k+l+m+n+g+i+j
 k+l+m+o+p+j
 k+l+q+p+j
 k+l+q+r
 k+s+t+r
 u+t+v
 a+k+u
 a+l+s+u
 a+l+t
 b+f+l+s+u
 b+f+l+t
 b+f+m+q+t
 b+f+n+o+q+t
 b+g+o+q+t
 b+g+q+r
 b+h+i+p+r
 b+h+j+r
 c+d+h+i+p+r
 c+d+h+j+r
 c+e+j+r
Next take 21 of these equations and set them equal to 1. You can not
just use any 21, I would suggest taking out four from various places among
the list as opposed to four in a row. After taking out four of the above equations
I was left with the following matrix:
1  1  1  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  1 
1  1  0  1  1  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  1 
1  0  0  0  1  1  1  1  0  0  0  0  0  0  0  0  0  0  0  0  0  1 
1  0  0  0  0  1  1  0  1  1  0  0  0  0  0  0  0  0  0  0  0  1 
0  0  0  0  0  0  1  0  1  1  1  1  1  1  0  0  0  0  0  0  0  1 
0  0  0  0  0  0  0  0  0  1  1  1  1  0  1  1  0  0  0  0  0  1 
0  0  0  0  0  0  0  0  0  1  1  1  0  0  0  1  1  0  0  0  0  1 
0  0  0  0  0  0  0  0  0  0  1  1  0  0  0  0  1  1  0  0  0  1 
0  0  0  0  0  0  0  0  0  0  1  0  0  0  0  0  0  1  1  1  0  1 
1  0  0  0  0  0  0  0  0  0  1  0  0  0  0  0  0  0  0  0  1  1 
1  0  0  0  0  0  0  0  0  0  0  1  0  0  0  0  0  0  1  0  1  1 
1  0  0  0  0  0  0  0  0  0  0  1  0  0  0  0  0  0  0  1  0  1 
0  1  0  0  0  1  0  0  0  0  0  1  0  0  0  0  0  0  1  0  1  1 
0  1  0  0  0  1  0  0  0  0  0  0  1  0  0  0  1  0  0  1  0  1 
0  1  0  0  0  1  0  0  0  0  0  0  0  1  1  0  1  0  0  1  0  1 
0  1  0  0  0  0  1  0  0  0  0  0  0  0  1  0  1  0  0  1  0  1 
0  1  0  0  0  0  1  0  0  0  0  0  0  0  0  1  0  1  0  0  0  1 
0  1  0  0  0  0  0  1  1  0  0  0  0  0  0  1  0  1  0  0  0  1 
0  0  1  1  0  0  0  1  1  0  0  0  0  0  0  1  0  1  0  0  0  1 
0  0  1  1  0  0  0  1  0  1  0  0  0  0  0  0  0  1  0  0  0  1 
0  0  1  0  1  0  0  0  0  1  0  0  0  0  0  0  0  1  0  0  0  1 
Then use Cramer's Rule to solve the equations. In Excel the MDETERM(a1:u21)
command will give you the determinant of the matrix bounded by a1 and u21,
making this process quick and mostly automated. If the determinant used for
the denominator is zero then choose a different set of 21 equations.
When solving for a..u ignore the deonominator determinant since it will
be the same for all 21 variables. After solving for all 21 looking for
common factors among the solutions and divide.
Thanks to Terry W. Ryder of San Leandro, CA and
New Scientist magazine for the problem and Terry for the graphic.
Michael Shackleford, A.S.A., February 23, 1999
  219.  23 prisoners and two light switches problem
   There is a prison with 23 prisoners. The warden brings all the prisoners to a meeting and explains that he set up a room with two switches, labeled A and B. Each can be in an on or off position only. Starting the next day, the warden will pick a prisoner at random, with replacement, once per day. That prisoner will be led to the switches room. He must then switch one, and only one, switch of his choice.
When any member of the group declares "We have all been to the switch room," the warden will check his records to see if it is true. If it is true, then each prisoner will be set free immediately. If not, they will all be given the death penalty immediately.
The warden explains they can have the rest of the day to discuss strategy. Then they will all be separated and will never have a chance to communicate again, except via how they set the switches.
One final detail, the warden will set the switches in their initial state at random and not inform the prisoners beforehand of the initial setting.
Answer Problem 219 Answer
There are various solutions to the problem but most are variants of the one below. I believe my answer will get the prisoners released in the least number of expected days.
 The prisoner called to the switch room on the first day will be designated the leader. If switch A is in the up position, then he will flip in the down position. If it is already down, he will flip switch 2 (the dummy switch). He will start his running count of visits to 1.
 Starting with the second day, do the following:
 If a nonleader is called, and switch A is down position, and he has never fipped switch A, then he will flip switch A to the up position. This is the way to announce that one more person has visited the switch room.
 Otherwise, if a nonleader is called into the switch room, and any of the conditions are not met, then he should flip switch B.
 If the leader is called back into the room and switch A is in the up position it means a new prisoner has entered. The leader will flip A down and add one to his running count.
 If the leader is called back into the room and switch A is in the down position then it will mean no new prisoner has entered the room since the last time he was there. He will flip switch B.
 When the leader reaches a running count of 23, he can safely declare that everyone has been to the switch room at least once.
This method requires an average of 22*23 + sum for i=1 to 22 of 23/(23i), which equals 591.8887 days.
Michael Shackleford, March 25, 2017
  221.  25 horses problem
   You have 25 horses and a track that can race five horses at a time. The only thing you learn from each race is the winning order, from 1 to 5. You don't have a watch. Each horse always runs at its own constant speed. What is the least number of races required to determine the fastest three horses in order and how should it be done?
Answer Problem 221 Answer
The answer is 7.
Michael Shackleford, June 11, 2017
Solution
Problem 221 Solution
 Divide the 25 horses into five groups of five.
 Races 15: Race each group and note the top three horses, in order. Call these the heats.
 Race 6: Race the winner of each race from step 2. The winner of this race will be the fastest horse.
 Race 7: Race the the following five horses:
 Second and third place horses from race 6.
 The two horses from the group of five second place horses from step 1 that finished second in its heat to the winning two horses in race 6.
 The horse from the third place group that finished third in its heat to the winning horse in race 6.
There is a good explanation at A Collection of Quant Riddles With Answers of why this solution works.
Michael Shackleford, ASA — April 5, 2017
  239.  3 hot dog vendors and a beach
   There is a beach one mile long. It is a nice day so a big crowd of people are expected, who will be equally distributed along the beach. When anyone gets hungry, he heads to the nearest hot dog stand. Three logicians operate hot dog carts and arrive to the beach one at a time before the crowds arrive. A is first, then B, and finally C. All three will locate their stand with the goal of maximizing sales for the day. Once a spot is picked, the vendor must stay there all day. The vendors don't trust each other, so collusion is out. B will note the location of A and choose his location accordingly. C will note the locations of A and B and pick his location accordingly. Where should A set up his stand? How much of the beach will each vendor get, assuming all follow optimal strategy. You may assume that if a vendor is indifferent between multiple locations, then he will pick randomly.
Answer
Problem 239 Answer
Problem 239 Answer
The first should go either a smidge to the left of the point 1/4 from the left end of the beach and a smidge to the right of the point 3/4 from the right end.
The second vendor will pick the spot from A's two choices that A didn't pick.
The third vendor will go in the middle.
This will give A and B 3/8 of the beach each, and C 1/4.
Michael Shackleford
Solution
Problem 239 Solution
Problem 239 Solution
It stands to reason that A and B will want C to go in the middle, least he be the closest to an edge and cut them off from that section of the beach. How far can they extend away from an edge and still keep C in the middle?
Let's call x the distance from either edge that would put C indifferent between going right next to A and B but closer to an edge and the middle.
At x, assuming C goes in the middle, A will get x + (0.5  x)/2 = x/2 + 1/4 of the beach. Same with B. C will bet (12x)/2 = 1/2  x of the beach.
If C located right next to A or B, but closer to an edge, he would get territory of length x. So, we want to equate x and (1/2  x) to find that indifference point for C.
x = 1/2  x
2x = 1/2
x = 1/4
To make sure C isn't indifferent, and barely prefers the middle, A and B should go just a smidge closer to the edges than the 1/4 and 3/4 points. This will give A and B a smidge less then 3/8 of the beach each and C a smidge more than 1/4.
Michael Shackleford
  240.  4 hot dog vendors and a beach
   Same problem as 239, except four vendors.
Answer
Problem 240 Answer
Problem 240 Answer
A will set is stand a smidge less than 1/6 of a mile from either end.
B will do the same as A, but at the opposite end.
C will set his stand exactly in the middle of the beach.
D will be indifferent between placing his stand between either (1) A and C and (2) B and C.
Michael Shackleford
Solution
Problem 240 Solution
Problem 240 Solution
A will want to pick a spot close to either edge, maximizing his space without giving another vendor the incentive to put his stand right next to his on the side close to the edge. It is good to be close to an edge, because you get every customer between you and that edge. Let's just say he picks the left side of the beach.
B will want to do the same, but on the right side.
C will pick a spot in the middle, putting D to an indifference point which side of C to go on.
D will arbitrarily pick a spot between A and C or B and C.
The question is how far from the edge should A and B pick?
We want D to pick on either side of C. Let's ask the question at what point would D be indifferent between going just to the left of A, between A and C, and between B and C.
If A sets his stand at point x, then D would get x miles just to the left of A or (1/2x)/2 by going between A and C or B and C. Let's set these two equations equal to each other and solve, to find the indifference point for x.
x = (0.5x)/2
2x = 0.5  x
3x = 0.5
x = 1/6
However, we don't want D to be indifferent, we want him to pick a spot on either side of C. So, we pick a spot a smidge to the left of 1/6, to get D to go next to C.
Same logic for B  a smidge to the right of 5/6.
C will want to make D indifferent to going on either side of him, which is obviously 1/2.
D will then arbitrarily picks between 1/3 and 2/3.
A will then get 1/4 if D picks 3/8 and 1/3 if he picks 5/8. The average of those two is 7/24.
B is in the same situation as A, so 7/24 on average to him too.
Let's say D flips a coin and goes in the 1/3 spot. Then C will get everything from the 5/12 to the 2/3 point, which is 1/4 mile.
With D and the 1/3 spot, he will everything from the 1/4 to the 5/12 points, which is 1/6 mile.
Michael Shackleford
  165.  64minute fuse problem
   You have a box of matches and six 64minute fuses. Each fuse burns exactly 64 minutes but the rate of burning is inconsistent. A half a fuse would not necessarily burn in 32 minutes. The fuses are also not equally inconsistent. Two cut fuses of the same length would not necessarily burn in the same time. Without the use of a clock, scissors, or anything other than the matches and the fuses how can you create a 60minute fuse?
Answer
Problem 165 Answer
Problem 165 Answers
Following is my original solution.
 At time t=0 light fuse 1 at both ends and fuses 2, 3, 4, and 5 at one end. Fuse 1 will have 32 minutes remaining, fuses 25 will have 64 minutes remaining.
 At time t=32 fuse 1 will have burned out. At this moment light fuse 2 at the other end. Fuse 2 will have 16 minutes remaining, fuses 35 will have 32 minutes remaining.
 At time t=48 fuse 2 will have burned out. At this moment light fuse 3 at the other end. Fuse 3 will have 8 minutes remaining, fuses 4,5 will have 16 minutes remaining.
 At time t=56 fuse 3 will have burned out. At this moment light fuse 4 at the other end. Fuse 4 will have 4 minutes remaining, fuse 5 will have 8 minutes remaining.
 At time t=60 fuse 4 will have burned out. At this moment start fuse 6. Fuse 5 will have 4 minutes remaining, fuse 6 will have 64 minutes remaining.
 At time t=64 fuse 5 will have burned out. At this moment snuff out fuse 6. Fuse 6 will have 60 minutes remaining when relit.
Next is another solution submitted by Michael Nolan.
Solution 2
 At time t=0 light fuse 1 at both ends and fuse 2 at one end. Fuse 1 will have 32 minutes remaining and fuse 2 will have 64 minutes remaining.
 At time t=32 fuse 1 will have burned out. At this moment light the other end of fuse 2 and fuse 3 at one end. Fuse 2 will have 16 minutes remaining and fuse 3 will have 64 minutes remaining.
 At time t=48 fuse 2 will have burned out. At this moment light fuse 3 at the other end and light fuse 4 at one end. Fuse 3 will have 24 minutes remaining and fuse 4 will have 64 minutes remaining.
 At time t=72 fuse 3 will have burned out. At this moment light fuse 5 from both ends. Fuse 4 will have 40 minutes remaining and fuse 5 will have 32 minutes remaining.
 At time t=104 fuse 5 will have burned out. At this moment light fuse 4 from the other end and light fuse 6 from one end. Fuse 4 will have 4 minutes remaining and fuse 6 will have 64 minutes remaining.
 At time t=108 fuse 4 will have burned out. At this moment snuff out fuse 6. Fuse 6 will have 60 minutes remaining.
Next is another solution submitted by Michael Nolan and Andy 'Deuce' Smith.
Solution 3
 Light #1 at both ends, and #2 and #3 at one end. Unlight #2 and #3 when #1 burns out. This makes #2 and #3 both 32minute fuses.
 Light #2 at both ends, and #4 at one end. Unlight #4 when #2 burns out. This makes #4 a 48minute fuse.
 Light #4 at both ends, and #3 at one end. Unlight #3 when #4 burns out. This makes #3 an 8minute fuse.
 Light #3 at both ends, and #5 at one end. Unlight #5 when #3 burns out. This makes #5 a 60minute fuse.
Next is another solution submitted by Michael Nolan and Kenneth R. Smith.
Solution 4
 At time t=0 light fuse 1 at both ends and fuses 2, 3, and 4 at both ends. Fuse 1 will have 32 minutes remaining, fuses 24 will have 64 minutes remaining.
 At time t=32 fuse 1 will have burned out. At this moment light fuse 2 at the other end. Fuse 2 will have 16 minutes remaining, fuses 3,4 will have 32 minutes remaining.
 At time t=48 fuse 2 will have burned out. At this moment light fuse 3 at the other end. Fuse 3 will have 8 minutes remaining, fuses 4 will have 16 minutes remaining.
 At time t=56 fuse 3 will have burned out. At this moment light fuse 4 at the other end and fuse 5 at one end. Fuse 4 will have 4 minutes remaining and fuse 5 will have 64 minutes remaining.
 At time t=60 fuse 4 will have burned out. At this moment snuff out fuse 5. Fuse 5 will have 60 minutes remaining when relit.
Credit and thanks for this problem goes to Craig Olson, Alan Goldberg, and the Nov/Dec 1999 Contingencies magazine.
Michael Shackleford, ASA  July 1, 2001
  97.  A dog, a chain, and a lighthouse problem
   A dog is chained to a point of the side of a circular lighthouse of radius 1. The length of the dog's chain is pi. The dog may not go inside the lighthouse. How much area can the dog cover?
Answer
Problem 97 Answer
Problem 97 Answer
The answer is 5/6 *pi ^{3}.
Michael Shackleford, A.S.A.
Solution
Problem 97 Solution
Problem 97 Solution
Lets call x the angle formed by (1,0), (0,0), and the point where dog's chain
leaves the edge of the lighthouse, assuming the dog is trying it extending as
far as possible causing the straight part of the chain to be tanjent to the
edge of the lighthouse. For now lets only concern ourselves with the area
to the right of x=1 and above y=0.
Since the radius of the lighthouse is 1 the length of the straight part of the
chain is the same as x. Through a little trigonomtry we can get the coordinates
of the dog given x as (cos(x)+x*sin(x),sin(x)x*cos(x)). The distance from
(0,0) to the dog, given x, conveniently works out to (1+x^{2})^{1/2},
using the length of a line segment formula.
Lets call y the angle fromed by (1,0), (0,0), and the location of the dog.
Through some more trigonomty (again left up to you) we can solve for y
in terms of x as y=xtan^{1}. Now we are ready to integrate.
The area in terms of polar coordinates is 1/2 * the integral over the range
of the angle of the radius squared. We can not simply take the integral from
0 to Π of 1+x^{2} because this is not the radius formed by the angle
x, but the radius formed by y. We could take the integral over the range of
the angle y but finding the coordinates of the dog in terms of y is very hard
(I couldn't do it). However we can take the ingregal as x goes from 0 to
Π of 1+x^{2} multiplied by the change in y given a change a change
in x. Since y=xtan^{1}(x). dy/dx = 11/(1+x^{2})!
So now the area becomes the integral as x goes from 0 to Π of 1/2 the product
of 1+x^{2} and 11/(1+x^{2}) which equals 1/2 the integral from
0 to Π of x^{2} which equals Π^{3}/6.
This integral, however, does not cover the area bounded by the triangle (0,0), (1,0) and
(1,Π). So we must add this area which is Π/2. Yet we must subtract the area
inside the lighthouse which the dog may not go in, which is also Π/2. So these
two modifications conventiently cancel each other out.
The area of the quarter circle to the left of x=1 and above y=0 is Π^{3}/4,
since the length of the chain is Π and the dog covers a semicircle to the left of the
lighthouse. So the total area above y=0 is 5/12 * Π^{3}. Finally double
this for the area under y=0 and the total area is 5/6 * Π^{3}
Here is another solution I received from Patrick.
I would like to thank Steven Lutz for sending me
this problem.
Michael Shackleford, A.S.A.
  32.  A fence, board, and a shadow problem
   You have a 6 foot fence and at 15 foot board. The sun is shining directly overhead. You want to lean your board against the fence so that as large a shadow as possible covers your neighbor's yard. What is the maximum length of the shadow you can form? The ladder must rest against the fence and touch the ground on your side of the fence.
Answer
Problem 32 Answer
Problem 32 Answer
The answer is (540 ^{2/3}36) ^{1/2} * (15/540 ^{1/3}1) =~ 4.635873 feet.
The shadow over your yard is (540^{2/3}36)^{1/2} =~ 5.505698 feet.
The angle formed is approximately 47.46 degrees.
Michael Shackleford, A.S.A.
Solution
Problem 32 Solution
Problem 32 Solution
Let x be the shadow over your yard.
Let y be the shadow over neighbor's yard.
By similar triangles: x/(36+x^{2})^{1/2}=(x+y)/15.
y=15x/(36+x^{2})^{1/2}x.
y^{'}=(((15*(36+x^{2})^{1/2})(15x^{2}*(36+x^{2})^{1/2}))
/ 36+x^{2})  1.
Then set y^{'}=0 and solve for x.
Finally subsitute x in the similar triangles formula to solve for y.
Michael Shackleford, A.S.A.
  72.  A jar and an amoeba problem
   A jar begins with one amoeba. Every minute, every amoeba turns into 0, 1, 2, or 3 amoebae with probability 25% for each case ( dies, does nothing, splits into 2, or splits into 3). What is the probability that the amoeba population eventually dies out?
Answer
Problem 72 Answer
The answer is sqr(2)1 =~ .414213562
Michael Shackleford, A.S.A.
Solution
Problem 72 Solution
Let p be the probability that just one amoba eventually dies out, and all its
decendants. The probablity of n amoba all dying out is p^{n}.
After the first turn there are four possibilities for the number of
amoba left, 0, 1, 2, 3 with probabilites of eventually dying out of
0, p, p^{2}, and p^{3}. With each outcome being
equally likely the probability of all amoba eventually dying out is
1/4*[ 1 + p + p^{2} +
p^{3} ]. So p = 1/4*[ 1 + p + p^{2} + p^{3} ] or
1 3p + p^{2} + p^{3} = 0. This reduces to
(p  1)(p^{2} + 2p  1) = 0.
The solutions for p are 1, (sqr(2)1), and (sqr(2)1). The only one
which satisfies the constraints of the problem is sqr(2)1.
Michael Shackleford, A.S.A.
  150.  ABCDE * 4 = EDCBA problem
   ABCDE * 4 = EDCBA. Solve for A,B,C,D, and E where each is a unique integer from 0 to 9.
Answer
Problem 150 Answer
Problem 150 Answer
ABCDE=21978.
Michael Shackleford, ASA, December 28, 1999
Solution
Problem 150 Solution
Problem 150 Solution
It is obvious that A can be no more than 2. If A were 3 then 3BCDE * 4 would be
at least 120,000 which is more than five digits. Also A must be an even number
because EDCBA is an even number since it is the product of at least one even
number (4). We can eliminate A=0 because E would have to be 5 (5*4=0) but BCDE*4
could not hope to reach 50,000. So A must be 2.
Next consider E. E*4 must end in the digit 2. The only numbers that works for are
3 and 8. However with A=2 EDCBA must be at least 80,000. So 8 is the only number
that satisifies both conditions.
Next consider B. We already know that 2BCD8*4 is at least 80000 and less than
90000. B can not be more than 2 because then 2BCD8 * 4 would be more than
80000. 2 is already taken so B must be 0 or 1. Lets consider the case that
B=0. Then D8 * 4 must end in the digit 02. However there is no D that satisfies
this condition. So B must be 1.
Next consider D. D8*4 must end in the digits 12. The only possiblity is D=7
(78*4=312).
Now solve for C:
21C78 * 4 = 87C12.
84312+400C = 87012 + 100C
2700 = 300C
C=2700/300=9.
So ABCDE=21978.
I'd like to thank "Marisa" for this problem.
Michael Shackleford, ASA, December 28, 1999
  179.  Acceleration problem
   A drag racer accelerates at a uniform rate from its starting point. It travels the last one fourth of the distance from the starting point to the finish line in 3 seconds. How long did it take to travel the entire distance from starting point to finish line?
Answer
Problem 179 Answer
Problem 179 Answer
The answer is 12 + 6*3 ^{1/2} =~ 22.39230485
Michael Shackleford, ASA  December 14, 2000
Solution
Problem 179 Solution
Problem 179 Solution
Let the total distance traveled be 1.
Let a be the acceleration.
Let t be the total time traveled.
Basic calculus tells us that the velocity at time t is a*t and the total distance traveled is a*t^{2}/2.
So we know:
(1) 1=a*t^{2}/2 and
(2) 0.75 = a*(t3)^{2}/2
Solving both sides for a we have :
(3) 2/t^{2} = 3/(2*(t3)^{2})
4*(t3)^{2} = 3*t^{2}
t^{2}  24t + 36 = 0
t = (24 +/ 432^{1/2})/2
(11) t = 6*3^{1/2} + 12
Add 3 for the total distance of 12 + 6*3^{1/2} =~ 22.39230485
Michael Shackleford, ASA  December 6, 2000
  27.  Alley and two ladders problem
   There is a 6 foot wide alley. Both walls of the alley are perpendicular to the ground. Two ladders, one 10 feet long, the other 12 feet, are propped up from opposite corners to the adjacent wall, forming an X shape. All four feet of each ladder are firmly touching either the corner or the wall. The two ladders are also touching each other at the intersection of the X shape. What is the distance from the point of intersection to the ground?
Answer
Problem 27 Answer
Problem 27 Answer
The answer is 108 ^{1/2}/(1+(108 ^{1/2}/8)) =~ 4.52 feet.
Michael Shackleford, A.S.A.
Solution
Problem 27 solution
Problem 27 Solution
Solution 1
Consider the two ladders two lines on a graph. Let the shorter ladder extend from
(0,0) to (6,8). Let the longer ladder extend from (6,0) to (0,108^{1/2}).
The square root of 108 can be found using the pythagorean formula.
Then solve for the slope and y intercept to find the equations of the two lines:
The shorter ladder have the equation y=(4/3)*x.
The longer ladder has the equation y=(108^{1/2}/6)*x + 108^{1/2}.
The lines meet where the ladders cross.
Use substitution to solve for y.
Acknowledgement: Thanks to Steve Lutz for suggesting this
method of solution.
Solution 2
Scott R. Walshon pointed out that if x and y are the heights where the ladders touch the wall, and p is the height of the intersection then 1/p = 1/x + 1/y.
To prove this let x be the height on the right wall and y the height on the left wall. Call a the distance from the left edge of the alley to the point on the alley directly below the intersection point of the ladders. Call b the distance from the right edge of the alley to the point on the alley directly below the intersection point of the ladders. From similar triangles we get:
p/x = a/(a+b)
p/y = b/(a+b)
Add the two equations:
p/x + p/y = a/(a+b) + b/(a+b)
py/xy + px/xy = (a+b)/(a+b)
p(x+y)/xy = 1
p = xy/x+y
1/p = x+y/xy
1/p = x/xy + y/xy
1/p = 1/y + 1/x
Thanks Scott for this observation.
Michael Shackleford, A.S.A.
  152.  Ant and spider on a piece of wood problem
   In a backyard lies a block of wood 9" by 9" by 22". The long edge of the board lies along a north/south direction. An ant is sitting on the south end of the board, half way up vertically, and 1" from the east edge. A spider is sitting on the north end of the board, half way up vertically, and 1" from the west edge. The spider can crawl at a rate of 1" per minute. The ant figures that it will take 1+22+8=31 minutes for the spider to reach him so he dozes off for a 30 minute nap. Just as the ant wakes up the spider kills him. By what route did the spider take to get to the ant in 30 minutes?
Answer Problem 152 AnswerVisualize the box folded out. The spider has to crawl 22"+1"+1"=24" vertically and 4.5"+9"+4.5"=18" horizontally. The total distance traveled is (24"^{2} + 18"^{2})^{1/2} = (576" + 324")^{1/2} = 900"^{1/2} = 30".
Thanks to Anil Rhemtulla for this problem. Michael Shackleford, ASA, March 11, 2000>   4.  Ant and spider problem
   An ant and a blind spider are on opposite corners of a cube. The ant is stationary and the spider moves at random from one corner to another along the edges only. What is the expected number of turns before the spider reaches the ant? Optional: Also solve for a square, octahedron, icosahedron, and dodecahedron.
Answer
Problem 4 Answer
Problem 4 Answer
The answer is 10.
If the spider started at a corner diagonally on the same face as
the ant the answer would be 9, and if the spider started at an
adjacent corner the answer would be 7.
Here are answers for other figures:
Square: 4
Octahedron: 6
Dodecahedron: 35
Icosahedon: 15
Michael Shackleford, A.S.A.
Solution
Problem 4 Solution
Let x=number of turns to reach ant from starting point.
Let y=number of turns to reach ant from diagonal corner on same face as
ant.
Let z=number of turns to reach ant from an adjacent corner to ant.
After one turn the spider will be on a diagonal corner of a common face as
the ant. So the mean number of turns from the x position is one more than
the mean number from the y position:
E(x)=1+E(y).
Once at a y position there is a 2/3 chance it will then move to a z
position, and a 1/3 chance back to an x position:
E(y)=(2/3)*(1+E(z))+(1/3)*(1+E(x)).
If the spider arrives at a z position there is a 1/3 chance it will move to
the ant, and a 2/3 chance it will move back to a y position:
E(z)=(1/3)*1+(2/3)*(1+E(y)).
With these three equations and three unknowns it is not difficult to solve
for E(x), E(y), and E(z).
Michael Shackleford, A.S.A.
  206.  Ant on a rubber band problem
   Imagine an infinitely elastic rubber band that is 1 km. long unstretched. It expands at a rate of 1 km. per second. Next, imagine an ant at one end of the rubber band. At the moment the rubber band starts expanding the ant crawls towards the other end at a speed, relative to his current position, of 1 cm. per second. Will the ant ever reach the other end? If so, when?
Answer Problem 206 Answer
Question
Imagine an infinitely elastic rubber band that is 1 km. long unstretched. It expands at a rate of 1 km. per second.
Imagine an ant at one end of the rubber band. At the moment the rubber band starts expanding the ant crawls towards the other end at a speed, relative to his current position, of 1 cm. per second.
Will the ant ever reach the other end? If so, when?
Answer
Yes. The ant will reach the other end after e^{100,000}  1 seconds.
Michael Shackleford, ASA — Dec. 10, 2010
Solution
Problem 206 Solution
Question
Imagine an infinitely elastic rubber band that is 1 km. long unstretched. It expands at a rate of 1 km. per second.
Imagine an ant at one end of the rubber band. At the moment the rubber band starts expanding the ant crawls towards the other end at a speed, relative to his current position, of 1 cm. per second.
Will the ant ever reach the other end? If so, when?
Solution #1
Let f(t)=distance traveled from origin at time t.
So f'(t)=speed of ant at time t.
f'(t)= 1 + 100,000×ratio of progress.
The ratio of progress is the distance the ant has covered divided by the length of the rubber band = f(t)/(100,000*(1+t))
So f'(t) = 1 + 100,000 * f(t)/(100,000*(1+t))
f'(t) = 1 + f(t)/(1+t)
So, for what f(t) is this true? Here is where you pretty much need to have some linear differential equations in your back pocket. What works is f(t)=(1+t) × ln(1+t)
Let g(t)=length of rubber band time t.
g(t)=100,000×(1+t)
The question is at what time t is f(t)=g(t)?
(1+t) × ln(1+t) = 100,000×(1+t)
ln(1+t) = 100,000
1+t = e^{100,000}
t = e^{100,000}1
My thanks to Doc for solution #1.
Solution #2
The ratio of the rubber band the ant covers at time t is 1/[100000×(1+t)]. To help visualize this, think of the ant's position as your relative point of reference, and both ends of the rubber band moving away from the ant. So the ratio of progress is the ant's speed of 1 cm/sec compared to the total length of the rubber band at time t of 100,000×(t+t).
The question is, at what time T will the sum of the progress be 1?
Integral from 0 to T of 1/[100000×(1+t)] dt = 1
10^{5}×ln(1+t) from 0 to T = 1
10^{5}×(ln(1+T)ln(0)) = 1
10^{5}×ln(T) = 1
ln(1+T)=100,000
1+T=e^{100,000}
T=e^{100,000}1
My thanks to PapaChubby for solution #1.
General Answer
Let:
a = ant's speed.
b = rubber band's expansion speed.
c = initial length of rubber band.
Then the ant will get to the end in (c/b)×(e^{b/a}1) units of time.
Michael Shackleford, ASA — Dec. 6, 2010   195.  Average of the least of n random numbers from 0 to 1?
   What is the average of the least of n random numbers from 0 to 1?
Answer
Problem 195 Answer
Problem 195 Answer
The answer is 1/(n+1).
Michael Shackleford, A.S.A.
Solution
Problem 195 Solution
Problem 195 Solution
In problem 194, we see the solution for the 2number case is 1/3. The following shows the solution for the 3number case is 1/4, and for the 4number case is 1/5. Following the pattern, the solution for the nnumber case is 1/(n+1).
If it isn't clear from the scan below, first I find the area of integration, and then multiply the expected average by the inverse of that area. In the 3number case, x is the largest number, and ranges from 0 to 1. y is the second largest, and ranges from 0 to x. z is the smallest and ranges from 0 to y. Same concept in the 4number case, but we start with w as the largest number.
Michael Shackleford, A.S.A.
  194.  Average of the lesser of two random numbers from 0 to 1?
   What is the average of the lesser of two random numbers from 0 to 1?
Answer
Problem 194 Answer
Problem 194 Answer
The answer is 1/3.
Michael Shackleford, A.S.A.
Solution
Problem 194 Solution
Problem 194 Solution
See the following scan. The reason for multiplying by 2, is the area of integration is 1/2, as shown by the shaded area. x is the larger of the two numbers, which can range from 0 to 1. y is the smaller number, which can range from 0 to x.
Michael Shackleford, A.S.A.
  56.  Babies in the nursery problem
   A baby is added to a hospital nursery. Before the baby was added there were two boys in the nursery and an uncounted number of girls. After the new baby is added a baby is selected at random among all the babys. The selected baby is a boy. What is the probability that the added baby was a girl?
Answer
Problem 56 Answer
Problem 56 Answer
The answer is 40%.
Michael Shackleford, A.S.A.
Solution
Problem 56 Solution
Problem 56 Solution
Bayes' Theorem:
Let events A_{1},...,A_{k} form a partition of the space
S such that Pr(A_{j})>0 for j=1,...,k, and let B be any
event such that Pr(B)>0. Then for i=1,...,k,
Pr(A_{i}B) = Pr(A_{i})*Pr(BA_{i}) / Sum for j=1 to k of
Pr(A_{j})*Pr(BA_{j}).
The source of this problem is the May 1997 issue of the Actuarial Review.
Applying this theorm:
B=boy selected
A_{1}=boy added
A_{2}=girl added
g=number of girls before baby is added.
Pr(A_{2}B) = (1/2)*(2/(3+g)) / [(1/2)*(2/(3+g)) + (1/2)*(3/(3+g))] = 0.4
Michael Shackleford, A.S.A.
  24.  Baby bottle problem
   You need to warm milk in a baby bottle from its initial temperature of 15 degrees centigrade to 25 degrees. You put the bottle in a pot of boiling water which stays at constant temperature of 100 degrees. The thickness and conductivity of the bottle are such that the initial rate of heat transfer is 85 degrees per minute. However heat transfer is proportional to the difference between the temperature of the milk and the water. How many minutes will it take to heat the bottle to 25 degrees?
Answer
Problem 24 Answer
Problem 24 Answer
ln(17/15) =~ 0.1252
The Wizard of Odds
Solution
Problem 24 Solution
Problem 24 Solution
dc/dt = 100c
dc = 100c dt
dc 1/(100c) = dt
ln(100c) = t + K_{1}
t = ln(K_{2})  ln(100c) (where K_{1} = ln(K_{2})
t = ln(K_{2}/(100c))
When t=0 c=15 so K_{2} must be 85
t = ln(85/(100c))
Solving for t when c=25:
t = ln(85/(10025)) = ln(85/75) =~ ln(17/15) = 0.1252
The Wizard of Odds
  19.  Balls and an urn problem #1
   You have an urn with four balls of different colors. Randomly you draw two at a time, then painting the first ball to match the second. What is the expected number of drawings before all balls are the same color?
Answer
Problem 19 Answer
Problem 19 Answer
The answer is 9.
Michael Shackleford, A.S.A.
Solution
Problem 19 solution
Problem 19 Solution
Call the four colors 1,2,3, and 4.
After the first turn you will have a configuration like 1,1,2,3.
Call this configuration 1.
From configuration 1 if you draw a 1 on the first ball you will end
up with the same or similar configuration. The probability of this
happening is 1/2.
From configuration 1 if you draw a 3 or 4 first and then a 1 you
will have a configuration like 1,1,1,2. The probability of this
happening is 1/3. Call this new pattern configuration 2.
From configuration 1 if you draw a 3 or 4 first and then draw
the other non 1 you will have a configuration like 1,1,2,2. The
probability of this is 1/6. Call this new pattern configuration 3.
From configuration 2 you will end up with the same pattern with
probability 1/2, configuration 3 with probability 1/4, and ending
the experiment with probability 1/4.
From configuration 3 you will end up with the same thing with
probability 1/3, and with configuration 2 with probability 2/3.
From this information you can draw the following equations, where
a is the expected number of turns from equation 1, b from equation 2,
and c from equation 3:
 a=(a+1)/2 + (b+1)/3 + (c+1)/6
 b=(b+1)/2 + (1)/4 + (c+1)/4
 c=(c+1)/3 + (2/3)*(b+1)
Finally simple matrix algebra will show a=8. Considering the first
turn from the original condition to equation 1 the expected number of
turns is 9.
Michael Shackleford, A.S.A.
  200.  Baseball playoffs question
   Prior to 2013, in Major League Baseball, the American League had three divisions, the East and Central divisions with 5 teams each, and the West division with 4 teams. Four teams in the the league will make the playoffs, the three division leaders, and a wild card team. The wild card team is the team with the best record in the league, not including the three division leaders. Assuming all teams are equally good, what is the probability of a team in each division making the playoffs? Assume any ties are broken randomly. For extra credit, also do the National League, in which the East and West divisions have 5 teams, and the central division has 6 teams.
Answer
Problem 200 Answer
Problem 200 Answer
American East 27.492507%
American Central 27.492507%
American West 31.268731%
National East 25.934066%
National Central 23.443223%
National West 25.934066%
Michael Shackleford, ASA — Nov 26, 2008
Solution
Problem 200 Solution
Problem 200 Solution
Question
In Major League Baseball, the National League has three divisions, the East and Central divisions with 5 teams each, and the West division with 4 teams. Four teams in the National League will make the playoffs, the three division leaders, and a wild card team. The wild card team is the team with the best record in the league, not including the three division leaders. Assuming all teams are equally good, what is the probability of a team in each division making the playoffs? Assume any ties are broken randomly.
Solution
The difficult part of this question is determing the probability that the Wild Card team will fall in any given division. Let's consider the American League East first. There are six ways the wild card team could be in the American League East, as follows.
American League East has both first and second place teams in the entire American League. The probability of this is (5/14)×(4/13) = 10.9890%.
American League East has both first and third place teams, but not the second, in the entire American League. The probability of this is (5/14)×(4/13)×(9/12) = 8.2418%.
American League East has both second and third place teams, but not the first, in the entire American League. The probability of this is (5/14)×(4/13)×(9/12) = 8.2418%.
American League East has both first and fourth place teams, in the entire American League, and the second and third place teams fall one each into the other two divisions. The probability of this is = (5/14)×(4/13)×((5/12)×(4/11)+(4/12)×(5/11)) = 3.3300%.
American League East has both second and fourth place teams, in the entire American League, and the first and third place teams fall one each into the other two divisions. The probability of this is = (5/14)×(4/13)×((5/12)×(4/11)+(4/12)×(5/11)) = 3.3300%.
American League East has both third and fourth place teams, in the entire American League, and the first and second place teams fall one each into the other two divisions. The probability of this is = (5/14)×(4/13)×((5/12)×(4/11)+(4/12)×(5/11)) = 3.3300%.
Add these probabilties together and you get 37.4625%. It is obvious the probability should be the same for the American League Central. The probability for the American League West thus has a probability of 100%  2×37.4625% = 25.074925% of having the wild card team.
The probability any given team will be a division leader is 20% in the East and Central divisions, and 25% in the West division. These are also the probabilities for the wild card spot, given that it is in the same division. So, the probability that a team from the East or Central divisions makes the playoffs is 20% + 37.4625%×20% = 27.492500%. The probability that a team from the West division makes the playoffs is 25% + 25.074925%×25% = 31.268731%.
Extra Credit
The probability the National League East or West division will get the wild card is 29.670330% each, and 40.659341% for the Central division. The probability any given team in the East or West will make the playoffs is 25.934066%, and 23.443223% for Central.
Summary
American East 27.492507%
American Central 27.492507%
American West 31.268731%
National East 25.934066%
National Central 23.443223%
National West 25.934066%
Michael Shackleford, ASA — Nov 26, 2008
  154.  Biased coin flipping problem #1
   A loaded coin has probability of landing on heads of 0.6 and for tail 0.4. Winning bets on heads pay even money. With an initial fortune of $10 and betting $1 at a time what is the probability of eventual ruin assuming you kept playing until being ruined?
Answer
Problem 154 Answer
Problem 154 Answer
The answer is (2/3) ^{10} =~ 0.017342 .
This problem is from the March 2000 issue of The Actuary, published
by the Society of Actuaries.
Michael Shackleford, ASA  April 5, 2000
Solution
Problem 154 Solution
Problem 154 Solution
Let p _{i} be the probability of ruin given a bankroll of $i.
p_{1} = .4 + .6*p_{2}
p_{2} = .4*p_{1} + .6*p_{3}
p_{3} = .4*p_{2} + .6*p_{4}
p_{4} = .4*p_{3} + .6*p_{5}
p_{5} = .4*p_{4} + .6*p_{6}
.
.
.
Now sum the equations.
Sum for i=1 to infinity of p_{i} = .4 + .4*(sum for i=1 to infinity of p_{i}) + .6*(sum for i=2 to infinity of p_{i})
Sum for i=1 to infinity of p_{i} = .4 + (sum for i=1 to infinity of p_{i}) .6*p_{1})
.6*p_{1}=.4
p_{1}=.4/.6 = 2/3
At this point you can use substitution to find that p_{i}=(2/3)^{i}.
So p_{10}=(2/3)^{10}=~0.017342
This problem is from the March 2000 issue of The Actuary, published
by the Society of Actuaries.
Michael Shackleford, ASA  April 5, 2000
  155.  Biased coin flipping problem #2
   Suppose before taking the first bet of the previous problem you consult your crystal ball which says you will eventually be ruined. Being a compulsive gambler you play anyway until ruin. What is the ratio of total heads to total flips?
Answer
Problem 155 Answer
Problem 155 Answer
The answer is 0.4 .
This problem is from the March 2000 issue of The Actuary, published
by the Society of Actuaries.
Michael Shackleford, ASA  April 5, 2000
Solution
Problem 155 Solution
Problem 155 Solution
Let p _{i} be the probability that the first flip with bankroll of $i will be heads.
p_{i} given eventual ruin = (p_{i}&pr(ruin at $i+1))/pr(ruin at $i)=
.6*(.4/.6)^{11}/(.4/.6)^{10} (see solution to problem 154 for probability of ruin) =
.6*(.4/.6) = 0.4
So the probability that any given flip will be heads is 0.4, thus the expected total flips
that are heads is also 0.4 .
This problem is from the March 2000 issue of The Actuary, published
by the Society of Actuaries.
Michael Shackleford, ASA  April 5, 2000
  202.  Blue and Red Amoeba
   Every generaion of red amoeba has a 5% chance of spawning 5 more red amoeba, a 10% chance of spawning 3 blue amoeba, a 1% chance of doing both, and a 84% chance of not spawning.
Every generaion of blue amoeba has a 3% chance of spawning 5 more blue amoeba, a 2% chance of spawning 10 red amoeba, a 1% chance of doing both, and a 94% chance of not spawning.
Your petri dish starts with one red amoeba. What is the expected number of red and blue amoeba observed before all amoeba die off?
Answer
Problem 202 Answer
Problem 202 Answer
Question
Every generaion of red amoeba has a 5% chance of spawning 5 more red amoeba, a 10% chance of spawning 3 blue amoeba, a 1% chance of doing both, and a 84% chance of not spawning.
Every generaion of blue amoeba has a 3% chance of spawning 5 more blue amoeba, a 2% chance of spawning 10 red amoeba, a 1% chance of doing both, and a 94% chance of not spawning.
Your petri dish starts with one red amoeba. What is the expected number of red and blue amoeba observed before all amoeba die off?
Answer
One red amoeba will spawn into 800/461 =~ 1.7354 on average.
One red amoeba will spawn into 330/461 =~ 0.7158 blue amoeba on average.
Michael Shackleford, ASA — Apr. 1, 2009
Solution
Problem 202 Solution
Problem 202 Solution
Question
Every generaion of red amoeba has a 5% chance of spawning 5 more red amoeba, a 10% chance of spawning 3 blue amoeba, a 1% chance of doing both, and a 84% chance of not spawning.
Every generaion of blue amoeba has a 3% chance of spawning 5 more blue amoeba, a 2% chance of spawning 10 red amoeba, a 1% chance of doing both, and a 94% chance of not spawning.
Your petri dish starts with one red amoeba. What is the expected number of red and blue amoeba observed before all amoeba die off?
Solution
Let r = expected number of red amoeba observed per initial red amoeba.
Let b = expected number of red amoeba observed per initial blue amoeba.
Next, from the first piece of information, "Every generaion of red amoeba has a 5% chance of spawning 5 more red amoeba, a 10% chance of spawning 3 blue amoeba, a 1% chance of doing both, and a 84% chance of not spawning," we can put in equation form:
r = 1 + 0.05×5×r + 0.1×3×b + 0.01×(5r+3b)
r = 1 + 0.25r + .3b + 0.05r + 0.03b
r = 1 + 0.3r + 0.33b
Next, from the second piece of information, "Every generaion of blue amoeba has a 3% chance of spawning 5 more blue amoeba, a 2% chance of spawning 10 red amoeba, a 1% chance of doing both, and a 94% chance of not spawning," we can put in equation form:
b = 0.03×5b + 0.02×10r + 0.01×(5b+10r)
b = 0.15b + 0.2r + .05b + 0.1r
b = 0.2b + 0.3r
Using basic algebra, we can solve for r and b , given the two equations, and get:
r = 800/461 =~ 1.7354
b = 300/461 =~ 0.6508
So, one initial red amoeba will spawn into 1.7354 on average. We also know that one blue amoea will spawn into 0.6508 red amoeba on average. However, that doesn't tell us how many blue amoeba a red amoeba will spawn into. To answer that question, let's redefine the variables as follows:
Let r = expected number of blue amoeba observed per initial red amoeba.
Let b = expected number of blue amoeba observed per initial blue amoeba.
Putting the initial conditions into equations:
r = 0.3r + 0.33b
b = 1 + 0.2b + 0.3r
Using algebra to solve the equations we get:
r = 330/461 =~ 0.7158
b = 700/461 =~ 1.5184
So, one red amoeba will spawn into 0.7158 blue amoeba on average.
Simulation
I did a random simulation of 74,016,000,000 initial states of one red amoeba. Following were the total counts observed:
Red amoeba: 128,445,045,020
Blue Amoeba: 52,983,662,357
That is an averoage of 1.735368637 red and 0.715840661 blue.
Michael Shackleford, ASA — Apr. 1, 2009
  149.  Burried treasure problem
   You find an old treasure map to the location of a buried treasure chest on a deserted island. The map states on this island are an oak tree, a palm tree, and a gallows. You are to start at the gallows, walk toward the palm tree while counting your steps. Then turn 90 degrees to the left and walk the same number of steps. At that spot drive a stake. Then walk from the gallows to the oak tree, again measuring your steps. When you get to the oak tree turn 90 degrees to the right and walk the same number of steps, then drive a stake at that spot. Then find the midpoint between the two stakes and dig.
When you get to the island you find the gallows are gone. Without digging at random is it possible to figure out the location of the treasure chest? If so where is it?
Answer
Problem 149 Answer
Problem 149 Answer
The treasure chest can be in one of two places. Find the midpoint of the two
trees and walk in either direction on the line perpendicular to the line that
contains both trees. Stop after walking half the distance between the two
trees and dig. If the treasue isn't there then walk the other direction from the
midpoint of the two trees and it must be there.
Thanks to the
Geometry Forum for this problem.
Michael Shackleford, ASA, October 21, 1999
Solution
Problem 149 Solution
Problem 149 Solution
Consider the oak tree to be at point (0,0), the palm tree to be at
(p,0), and the gallows to be at (x,y). The location of the first stake is
(p+y,px). The location of the second stake is (y,x). The midpoint
is ((p+yy)/2,(px(x))/2) = (p/2,p/2).
Thanks to the
Geometry Forum for this problem.
Michael Shackleford, ASA, October 21, 1999
  110.  Cable in the field problem
   There is a straight cable buried under a unit square field. You must dig one or more ditches to locate the buried cable. Where should you dig to guarantee finding the cable and to minimize digging? For example you could dig an X shape for total ditch length of 2*sqr(2) but there is a better answer.
Answer
Problem 110 Answer
Problem 110 Answer
Let the corners of the square be at (0,0),(1,0),(1,1), and (0,1).
The solution is to dig the following ditches:
 (0,0) to (x,x)
 (1,0) to (x,x)
 (0,1) to (x,x)
 (1/2,1/2) to (1,1)
Where x = (3sqr(3))/6 =~ 0.2113248654
The total length of all ditches is aproximately 2.6389584338
I'd like to thank the
Macalester College Problems of the Week for this problem.
Michael Shackleford, A.S.A.
  99.  Camel and grain problem
   It is your task to deliever as much grain as possible from city A to city B. The cities are 1,000 miles apart. You initially have 10,000 pounds of grain. Your camel may carry up to 1,000 pounds and eats 1 pound of grain per mile traveled. You may leave grain along the way and return to it later. How much grain can you deliever to city B?
Answer
Problem 99 Answer
Problem 99 Answer
You can deliver 1399.77 pounds of grain.
Michael Shackleford, A.S.A., January 14, 1999
Solution
Problem 99 Solution
Problem 99 Solution
The way to deliver as much grain as possible is not not waste the camel's movement with
light loads. However moving heavy loads too small of a distance can often leave a small remainder
in the last trip, which is a waste of grain.
From the beginning you have enough grain to make 19 oneway trips to some cache point
along the way. The first question is, far out should you transport that grain? The answer is far enough so that you leave a cache somewhere of
9000 pounds of grain. Why 9000? Every cache point should have some amount of grain
evenly divisible by 1000. This way the camel can start out every trip with 1000 pounds, and not
have a remainder on the last trip. So you have 1000 pounds of grain to let the camel eat, divided by 19 oneway trips. 1000 pounds divided by 19 trips is 1000/19=~52.63 pounds, or 52.63 miles per trip. Each trip you will eat 52.63 miles along the way there, deposit
894.73 pounds of grain, and take back 52.63 pounds of grain to eat on the way home, except
the last trip in which you don't need to return. The total deposit works out to (10*894.73)+52.63 = 9000 pounds, as arranged. So you will be
able to cache 9000 pounds of grain 52.63 miles.
Next repeat the above but only allowing 17 oneway trips. 1000 divided by 17 is 1000/17 =~ 58.82 . So you will be able to cache 8000 pounds of grain 111.46 miles
from city A.
Below is a summary of the entire journey:
 Cache 9000 pounds over 19 oneway trips, moving 52.63 miles, or 52.63 miles from city A
 Cache 8000 pounds over 17 oneway trips, moving 58.82 miles, or 111.46 miles from city A
 Cache 7000 pounds over 15 oneway trips, moving 66.67 miles, or 178.12 miles from city A
 Cache 6000 pounds over 13 oneway trips, moving 76.92 miles, or 255.04 miles from city A
 Cache 5000 pounds over 11 oneway trips, moving 90.91 miles, or 345.95 miles from city A
 Cache 4000 pounds over 9 oneway trips, moving 111.11 miles, or 457.07 miles from city A
 Cache 3000 pounds over 7 oneway trips, moving 142.86 miles, or 599.92 miles from city A
 Cache 2000 pounds over 5 oneway trips, moving 200.00 miles, or 799.92 miles from city A
 Take 1000 pounds directly to city B, comsuming 200.08 pounds on the way, leaving 599.84 pounds
and take back 200.08 pounds to eat on the way back.
 Take the last 1000 pounds directly to city B, comsuming 200.08 pounds on the way, and
depositing an additional 799.92 pounds of grain, for a total deposit of 1399.77 pounds.
I would like to thank Scott Morris for this problem.
Michael Shackleford, A.S.A., January 14, 1999
  64.  Cardiod problem
   Two circles of radius one are adjancent at point (0,0). Let the center of the first circle be at point (1,0). Label the point on the edge of the first circle at (0,0) x. The second circle is centered at (1,0). Rotate the first circle around the second, like a gear, keeping the second circle stationary. If you traced the path of point x as the first circle traveled around the second what would be the length of the path point x traced? The figure above shows the path in red.
Answer
Problem 64 Answer
Problem 64 Answer
The answer is 16.
Michael Shackleford, A.S.A.
Solution
Problem 64 Solution
Problem 64 Solution
Let a be the distance from the point (0,0) to a point on the figure (known as
a cardoid). Next consider the angle t formed from the points (1,0), (0,0) and a at (0,0).
After working through the trigonomtry (this step is left to you) the length of a in terms of
t is 2*(1cos(t)).
The formula for arc length is the integral of
( a^{2} + (da/dt)^{2} )^{1/2}.
In this case it is the integral from 0 to 2*pi of:
( 4*(1cos(t))^{2} + 4*sin^{2}(t) )^{1/2}. =
8^{1/2} * (1  cos(t))^{1/2}.
After consulting a table of integrals we find that the integral of
(1  cos(t))^{1/2} is 2*2^{1/2}*cos(t/2).
Taking the integral the answer works out to 16.
I'd like to thank Andrea for correcting an earlier mistake in my answer.
Michael Shackleford, A.S.A.
  57.  Chessboard problem #1
   Given an 2^{n} by 2^{n} chessboard, with one corner removed, show that you can cover it completely without overlapping or going off the board using only triominoes of the shape below.
Rotations are allowed.
Answer
Problem 57 Answer
Problem 57 Answer
For n=2 (4 by 4) you can cover the board like this:
For n=3 (8 by 8) use four sets of the 4 by 4 configuration. Use three of them
in such a way that the three extra squares form one extra triominoe, the fourth
one should have the extra square in the corner.:
For each additional n repeat the n1 by n1 configuration four times, rotating
in such as way as to create one extra triominoe.
Here is a nice graphic from Chris Johnson that explains the solution more clearly.
Thanks to Guy de Kindler
for this one.
Michael Shackleford, A.S.A.
  58.  Chessboard problem #2
   How many subsquares can you form on chessboard with n*n unit squares? Subsquares must be an integer squared.
Answer
Problem 58 Answer
Problem 58 Answer
The answer is n*(n+1)*(2n+1)/6.
Michael Shackleford, A.S.A.
Solution
Problem 58 solution
You can form n^{2} unit squares.
You can form (n1)^{2} 2 by 2 squares.
You can form (n2)^{2} 3 by 3 squares.
You can form (n3)^{2} 4 by 4 squares.
.
.
.
You can form 1^{2} n by n square.
The sum of all these sums is n*(n+1)*(2n+1)/6
Thanks to Guy de Kindler
for this one.
Michael Shackleford, A.S.A.
  9.  Chicken McNugget problem
   At McDonalds you can order Chicken McNuggets in boxes of 6, 9, and 20. What is the largest number such that you can not order any combination of the above to achieve exactly the number you want?
Answer
Problem 9 Answer
Problem 9 Answer
The answer is 43.
Michael Shackleford, A.S.A.
Solution
For any desired number if it is divisible
by 3 it can easily be made with 6 and 9 packs,
except if the number is 3 itself. If you can't
use all six packs then use one 9 pack and you
can do the rest with six packs.
If the number is not divisible by 3 then
use one 20 pack. If the remaining number is
divisible by 3 then use the above method for
the rest.
If the number still isn't divisible by 3 use
a second 20 pack. The remainder must
be divisible by 3, in which case use the 6 and 9
packs as above.
The largest impossible number would be such that
you would have to subtract 20 twice to get a
remainder divisible by 3. However, you can't
make 3 itself with 6 and 9 packs. So the
largest impossible number is 2*20+3=43.
Michael Shackleford, A.S.A.
  189.  Circle and triangle problem
   The radius of the circle is 1. The triangle is equilateral. Find the area of each colored region.
Answer
Problem 189 Answer
Problem 189 Answer
Yellow = (5*3 ^{1/2}  2*pi)/6 =~ 0.39618
Orange =(3*3^{1/2}+2*pi)/6 =~ 1.91322
Green = (4*pi  3*3^{1/2})/6 =~ 1.22836
Thanks to Eugene Wilson for this problem and image.
Michael Shackleford, A.S.A.
Solution
Problem 189 Answer
Problem 189 Answer
Consider the subtriangle formed by the three points where the circle touches the large triangle. This is also an equilateral traingle because each angle is 60 degrees. Once you realize that is it just basic trigonomtry to get the rest as shown in this diagram.
Yellow = 2c = (5*3 ^{1/2}  2*pi)/6 =~ 0.39618
Orange = 6a+2b = (3*3^{1/2}+2*pi)/6 =~ 1.91322
Green = 4b = (4*pi  3*3^{1/2})/6 =~ 1.22836
Michael Shackleford, A.S.A.
  125.  Clock problem
   On average how often does the minute hand pass the hour hand on an ordinary clock?
Answer
Problem 125 Answer
The answer is every 60/55 hours, or every 1 hour, 5 minutes, and 27+3/11 seconds.
Michael Shackleford, A.S.A.
Solution
Problem 125 Solution
Problem 125 Solution
Let both hands be at 12. It will take the hour hand 1 hour to return to where the minute hand started. However at that time the minute hand will have advanced 1/12 of an hour. When the hour hand gets there the minute hand will have advanced (1/12) ^{2} or an hour. When the hour hand gets there the minute hand will have advanced (1/12) ^{3} or an hour. So the total distance the hour hand will cover until it reaches the minute hand is the sum for i= 0 to infinite of (1/12) ^{i}. Everyone should know that the infinite series for x ^{i}, where x<1 and i from 0 to infinity is 1/(1x). In this case the answer is 1/(11/12) = 1/(11/12) = 12/11 = 65.45 minutes.
Björn from Germany suggested another solution. Let both hands be at 12. Over the next 12 hours they will cross 11 times. Thus they every 12/11 hours.
Michael Shackleford, A.S.A.
  66.  Closest distance between two points on a sphere problem
   Assuming the earth earth is spherical with radius r. What is the shortest distance along the surface from point A to point B. Let p_{1} and p_{2} be the two longitudes, measured from 0 to 2*pi. Let q_{1} and q_{2} be the two latitudes, measured from 0 to pi, where 0=north pole and pi=south pole.
Answer
Problem 66 answer
Problem 66 Answer
The answer is
cos ^{1}[(sin(q _{1})*cos(p _{2}p _{1})*sin(q _{2}) +
cos(q _{1})*cos(q _{2}))] * r.
Michael Shackleford, A.S.A.
Solution
Problem 66 Solution
Problem 66 Solution
First, assume the earth has radius 1 until the end when the answer will be
multiplied by r.
Next order the two points such that p_{1} <= p_{2}.
Next put the two points in cartesial coordinates (x,y,z).
Through a little geometry it can be found that:
(x_{1},y_{1},z_{1}) =
(cos(p_{1})*sin(q_{1}), sin(p_{1})*sin(q_{1}), cos(q_{1})) and
(x_{2},y_{2},z_{2}) =
(cos(p_{2})*sin(q_{2}), sin(p_{2})*sin(q_{2}), cos(q_{2})).
The angle between two vectors is defined as cos(t)=(v dot w)/(v * w).
Let v donate the vector from (0,0,0) to (x_{1},y_{1},z_{1}) and
let w donate the vector from (0,0,0) to (x_{2},y_{2},z_{2}).
The dot product of v and w is:
cos(p_{1})*sin(q_{1})*cos(p_{2})*sin(q_{2}) +
sin(p_{1})*sin(q_{1})*sin(p_{2})*sin(q_{2}) +
cos(q_{1})*cos(q_{2}).
If we rotate the longitude of both points by p_{1} then p_{1} becomes 0 and
p_{2} becomes p_{2}p_{1}, in which case the dot product is:
cos(0)*sin(q_{1})*cos(p_{2}p_{1})*sin(q_{2}) +
sin(0)*sin(q_{1})*sin(p_{2}p_{1})*sin(q_{2}) +
cos(q_{1})*cos(q_{2}). =
sin(q_{1})*cos(p_{2}p_{1})*sin(q_{2}) +
cos(q_{1})*cos(q_{2}).
Now, both u and v must be 1 since they lie on the surface of a unit sphere. Thus the
angle between the two points must be cos^{1}[sin(q_{1})*cos(p_{2}p_{1})*sin(q_{2}) +
cos(q_{1})*cos(q_{2}) / (1 * 1). = sin(q_{1})*cos(p_{2}p_{1})*sin(q_{2}) +
cos(q_{1})*cos(q_{2})].
This angle is the ratio of the distance along the surface between the
two points to the circumference of the earth. Thus the distance is the angle times
the circumference of the earth which equals:
cos^{1}[(sin(q_{1})*cos(p_{2}p_{1})*sin(q_{2}) +
cos(q_{1})*cos(q_{2}))] * r.
Michael Shackleford, A.S.A.
  184.  Coconuts and monkey problem #2
   Ten people land on a deserted island. There they find lots of coconuts and a monkey. During their first day they gather coconuts and put them all in a community pile. After working all day they decide to sleep and divide them into two equal piles the next morning. That night one castaway wakes up hungry and decides to take his share early. After dividing up the coconuts he finds he is one coconut short of ten equal piles. He also notices the monkey holding one more coconut. So he tries to take the monkey's coconut to have a total evenly divisible by 10. However when he tries to take it the monkey conks him on the head with it and kills him. Later another castaway wakes up hungry and decides to take his share early. On the way to the coconuts he finds the body of the first castaway, which pleases him because he will now be entitled to 1/9 of the total pile. After dividing them up into nine piles he is again one coconut short and tries to take the monkey's coconut. Again, the monkey conks the man on the head and kills him. One by one each of the remaining castaways goes through the same process, until the 10th person to wake up gets the entire pile for himself. What is the smallest number of possible coconuts in the pile, not counting the monkeys?
Answer
Problem 184 Answer
Problem 184 Answer
The answer is 2519.
Michael Shackleford, A.S.A, January 16, 2002
Solution
Problem 184 Solution
Problem 184 Solution
We know from the first man that the total must be one less than a number divisible by 10.
We know from the second man that that total must be one less than a number divisible by 9.
After considering all 10 people the final number must be one less than a number divisible by every number from 1 to 10. Certainly 10!=3628800 works but there is smaller solutions. The smallest answer is LCM(1,2,3,4,5,6,7,8,9,10)1 = 7*2*2*2*3*3*51 = 2519.
Note that LCM stands for Least Common Multiple.
Thanks to David Brown for suggesting this problem.
Michael Shackleford, A.S.A, January 16, 2002
  11.  Coin toss problem #1
   Given a coin with probability p of landing on heads after a flip, what is the probability that the number of heads will ever equal the number of tails assuming an infinite number of flips?
Answer
Problem 11 Answer
Problem 11 Answer
The answer is 2*min(p,1p).
Michael Shackleford, A.S.A.
Solution
Problem 11 Solution
Problem 11 Solution
Think of this problem instead as a random walk along the number line, starting at zero. If the walker ever returns to zero, then success is achieved. Let x_{i} be the probability of success from point i on the number line. Let p be the probability that the walker moves to the right. Let's also let p be the lesser of the probabilities of flipping heads and tails.
I hope I won't have to prove that x_{1}=1. In other words, the gravity of the greater probability of q will pull the walker to the right, over the long run. So if his first move is to the left, he will eventually make it back to point 0. For the same reason, given a casino game with a house advantage, the house always wins in the long run. If the player wins his first bet, if he plays indefinitely, he will eventually lose it back.
That said, we can set up the following equations:
x_{0} = p + q×x_{1}
x_{1} = p + q×x_{2}
x_{2} = p×x_{1} + q×x_{3}
x_{3} = p×x_{2} + q×x_{4}
.
.
.
Next, take the sum of both sides.
Let E = x_{0} + x_{1} + x_{2} + ...
E = 2p + pE p×x_{0} + qE  q×x_{0}
E = 2p + E×(p+q)  x_{0}×(p+q)
E = 2p + E  x_{0} (because p+q=1)
0 = 2p  x_{0}
x_{0} = 2p
So the answer is 2p.
Below are the results of computer trials which have verified the answer
above. For each probability p 100,000 trials were conducted, in which a
trial ended when either the number of heads equalled the number of tails,
or the difference was more than 25. It seems reasonable to assume that for
p<=.4 if the difference ever equals 25 it is very unlikely that the gap
ever be closed. A 'win' means the number of heads was ever equal to the
number of tails. The number of losses would be 100,000 minus the number of
wins.
p wins
 
0.4 80058
0.3 59799
0.2 39957
0.1 19803
If you don't buy my solution, I have posted altnernate solutions by Richard Tucker and David Beim.
Thanks to Guy de Kindler
for providing the problem and Richard Tucker and David Beim for the alternate solutions.
Michael Shackleford, A.S.A.
  44.  Coin toss problem #1
   You are given an initial prize of one dollar. A coin will be tossed indefinitely until a tails occurs. Every time a heads is tossed your prize will be doubled. When a tails is tossed you walk away with whatever your prize has grown to. For example if you toss three heads and then tails you would get $1*2*2*2=$8 dollars. What is the expected return of this game?
Answer
Problem 44 Answer
Problem 44 Answer
The answer is infinity dollars.
Michael Shackleford, A.S.A.
Solution
Problem 44 Solution
Problem 44 Solution
The probability of getting x consecutive heads followed by a tails is (1/2) ^{x+1} = (1/2)*(1/2) ^{x}.
The payoff for getting x heads followed by a tails is $2^{x}.
The expected return is the sum of the products of the probabilities and
the payoffs is the sum for x=0 to infinity of (1/2) * (1/2)^{x} * 2^{x}.
(1/2)^{x} * 2^{x} = 1. So the answer is an infinite sum of 1/2's, or
infinity.
This problem is a well known paradox in mathematics. The final result must be finite yet
the expected value is infinite.
Michael Shackleford, A.S.A.
  46.  Coin toss problem #3
   Assuming you flipped a coin until over and over, keeping track of the number of both heads and tails flipped, what is the expected number of flips before the two totals were equal?
Answer
Problem 46 Answer
Problem 46 Answer
The answer is infinity.
Michael Shackleford, A.S.A.
Solution
Problem 46 Solution
Problem 46 Solution
Think of this problem instead as a random walk along a number
line starting at the point 0, moving one unit to the right or
left each turn. What if the there were barriers at points x+1 and
x1, such that x and x were as far as you could go, in which case
your next turn would be one step closer to the origin?
Next determine the expected number of turns given such barriers
using the same method of interlocking expected values as in the
ant and spider problem. The
answer turns out to be 2x.
Thus, when x is infinity the expected number of turns must be
2*infinity=infinity.
Related question: What is the expected number of
turns it would take a random walker on a number line to revist any
point? The answer is here.
Michael Shackleford, ASA
  61.  Coin toss problem #3
   Suppose you have a coin in which the probability of flipping a heads is p, where p>=0.5 . What is the expected number of flips it will take for the number of heads to equal the number of tails, assuming the first flip is a tails?
Answer
Problem 61 Answer
Problem 61 Answer
The answer is 1/(2p1) .
Michael Shackleford, A.S.A.
Solution
Problem 61 solution
Problem 61 Solution
Think of the problem as a random walk where the probability is p of moving to the left
and 1p of moving to the right. The expected change in the position on the number
line per move is p*(+1) + (1p)*(1) = 2p1. The first flip resulted in moving one
space to the right. The expected number of moves to move one space to the left is the
inverse of the expected movement, which equals 1/(2p1).
It may not seem reasonable to take the last step on faith but millions of computer
trials have verified the answer.
Michael Shackleford, A.S.A.
  128.  Coin toss problem #4
   What is the expected number of flips of a fair coin until you get two heads in a row? What is the expected number until you get a head followed by a tail? Note: The answers are not the same.
Answer
Problem 128 Answer
The expected number until you obtain two heads is 6, the
expected number until you obtain a head followed by a
tail is 4.
Michael Shackleford, A.S.A., 11/10/1998
Solution
Problem 127 Solution
Problem 128 Solution
First lets answer the question of the expected number until
you obtain two heads. There are two possible states, either
the last flip was a tails (state x) or a heads (state y).
The first flip is also in state x. From state x you have a
50% chance of going to state y and 50% of staying in state x.
From state y you have a 50% chace of ending the experiment and
a 50% chance of dropping back to state x. Now set up these
statements in equation form and solve, call E(x) the
expected number of future flips from state x and likewise
for E(y) from state y:
E(x)=1 + (E(x)+E(y))/2
E(y)=1 + (E(x))/2
Solving for E(x) and E(y) yields E(x)=6 and E(y)=4. Since the first state
is x the answer is 6.
Approach the second problem in the same way, however this time from state
y you either end or stay in state y, thus the equations are:
E(x)=1 + (E(x)+E(y))/2
E(y)=1 + (E(y))/2
In this case E(x)=4 and E(y)=2, the answer being E(x) or 4.
Thanks to Alon Amit for this problem.
Michael Shackleford, A.S.A., 11/10/1998
  246.  Common birthday for three or more people
   In an office of 200 people, what is the probability any five or more people will share a common birthday?
Answer
Problem 246 Answer
Problem 246 Answer
The answer is 0.0880680413.
Michael Shackleford
Solution
Problem 246 Solution
Problem 246 Solution
 First a note about terminology, let C(x,y) equal the number of ways to choose y items out of x. The formula is x!/(y! × (xy)!). For example, the number of way to choose five cards out of 52 is C(52,5) = 52!/(5! × 47!) = 2,598,960.
 Rather than keep using the word "birthday," let's say each person will get one "day."
 Next, let's consider the case of five people sharing a common birthday. The logic will work for any number of people sharing a common birthday.
 To make a specific example, consider the probability that in an office of 200 people there will be at least one day of the year shared by five or more people.
 To find that answer, first find the probability that in 200 people nobody day will be shared by five or more people. Subtracting that answer from 1 will give us the probability of a common birthday to five or more people.
 There are 365^{200} possible sequences.
 To find the number of sequences where there is no day celebrated by five or more people, consider all the ways to clump the 200 people into days shared by 4 people, shared by 3, shared by 2, and shared by 1.
 Let a = the number of days shared by 4 people. This number can be 0 to 50.
 Let b = the number of days shared by 3 people. This number can be 0 to int((2004a)/3).
 Let c = the number of days shared by 2 people. This number can be 0 to int((2004a3b)/2).
 Let d = the number of days occupided by just one person. This number must be 2004a3b2c.
 The number of ways you can assign specific days to each of these days with at least one birthday is C(365,a)×C(365a,b)×C(365ab,c)×combin(365abc,d). Let's call that number p.
 The number of ways to order the 200 birthdays is 200!. Multiply that by p.
 If we stopped here, p would be too big. Why? Let's say two people share May 23. Let's go onto say that if you lined up all 200 people they took spots 81 and 144. It wouldn't make any difference which May 23 person got 81 and who got 144. The sequence of birthdays would still look the same.
 Likewise, we would divide by 3!=6 for each group of 3 people, and 4!=24 for each group of 4 people.
 To adjust for the c groups of 2 people, divide p by 2^{c}.
 To adjust for the b groups of 3 people, divide p by 6^{b}.
 To adjust for the a groups of 4 people, divide p by 24^{a}.
 That will give us the correct value of p, the number of sequences of birthdays where no day is shared by 5 or more people.
Thus, the probability of no common birthday is C(365,a)×C(365a,b)×C(365ab,c)×combin(365abc,d)/(2^{c} × 6^{b} × 24^{a})/365^{200}, for all possible combinations of a, b, c, and d. This works out to 0.9119319587.
So, the probability of at least one day common to five or more people is 1  0.9119319587 = 0.0880680413.
As a practical note, there is no easy way to get at this number without a computer. Here is my code that shows how to calculate the probability of no common birthday, where p is the number of 200:
tot_pr=0;
for (a=0; a<=(int)p/4; a++)
{
for (b=0; b<=(int)((p4×a)/3); b++)
{
for (c=0; c<=(int)((p4×a3×b)/2); c++)
{
d=p4×a3×b2×c;
pr=combin(365,a);
pr×=combin(365a,b);
pr×=combin(365ab,c);
pr×=combin(365abc,d);
pr/=(double)power(24,a);
pr/=(double)power(6,b);
pr/=(double)power(2,c);
for (i=1; i<=p; i++)
{
pr×=(double)i;
pr/=365.0;
}
tot_pr+=pr;
}
}
}
printf("%i\t%12.10f\n",p,tot_pr);
cerr << p <<"\t" << tot_pr<< "\n";
Michael Shackleford
  37.  Common birthday problem
   What is the minimum number of people do you need, chosen at random, so that there is at least a 50% chance that at least two have the same birthday. Assume that people are born randomly throughout the year. You may ignore leap day.
Answer
Problem 37 Answer
Problem 37 Answer
The answer is 23.
Michael Shackleford, A.S.A.
Solution
Problem 37 Solution
Problem 37 Solution
The probability that n people have all different
birthdays is 1  (Product for i=1 to n of (365n)/365.)
One minus this figure is the probability that there is at
least one common birthday. The probability that there is
at least one common birthday in 22 people is 47.57%, for
23 people it is 50.73%. Since the percentage must be at
least 50% the answer is 23.
Michael Shackleford, A.S.A.
  31.  Compound interest problem #1
   What is the value of $1, invested for one year at 100% interest, compounded infinitely?
Answer
Problem 31 Answer
Problem 31 Answer
The answer $e =~ $2.72
Michael Shackleford, A.S.A.
Solution
Problem 31 Solution
Problem 31 Solution
Let f(t) equal the money in the account at time t. Since the
interest rate is 100% (compounded infinitely) f'(t) = f(t), in
other words at any moment in time the money is doubling (which is
not the case with simple interest). So for what function
does f'(t)=f(t)? This answer is e ^{x}. So the value
of the account after one year is f(1)=e ^{1}=e.
Below is another approach:
The value of $1 invested for n years at interest rate i, compounded x
times per year is (1+(i/x))^{nx}.
In this case f(x) = the value of $1 invested at 100% interest
compounded n times per year = (1+(1/x))^{x}, where x
approaches infinity.
The following table shows values of f(x) for various values of x:
x f(x)
 
1 2.00000000
4 2.44140625
12 2.61303529
52 2.69259695
365 2.71456748
1,000 2.71692393
10,000 2.71814593
100,000 2.71826824
1,000,000 2.71828047
10,000,000 2.71828169
100,000,000 2.71828180
As x approaches infinity it can be seen that f(x) approaches e =~ 2.71828182846.
Michael Shackleford, A.S.A.
  49.  Compound interest problem #2
   This situation actually happened to me at Toyota of Garden Grove, although the numbers were changed for the benefit of simplicity: You are about to pay $10,000 cash for a new car. The finance manager, wanting to loan you money, asks you what interest rate you could earn on that money if you didn't spend it. You answer 6%, compounded monthly. The finance manager then says that he will give you a loan at 7.5% interest, compounded monthly, with a five year term. He then shows you on his computer that if you invested the $10,000 on your own you would earn $3,488.50 in interest in five years. Then he shows you that your payments on the car would be $200.38 a month, making total interest paid on the car $2,022.77. You do the math on your own and you realize it is correct. He then argues that you would come out ahead by taking out a loan since you would make more in interest than you would pay. Is this argument valid?
Answer
Problem 49 Answer
Problem 49 Answer
The answer is no, in other words you should refuse the loan.
Michael Shackleford, A.S.A.
Solution
Problem 49 Solution
Problem 49 Solution
One way of looking at the flaw in his logic is that he is assuming
that you don't withdraw money from your investment to make the car
payments. If you did you would have only $110.47 in the account
after the 57th payment, with three full payments to go.
However, if you were willing to not touch your investment and make
the payments out of your pocket there is still a problem. You have to consider the time
value of money. Lets assume inflation equal to the 6% interest
rate, compounded monthly. The time value of the car payments
made would be $10,364.77, however the time value of your $13,488.50
after five years would only be $10,000. In other words after five years
your initial investment of $10,000 has gone down in value more than the
interest savings.
Note: The payments on a loan of $x, at intrest rate
i, over n periods is $xi/(1v^{n}), where v=1/(1+i). In this
case x=$10,000, i=.075/12=.00625, and n=60.
Michael Shackleford, A.S.A.
  86.  Compound interest problem #3
   This problem has been removed.
Answer
Problem 86 Answer
Problem 86 Answer
The answer is e ^{(x2/2)}, or Exp(x ^{2}/2).
Michael Shackleford, A.S.A.
Solution
Problem 86 Solution
Problem 86 Solution
Let f(t)=account value at time t.
f'(t)=t*f(t).
Let y=f(t).
dy/dt=ty.
dy=ty dt.
dy/y = t dt.
Integrating both sides:
ln(y) = t^{2}/2 + C.
y = exp(t^{2}/2 + C).
y = exp(t^{2}/2). Since y=1 when t=0
Michael Shackleford, A.S.A.
  244.  Connecting four points on a square
   There is a squareshaped state with a town at each corner. Your goal is to build a network of roads linking the four towns together, minimizing the total length of the roads. For example, this could be done with an X shape, but there is a better way. Assuming the side of the square is 200 kilometers, what is the minimum length of the roads? (Hint on the shape of the road network.)
Answer
Problem 244 Answer
Problem 244 Answer
Here is the general shape of the road network:
The minimum road distance is 200 * (3*sqrt(3)/3 + 1) =~ = 546.4102 kilometers.
Michael Shackleford
Solution
Problem 244 Solution
Problem 244 Solution
Here is the general shape of the road network:
The total length of the roads will be 4x+z.
We're told the length of the side of the state is 200, However, let's assume a side length of 1, to be more elegant, and multiply by 200 at the end.
If a side of the square is 1, then 2y+z = 1.
Rearranging, z = 1  2y.
So, the total distance can be expressed as 4x + 1  2y.
From the Pythagorean formula, 0.5^2 + y^2 = x^2.
Rearranging, y^2 = x^2  0.25.
y = (x^2  0.25)^0.5
To put the total distance as a function of just x:
f(x) = 4x + 1  2*(x^2  0.25)^0.5.
Next, set the derivative to equal zero, to find the value of x that maximizes f(x).
f'(x) = 4  2*0.5*2x*(x^2  0.25)^0.5 = 0.
2*0.5*2x*(x^2  0.25)^0.5 = 4
x*(x^2  0.25)^0.5 = 2
x = 2*(x^2  0.25)^0.5
x^2 = 4 * (x^2 0.25)
3x^2 = 1
x^2 = 1/3
x = sqrt(3)/3
Our total road length, based on a side of 1, is:
4*sqrt(3)/3 + 1  2*(1/3  1/4)^0.5 =
4*sqrt(3)/3 + 2*(1/12)^0.5 =
4*sqrt(3)/3 + 1 + 2*0.5*sqr(3)/3 =
3*sqrt(3)/3 + 1 =~ 2.7321
So, for a side of length 200 kilometers, the answer is 200 * (3*sqrt(3)/3 + 1) =~ = 546.4102 kilometers.
Michael Shackleford
  224.  Cooling beer problem
   A refrigerator is kept at a constant temperature of 5 degrees.
A can of beer is placed in it with initial temperature of 25 degrees.
In one hour it cools to 15 degrees.
Assume that that the penguin that lives in the refrigerator not only turns the light on and off when you open the door but shakes the beer so that it maintains a constant temperature (and explodes in your face when you open it).
Hint: The rate of heat transfer is proportional to the difference between the temperature of object to the ambient temperature.
How long will it take to cool the beer to 5.1 degrees from the time it is placed inside the refrigerator?
Answer Problem 224 Answer
ln(200)/ln(2) =~ 7.6439 hours.
Michael Shackleford, July 4, 2017
Solution Problem 224 Solution
Let: <
t = time (in hours) since beer was put in the refigerator.
T = temperature
k = constant of heat transfer
C = constant of integreation
We're given than Dt/dt = k(T5). It is negative because the beer can only decrease in temperature.
Dt = k(T5) dt
dt = 1/k(T5) Dt
Integrate both sides:
t + C = (1/k)*ln(T5)
We're given that the temperature of the beer is 25 at t=0, so plug that in:
C = (1/k)*ln(20)
We're also given that the temperature of the beer is 15 at t=1, so plug that in:
1 + C = (1/k)*ln(10)
We have two equations and two unknowns so we can solve for k and C:
k = ln(2)
C = ln(20)/ln(2)
So, the equation for time and temperature is:
t = ln(20)/ln(2)  ln(T5)/ln(2)
So, if T=5.1, then
t = ln(20)/ln(2)  ln(0.1)/ln(2) = [ln(20)  ln(0.1)]/ln(2) = ln(200)/ln(2) =~ 7.6439 hours.
Michael Shackleford, July 4, 2017
  168.  Cows eating grass problem
   Five cows can eat 2 acres of grass in 10 days. Seven cows can eat 3 acres of grass in 30 days. Each cow eats at a constant rate. The length of the grass before the cows begin grazing is constant. How many days will it take 16 cows to eat 7 acres of grass?
Answer
Problem 168 Answer
Problem 168 Answer
The answer is 70 days
Michael Shackleford, ASA  August 10, 2001
Solution
Problem 168 Solution
Problem 168 Solution
The problem again is, "5 cows can eat 2 acres of grass in 10 days. 7 cows can eat 3 acres of grass in 30 days. The grass grows at a constant rate and each cow eats at a constant rate. The length of the grass before the cows begin grazing is constant. How many days will it take 16 cows to eat 7 acres of grass?"
By the time the cows have eaten all the grass total grass consumed will equal total initial grass plus total grass regrowth. Next lets define some terms.
x=initial amount of grass in an acre.
y=amount of grass grown in one acre in one day.
Putting the given information in the form of equations we get:
50=2x+20y
210=3x+90y
To put the first equation in more simple English, 5 cows eating for 10 days results in a consumption of 50 units of grass. This is equal to the sum of 2x initial units of grass and 20y units of grass growth (10 days times 2 acres).
Next we must solve for x and y. Rewriting the above equations we get:
150=6x+60y (multiplying by 3)
420=6x+180y (multiplying by 2)
Subtracting the first equation from the second we get 270=120y, so y=9/4. Plugging this into either equation we get x=5/2.
The question to be answered is "How many days will it take 16 cows to eat 7 acres of grass?" Let's let d be the number of days. So setting this up as an equation we get.
16d=7x + 7dy
16d=35/2 + 63d/4
d/4=35/2
d=70.
So it will take 70 days for 16 cows to eat 7 acres of grass.
Michael Shackleford, ASA  August 10, 2001
  107.  Create 24 with 1, 3, 4, and 6 problem
   Create the number 24 using only a 1, 3, 4, and 6. You may only use +, , /, and *. Parenthesis are allowed. For example if I asked for 23 an answer would be ((61)*4)+3. This is not a trick question, for example the answer does not involve a number system other than base 10 and does not allow for decimal points.
Answer
Problem 107 Answer
Problem 107 Answer
The answer is 6/(1(3/4)).
Michael Shackleford, A.S.A.
  103.  Create 24 with two 3s and two 7s
   Create the number 24 using only these numbers once each: 3, 3, 7, 7. You may use only the following functions: +, , *, /. This is not a trick question, for example the answer does not involve a number system other than base 10 and does not allow for decimal points.
Answer
Problem 103 Answer
Problem 103 Answer
The answer is [3+(3/7)] x 7.
Thanks go to Stephanie Mullen for this one.
Michael Shackleford, A.S.A.
Solution
Problem 102 Solution
Problem 102 Solution
The density function of the distance to the center is the circumferance of the
circle, centered about the center of the dart board, that the dart hits divided
by the area of the dart board. Let x be the distance to the center and f(x)
the density function.
f(x)=2*pi*x/pi = 2x.
The mean distance to the center is the integral from 0 to 1 of f(x)*x:
Integral from 0 to 1 of 2x*x = 2x^{3}/3 from 0 to 1 = 2/3.
Michael Shackleford, A.S.A.
  130.  Cutting a pizza problem
   What is the most pieces you form with n cuts of a pizza?
Answer
Problem 130 Answer
The answer is n*(n+1)/2 + 1.
Michael Shackleford, A.S.A., 11/10/1998
Solution
Problem 130 Solution
With every cut you should cut through every previous cut. As long as you
keep doing this you will always be able to do so again your next cut. The number
of additional pieces formed on cut x will always be x. Everyone should know that
sum of all integers from 1 to n is n*(n+1)/2. So the sum of additional pieces cut will
be n*(n+1)/2. Finally add 1 because we start with 1 piece: n*(n+1)/2 + 1.
Michael Shackleford, A.S.A., 11/10/1998
  3.  Cycloid problem
   If you drew a dot on the edge of a wheel and traced the path of the dot as the wheel rolled one complete revolution along a line, then the path formed would be called a cycloid, combining both forward and circular motion. What is the length of the path formed by one complete revolution? Assume the wheel has a radius of r. Hint: (1cos(t))^{0.5} = 2^{0.5}×sin(t/2)
Answer
Problem 3 Answer
Problem 3 Answer
The answer is 8r.
Michael Shackleford, A.S.A.
Solution
Problem 3 Solution
Problem 3 Solution
Imagine the circle resting on coordinate (0,0) and moving east.
Also magine the point on the circle to start at (0,0).
Let t be the angle between the point on the circle and the center of the circle.
The position of the point on the circle, relative to t, is:
x = rt  r×sin(t)
y = r  r×cos(t)
Taking the derivatives:
dx/dt = r  r×cos(t)
dy/dt = r×sin(t)
The change in arc length can be defined as
( (dx/dt) ^{2} + (dy/dt) ^{2} ) ^{1/2}.
So the total arc length is the integral from 0 to 2pi of ( (dx/dt)^{2} + (dy/dt)^{2} ) ^{1/2}.
After a few steps this integral becomes:
r×2^{1/2} × (1cos(t))^{1/2}.
Using the hint:
r×2^{1/2} × 2^{1/2} × integral of sin(t/2) dt from 0 to 2×pi
= 2×r × (2×cos(t/w) from 2×pi to 0)
= 8r
Reference: Example 3, page 550, Calculus and Analytic Geometry, 1982 edition.
Michael Shackleford, A.S.A.
  223.  Cylinder in a sphere problem #1
   What is the radius of the cylinder of maximum volume, which can be inscribed in a unit sphere?
Answer Problem 223 Answer
The radius which achieves the maximum volume of the cylinder is (2/3)^0.5 =~ 0.816496581.
Michael Shackleford, July 3, 2017
Solution Problem 223 Solution
Call the radius of the cylinder 1. The distance from the center of the clinder to the edge is 1, because it is in a unit sphere. By phythagerous, we know the height of the cylinder must be (1r^2)^0.5.
The volume of the clinder V is equal to pi*r^2*(1r^2)^0.5.
Take the derivative and set it equal to zero:
pi*[2r*(1r^2)^0.5  r^3*(1r^2)^0.5] = 0.
2r*(1r^2) = r^3
2/3 = r^2
r = (2/3)^0.5 =~ 0.816496581. The radius which achieves the maximum volume of the cylinder is (2/3)^0.5 .
The volume of said cylinder is 2*pi*sqrt(3)/9 =~ 1.209199576.
Michael Shackleford, July 3, 2017
  228.  Cylinder in a sphere problem #2
   A drill drills through a section of a sphere of radius r right throuh the middle, leaving a hole 6" long. What is the volume of the remaining part of the sphere as a function of r?
Answer Problem 228 Answer
The answer is 36 × pi.
Michael Shackleford, Dec. 1, 2017
Solution Problem 228 Solution
Let r = radius of the sphere.
To find the volume of the sphere left after you drive the hole through, consider the sphere centered at (0,0). Then find the volume of the part of the sphere ranging in x value from 3 to +3. Then we'll subtract out the area of the cylinder in the middle that the drill removed.
The area of the sphere from x = 3 to +3 equals:
Integral from 3 to +3 of (pi * (r^2x^2)) dx =
pi * (r^2 * x  x^3/3) from 3 to 3 =
pi * (3r^2  9 + 3r^2  9) =
pi * (6r^2  18)
The volume of the missing cylinder is 6*pi*(r^2  9)
So, the volume of the part that is left is pi * (6r^2  18)  6*pi*(r^2  9) =
36 * pi
So, it makes no difference what the radius of the sphere is, there will 36*pi left over after drilling through it.
Michael Shackleford, Dec. 1, 2017
  17.  Dartboard problem #1
   A dart is thrown at a circular dart board of radius one. The dart can land at any place on the dartboard with equal probability. What is the mean distance between where the dart hits and the center of the board?
Answer
Problem 17 Answer
Problem 17 Answer
The answer is 2/3.
Michael Shackleford, A.S.A.
Solution
Problem 17 Solution
Problem 17 Solution
Lets consider the dartboard in polar coordinates. The dart can land at any angle relative
to the center with equal probability. Next take the intregral from the least to the greatest
distance, 0 to 1. The density function at distance d is 2*pi*d/pi = 2d. So the answer is:
Integral from 0 to 1 of d*2d = 2d^{3}/3 from 0 to 1 = 2/3.
Michael Shackleford, A.S.A.
  102.  Dartboard problem #2
   At dark is thrown at a dart board of radius 1. The dart can hit anywhere on the board with equal probability. What is the mean distance between where the dart hits and the center?
Answer
Problem 102 Answer
Problem 102 Answer
The answer is 2/3.
Michael Shackleford, A.S.A.
Solution
Problem 102 Solution
Problem 102 Solution
The density function of the distance to the center is the circumferance of the
circle, centered about the center of the dart board, that the dart hits divided
by the area of the dart board. Let x be the distance to the center and f(x)
the density function.
f(x)=2*pi*x/pi = 2x.
The mean distance to the center is the integral from 0 to 1 of f(x)*x:
Integral from 0 to 1 of 2x*x = 2x^{3}/3 from 0 to 1 = 2/3.
Michael Shackleford, A.S.A.
  174.  Dating service problem
   The puzzleville dating service has m men and w women, such that m>w. Introductions are made by randomly drawing any two names (without replacement) out of the total membership. If one man and one woman is drawn then a blind date is arranged. The probability that a draw will result in one of each gender is exactly 50%. After an advertising campaign to attract women to the club 100 more members join the service. After the increase in membership both properties still hold true: m>w and a 50% probability of a successful drawing. After the increase how many men and how many women are in the service?
Answer
Problem 174 Answer
Problem 174 Answer
The answer is 351 men and 325 women.
Michael Shackleford, A.S.A.
Solution
Problem 174 Solution
Problem 174 Solution
First let's define some terms:
m = men before increase
w = women before increase
m' = men after increase
w' = women after increase
d = mw
d' = m'w'
The fact that a random drawing will result in a date 50% of the time tells us that 2*(m/(m+w))*(w/(m+w1)) = 0.5
(m/(m+w))*(w/(m+w1)) = .25
m/(m+w) = (m+w1)/(4w)
4mw = (m+w)*(m+w1)
4mw = m^{2} + mw m + mw + w^{2}  w
4mw = m^{2} + w^{2} 2mw  m  w
(mw)^{2} = m + w, or d^{2} = m+w
By the same reasoning:
(m'w')^{2} = m' + w', or d'^{2} = m'+w'
We are also given:
m'+w'mw=100
d'^{2}  d^{2} = 100
(d'+d)*(d'd) = 100
We know that d and d' must be integers so the only possible values for d'+d and d'd are: (100,1), (50,2), (25,4), (20,5), (10,10).
Let's consider the first possibility of (100,1). If this were the solution then d'+d=100 and d'd=1. Adding the two equations together yields 2d'=101 > d'=101/2. However since m' and w' are integers then d' is also an integer and can not be 101/2. For the same reason (25,4) and (20,5) are not possible because the sum of the two factor is odd.
Let's consider (10,10). Then d'+d=10, d'd = 10 > d=0 > mw=0. However the problem stated m>w, so (10,10) doesn't work.
That leaves only (d'+d,d'd)=(50,2). Adding the two equations gives us d'=m'w'=26.
Remember, d'^{2} = m'+w', so m'+w'=26^{2} = 676.
Remember also that m'w'=26
Adding the two equations give us 2m'=702 > m'=351 > w'=325.
So after in increase there are 325 women and 351 men. From here it can also be easily found that before the increase there were 300 men and 276 women.
Thanks to Ady Tzidon for suggesting a very similar problem and to Rob Pratt for this solution (which is better than how I initially solved it).
Michael Shackleford, A.S.A.
  141.  Dice game #1
   Player A and B bet on the total roll of two normal dice. Player A bets that a 12 will be rolled first. Player B bets that two 7s will be rolled consecutively first. They keep rolling until one person wins? What is the probability A will win?
Answer
Problem 141 Answer
Problem 141 Answer
The probability that A will win is 7/13 =~ 0.538462 .
Thanks to
Extra Stuff: Gambling Rambling by Peter Griffin for this problem. See chapter 6.
Michael Shackleford, ASA, August 19 1999
Solution
Problem 141 Solution
Problem 141 Solution
Let p be the probability that A wins. Consider the following four events:
 If the first roll is a 12 then A wins.
 If the first roll is a anything other than a 7 and 12
then neither side has improved their odds and the probability of A winning is still p.
 If the first roll is a 7 and the second roll is a 12 then A wins.
 If the first roll is a 7 and the second roll is anything other than a 7 and 12
then neither side has improved their odds and the probability of A winning is still p.
We can now express p as follows:
p = 1/36 + (29/36)*p + (1/6)*(1/36) + (1/6)*(29/36)* p
p = 7/216 + (203/216)*p
13p = 7
p = 7/13
So the probability that A will win is 7/13 =~ 0.538462 .
Thanks to
Extra Stuff: Gambling Rambling by Peter Griffin for this problem. See chapter 6.
Michael Shackleford, ASA, August 19 1999
  142.  Dice game #2
   Suppose the two players from problem 141 do not wish to stop with one bet resolved but that they play through millions of rolls of the dice. All rolls are one continuous stream, in other words they don't start over after a bet is resolved. If you were to jump ahead x rolls (where x>0) what is the probability A would win the next bet?
Answer
Problem 142 Answer
Problem 142 Answer
The answer is 41/78 =~ 0.525641 .
Thanks to
Extra Stuff: Gambling Rambling by Peter Griffin for this problem. See chapter 6.
Michael Shackleford, ASA, August 19 1999
Solution
Problem 142 Solution
Problem 142 Solution
Recall from problem 141 that the probability that A wins the first time is 7/13.
Thus the probability B will win the first time is 6/13. In fact if the last roll was
not a 7 the probability that A and B will win are still 7/13 and 6/13 respectively. If
the last roll was a 7 the probability B will win is:
1/6 + (29/36)*(6/13) = 252/468 = 7/13.
So the probability that A will win if the last roll was a 7 is 6/13, otherwise it is
7/13. Likewise the probability that B will win if the last roll was a 7
is 7/13, and 6/13 otherwise.
By jumping ahead x rolls the probability that the (x1)^{th} roll was a 7
is 1/6 and the probability of a non 7 is 5/6. Thus the probability of of A winning
the next bet resolved is:
pr((x1)^{th} roll = 7)*(6/13) + pr((x1)^{th} roll != 7)*(7/13) =
(1/6)*(6/13) + (5/6)*(7/13) = 41/78 =~ 0.525641 .
Thanks to
Extra Stuff: Gambling Rambling by Peter Griffin for this problem. See chapter 6.
Michael Shackleford, ASA, August 19 1999
  143.  Dice game #3
   Consider the contest posed in problem 142. What is the probability A wins a bet on the x^{th} roll (where x>0), what is the probability for B?
Answer
Problem 143 Answer
Problem 143 Answer
The answer is 1/36 for both A and B.
Thanks to
Extra Stuff: Gambling Rambling by Peter Griffin for this problem. See chapter 6.
Michael Shackleford, ASA, August 19 1999
Solution
Problem 143 Solution
Problem 143 Solution
The probability of a win for A, or of rolling a 12 on any given roll is (1/6)*(1/6) = 1/36.
The probability that the (x1)^{th} roll will be a 7 is 1/6 and the
probability the x^{th} roll will be a 7 is 1/6, thus the probability of
the x^{th} roll being a win for B is also (1/6)*(1/6) = 1/36.
Thanks to
Extra Stuff: Gambling Rambling by Peter Griffin for this problem. See chapter 6.
Michael Shackleford, ASA, August 19 1999
  144.  Dice game #4
   The answer to problem 143 seems to imply that A and B have equal long term odds of winning. However, the answer to problem 142 seems to imply that A always has the advantage. What is the explanation behind this apparant paradox?
Answer
Problem 144 Answer
Problem 144 Answer
The answer is a bit hard to explain and appears paradoxical on the surface.
Consider the fact that whenever a bet is resolved the probability of the same
player winning the next one is 7/13. So in this game there will be longer strings
of the same player winning in a row than in a game where every trial is independent of
the last, like flipping a coin. Player B can expect winnings even closer together
than player A because the probability of B winning the roll after a win is just 1/6,
as opposed to the probability of 1/36 of A winning one roll after the last win.
So with Bs wins being more clumped together picking a flip at random is more likely
to preceed an A win, since As wins are more spaced apart.
Thanks to
Extra Stuff: Gambling Rambling by Peter Griffin for this problem. See chapter 6.
Michael Shackleford, ASA, August 19 1999
  145.  Dice game #5
   If A and B, from problems 141, play through one million rolls how much can A expect to be ahead if they each wager $1 for each bet resolved. If they play through one million bets resolved how much can A expect to be ahead?
Answer
Problem 145 Answer
Problem 145 Answer
A's expected gain over a million rolls is 2.78 cents.
A's expected gain over a million bets resolved is 8.33 cents.
Thanks to
Extra Stuff: Gambling Rambling by Peter Griffin for this problem. See chapter 6.
Michael Shackleford, ASA, August 19 1999
Solution
Problem 145 Solution
Problem 145 Solution
The expected gain by A on the first roll is (1/36) * $1 =~ 2.78 cents. On all future
rolls they have an equal chance of winning so there is no gain on the next 999,999.
Thus A's expected gain over any number of rolls is 2.78 cents.
The probability A wins the first bet is 7/13.
The probability A wins the second bet is (7/13)*(7/13)+(6/13)*(6/13) = 85/169, using
logic from earlier problems.
The probability A wins the second bet is (85/169)*(7/13)+(84/169)*(6/13) = 1099/2197.
Continuing this pattern, the probability A wins the n^{th} bet is
(((13^{n}+1)/2)/13^{n})*(7/13)+(((13^{n}1)/2)/13^{n})*(6/13) =
((13^{n}+1)/2)/13^{n+1}.
The expected gain by A on the first bet is (7/13)*(+1) + (6/13)*(1) = 1/13.
The expected gain by A on the second bet is (85/169)*(+1) + (84/169)*(1) = 1/169.
The expected gain by A on the n^{th} bet is 1/13^{n}.
Taking the sum from 1 to one million the total gain by A is:
$\sum $(for i=1 to 1,000,000) 1/13^{i} =~ (1/13)/(12/13) = 1/12 =~ 8.33 cents.
Thanks to
Extra Stuff: Gambling Rambling by Peter Griffin for this problem. See chapter 6.
Michael Shackleford, ASA, August 19 1999
  146.  Dice game #6
   How can the contest described in 141 end fairly so that neither player has an advantage?
Answer
Problem 146 Answer
Problem 146 Answer
The game should end immediately after B wins the first time, regardless of how many
A wins preceed it.
Thanks to
Extra Stuff: Gambling Rambling by Peter Griffin for this problem. See chapter 6.
Michael Shackleford, ASA, August 19 1999
  34.  Drug test problem
   10% of the people in a certain population use an illegal drug. A drug test yields the correct result 90% of the time, whether the person uses drugs or not. A random person is forced to take the drug test and the result is positive. What is the probability he uses drugs?
Answer
Problem 34 Answer
Problem 34 Answer
The answer is 1/2.
Michael Shackleford, A.S.A.
Solution
Problem 34 Solution
Problem 34 Solution
The probability of event A given event B is Pr(A and B)/Pr(B).
In this specific case this is [(.1)(.9)]/[(.1)(.9)+(.9)(.1)] = 1/2.
Michael Shackleford, A.S.A.
  42.  Drunken walk problem #1
   A drunk is on one point of an ngon, such that n is an even number. The drunk moves along the perimeter of the ngon. Each steps takes him to an adjacent point, and every step is chosen at random. What is the expected number of steps before he arrives at the point directly opposite of his starting point (as a function of n)?
Answer
Problem 42 Answer
Problem 42 Answer
The answer is n ^{2}/4.
Michael Shackleford, A.S.A.
Solution
Problem 42 Solution
Problem 42 Solution
See the ant and spider problem for the method to computer any given
expected number. Use telescoping sums to find that the expected number
at either point next to the final point is (n1). Use telescoping sums
again to determine the expected number of the starting point as a function
of the next to last point.
Michael Shackleford, A.S.A.
  222.  Eight pieces of bread problem
   Two men sit down to have dinner. One contributed three loaves of bread and the other five. As they were about to start eating, a third man came along, who asked to join. They said "yes," and all three ate until the eight loaves were gone. As the third man was leaving he reached into his pocket and produced eight coins. He left them on the table for the two other men to split as they saw fit. What is a fair way to divide the eight coins?
Answer Problem 222 Answer
The man who contributed five loaves should get seven coins and the man who contributed three loaves should get one coin.
Michael Shackleford, July 3, 2017
Solution Problem 222 Solution
Think about who contributed what to what the third man ate. Assuming each man ate the same amount, each ate 8/3 loaves. The man who contributed 5 loaves ate 8/3 of them, leaving 7/3 loaves to the third man. The man who contributed 3 loaves ate 8/3 of them, leaving 1/3 loaves to the third man.
So, of the 8/3 loaves eaten by the third man, 7/3 were from the man with five loaves and 1/3 from the man with three loaves. Thus split the coins according to whose bread you ate  7 coins to the man with five loaves and one coin to the man with three loaves.
Michael Shackleford, July 3, 2017
  147.  Expected number of random numbers, uniformally distributed from 0 to 1, needed for the sum to be greater than 1?
   What is the expected number of random numbers, uniformally distributed from 0 to 1, needed for the sum to be greater than 1?
Answer
Problem 147 Answer
Problem 147 Answer
The answer is e =~ 2.7182818
Thanks to
Extra Stuff: Gambling Rambling by Peter Griffin for this problem. See chapter 11.
Michael Shackleford, ASA, August 20 1999
Solution
Problem 147 Solution
Problem 147 Solution
Lets rephrase the question to a more general case, what is the expected number needed
for the sum to exceed x, where x<=1?
Obviously at least one number will be needed. If the first number is less than x then
another will will need to be drawn with a lesser total needed to finish.
So E(x) = 1 + integral from 0 to x of E(y) dy.
Take the derivative of both sides...
E'(x) = E(x)
What function is a derivative of itself? e^{x} of course!
So the answer is e^{1} = e.
Thanks to
Extra Stuff: Gambling Rambling by Peter Griffin for this problem. See chapter 11.
Michael Shackleford, ASA, August 20 1999
  218.  Expected random numbers from [0,1] for the sum to exceed 1?
   What is the expected number of random numbers drawn from a uniform distribution from 0 to 1 so that the sum is greater than 1?
Answer Problem 218 Answer
The answer is e, or approximately 2.7182818.
Michael Shackleford, ASA — April 10, 2016
Solution   216.  Expected rolls with two dice
   What is the expected number of rolls to attain every total from 2 to 12 with two sixsided dice?
Answer Problem 216 Answer
The answer is 61.217385.
Michael Shackleford, ASA — July 17, 2013
Solution Problem 216 Solution
Solution
This question was asked at TwoPlusTwo.com, and was answered correctly by BruceZ. The following solution is the same method as that of BruceZ, who deserves proper credit. It is a difficult answer, so pay attention.
First, consider the expected number of rolls to obtain a total of two. The probability of a two is 1/36, so it would take 36 rolls on average to get the first 2.
Next, consider the expected number of rolls to get both a two and three. We already know it will take 36 rolls, on average, to get the two. If the three is obtained while waiting for the two, then no additional rolls will be needed for the 3. However, if not, the dice will have to be rolled more to get the three.
The probability of a three is 1/18, so it would take on average 18 additional rolls to get the three, if the two came first. Given that there is 1 way to roll the two, and 2 ways to roll the three, the chances of the two being rolled first are 1/(1+2) = 1/3.
So, there is a 1/3 chance we'll need the extra 18 rolls to get the three. Thus, the expected number of rolls to get both a two and three are 36+(1/3)×18 = 42.
Next, consider how many more rolls you will need for a four as well. By the time you roll the two and three, if you didn't get a four yet, then you will have to roll the dice 12 more times, on average, to get one. This is because the probability of a four is 1/12.
What is the probability of getting the four before achieving the two and three? First, let's review a common rule of probability for when A and B are not mutually exclusive:
pr(A or B) = pr(A) + pr(B)  pr(A and B)
You subtract pr(A and B) because that contingency is double counted in pr(A) + pr(B). So,
pr(4 before 2 or 3) = pr(4 before 2) + pr(4 before 3)  pr(4 before 2 and 3) = (3/4)+(3/5)(3/6) = 0.85.
The probability of not getting the four along the way to the two and three is 1.0  0.85 = 0.15. So, there is a 15% chance of needing the extra 12 rolls. Thus, the expected number of rolls to get a two, three, and four is 42 + 0.15×12 = 43.8.
Next, consider how many more rolls you will need for a five as well. By the time you roll the two to four, if you didn't get a five yet, then you will have to roll the dice 9 more times, on average, to get one, because the probability of a five is 4/36 = 1/9.
What is the probability of getting the five before achieving the two, three, or four? The general rule is:
pr (A or B or C) = pr(A) + pr(B) + pr(C)  pr(A and B)  pr(A and C)  pr(B and C) + pr(A and B and C)
So, pr(5 before 2 or 3 or 4) = pr(5 before 2)+pr(5 before 3)+pr(5 before 4)pr(5 before 2 and 3)pr(5 before 2 and 4)pr(5 before 3 and 4)+pr(5 before 2, 3, and 4) = (4/5)+(4/6)+(4/7)(4/7)(4/8)(4/9)+(4/10) = 83/90. The probability of not getting the four along the way to the two to four is 1  83/90 = 7/90. So, there is a 7.78% chance of needing the extra 7.2 rolls. Thus, the expected number of rolls to get a two, three, four, and five is 43.8 + (7/90)×9 = 44.5.
Continue with the same logic, for totals of six to twelve. The number of calculations required for finding the probability of getting the next number before it is needed as the last number roughly doubles each time. By the time you get to the twelve, you will have to do 1,023 calculations.
Here is the general rule for pr(A or B or C or ... or Z)
pr(A or B or C or ... or Z) =
pr(A) + pr(B) + ... + pr(Z)
 pr (A and B)  pr(A and C)  ...  pr(Y and Z) Subtract the probability of every combination of two events
+ pr (A and B and C) + pr(A and B and D) + ... + pr(X and Y and Z) Add the probability of every combination of three events
 pr (A and B and C and D)  pr(A and B and C and E)  ...  pr(W and X and Y and Z) Subtract the probability of every combination of four events
Then keep repeating, remembering to add probability for odd number events and to subtract probabilities for an even number of events. This obviously gets tedious for large numbers of possible events, practically necessitating a spreadsheet or computer program.
The following table shows the the expected number for each step along the way. For example, 36 to get a two, 42 to get a two and three. The lower right cell shows the expected number of rolls to get all 11 totals is 61.217385.
Expected Number of Rolls Problem 
Highest Number Needed 
Probability 
Expected Rolls if Needed 
Probability not Needed 
Probability Needed 
Expected Total Rolls 
2  0.027778  36.0  0.000000  1.000000  36.000000 
3  0.055556  18.0  0.666667  0.333333  42.000000 
4  0.083333  12.0  0.850000  0.150000  43.800000 
5  0.111111  9.0  0.922222  0.077778  44.500000 
6  0.138889  7.2  0.956044  0.043956  44.816484 
7  0.166667  6.0  0.973646  0.026354  44.974607 
8  0.138889  7.2  0.962994  0.037006  45.241049 
9  0.111111  9.0  0.944827  0.055173  45.737607 
10  0.083333  12.0  0.911570  0.088430  46.798765 
11  0.055556  18.0  0.843824  0.156176  49.609939 
12  0.027778  36.0  0.677571  0.322429  61.217385 
This question was raised and discussed in the forum of my companion site Wizard of Vegas.
Michael Shackleford, ASA — July 17, 2013
  158.  Final Jeopardy problem
   Two players have made it to Final Jeopardy. Player A has $7000 and player B has $5000. Each player has a 60% of answering correctly and both know this to be true of their opponent. For the sake of simplicity assume a tie goes to player A. What should be the betting strategy of each player? You may assume that each player has a way to draw a random number if needed.
Answer
Problem 158 Answer
Problem 158 Answer
Player A should bet $3001 with probability 60/76 and $0 with probability 16/76. Player B should bet $5000 with probability 40/76 and $0 with probability 30/76.
Michael Shackleford, ASA  November 25, 2000
  175.  Five ants on a pentagon problem
   Five ants are on the corners of an equilateral pentagon with side of length 1. They each crawl directly towards the next ant, all at the same speed and traveling in the same orientation. How long will each ant travel before they all meet in the center?
Answer
Problem 175 Answer
Problem 175 Answer
The answer is 1/(1cos(72)) = 1+5 ^{1/2}.
Michael Shackleford, A.S.A.
Solution
Problem 175 Solution
Problem 175 Solution
This is a twist on the classic four ants problem. I have also seen dogs and heat seeking missles used instead of ants. I think it is helpful to first understand how to solve this problem for the case of four ants at the corners of a square with side of length 1. There is an easy and a hard way to solve this problem. I'll start with the hard way.
It is intuitive that at all times the ants will form the corners of a square, ever decreasing in size and rotating about the center.
Let C be the center of the square at point (0,0). Let the ants be at initial points (.5,.5), (.5,.5), (.5,.5), and (.5,.5). Let ant 1 bet at (.5,.5) who is chasing ant 2 at (.5,.5). Consider the line from C to ant 1. Next consider the line from ant 1 to ant 2. The angle between these lines is 135 degrees.
Theorem: Let C be a fixed point, D a point on a curve, r the distance from C to D, x the polar angle formed by r, and y the angle formed by the angle between CD and the tangent line to the curve at point D. Then tan(y)=r/(dr/dx).
So the line of sight from the center to the initial point of ant 1 would be be at a 270 degree angle. The line of sight from ant 1 to ant 2 would be a vertical line of 90 degrees. The angle between these lines is 270 degrees. The tangent of 270 is 1. So we have r/(dr/dx) = 1. Thus r(x)=c*e^{x}.
At the moment the ants start walking x equals 0. At this moment the distance from the center to any of the ants is 2^{1/2}. So r(x)=2^{1/2}*e^{x}.
The arc length is given by the formula:
Integral from negative infinity to 0 of ((f(x))^{2}+(f^{'}(x))^{2})^{1/2} =
Integral from negative infinity to 0 of ((2^{1/2}*e^{x})^{2} + (2^{1/2}*e^{x})^{2})^{1/2} =
Integral from negative infinity to 0 of (2^{1}*e^{2x} + 2^{1}*e^{2x})^{1/2} =
Integral from negative infinity to 0 of e^{x} =
e^{x} from negative infinity to 0 =
=0  (e^{0}) =
0 (1) = 1.
The easy way to solve this problem is to first note that the direction of one ant will always be perpendicular to the path of the ant he is chasing. If ant A is chasing ant B then B has zero velocity in the direction of A's path. In other words B neither approaches nor
recedes from A during the walk. So it is like A is walking directly towards an unmoving ant B, in which case the total distance is simply the intial separation of 1.
No we're ready to consider the pentagon case. Whether you solve the square case the easy or the hard way it should be clear that as ant A chases ant B what is important is the velocity of ant B relative to the path of ant A.
Before we go further we have to do some trigonometry. Consider an equilateral pentagon of side 1 with the first vertice at (0,0) and the second at (1,0). Where is the third vertice? Given that the interior angles of a pentagon are 108 degrees the next vertice is at (1+cos(72),sin(72)).
So as ant A inches 1 unit towards ant B, ant B will move cos(72) units directly away from ant A in the same direction and sin(72) in a perpenticular direction. We know form the four corners problem that the perpenticular direction can be ignorred. So for every micron ant A moves towards ant B, ant B will move away from ant A by cos(72) microns.
The total distance covered by A, or any ant, is the sum for i=0 to infinity of cos(72)^{i}. We know from infinite sums that this equals 1/(1cos(72)).
If you prefer to express the answer another way cos(72)=1/(1+5^{1/2}). So 1/(1cos(72)) = 1/(11/(1+5^{1/2})) = 1/((5^{1/2})/(1+5^{1/2})) = (1+5^{1/2})/5^{1/2} = 1 + 1/5^{1/2} =~ 1.4472 .
For more information on this problem please see the November 1957 and July 1965 issues of Scientific American. The July 1965 issue notes that for the general case of an ngon the total distance is as follows. Let A and B be the initial points of any two consecutive ants, where A is chasing B. Let C be the center of the ngon. Consider the line segment AB extended to a point X, such that the angle XCA is a right angle. The total distance any ant will travel is the same as the distance from A to X.
Thus, if the interior angle of the ngon is x then the total distance is 1/(1+cos(x)). So for an ngon the total distance is 1/[1+cos(180360/n)].
Michael Shackleford, A.S.A.
  230.  Five coins problem
   Two players, Sam and Dan, each have five coins. Both must choose to place one to five coins in his hand. At the same time, each must reveal the number of coins played. If both choose the same number of coins, then Sam will win collect all coins played. If both choose different numbers of coins, then Dan will collect all coins played. Assuming both players are prefect logicians, what is the optimal strategy for Dan?
Answer Problem 230 Answer
Dan should pick x coins with probability pr(x), as follows:
pr(1) = 77/548
pr(2) = 107/548
pr(3) = 117/548
pr(4) = 122/548
pr(5) = 125/548
Michael Shackleford, Jan. 15, 2018
Solution Problem 230 Solution
It stands to reason that Sam's expected win should be the same regardless of whatever he picks.
Let p_{1} = Probability Dan picks 1 coin
Let p_{2} = Probability Dan picks 2 coins
Let p_{3} = Probability Dan picks 3 coins
Let p_{4} = Probability Dan picks 4 coins
Let p_{5} = Probability Dan picks 5 coins
If Sam picks one coin, then Sam's expected value is p_{1}×2  p_{2}×3  p_{3}×4  p_{4}×5  p_{5}×6
If Sam picks two coins, then Sam's expected value is p_{1}×3 + p_{2}×4  p_{3}×5  p_{4}×6  p_{5}×7
If Sam picks three coins, then Sam's expected value is p_{1}×4  p_{2}×5 + p_{3}×6  p_{4}×7  p_{5}×8
If Sam picks four coins, then Sam's expected value is p_{1}×5  p_{2}×6  p_{3}×7 + p_{4}×8  p_{5}×9
If Sam picks five coins, then Sam's expected value is p_{1}×6  p_{2}×7  p_{3}×8  p_{4}×9 + p_{5}×10
Setting the expected value of Sam picking one to Sam picking two:
p_{1}×2  p_{2}×3  p_{3}×4  p_{4}×5  p_{5}×6 = p_{1}×3 + p_{2}×4  p_{3}×5  p_{4}×6  p_{5}×7
p_{1}×5  p_{2}×7 + p_{3}×1 + p_{4}×1 + p_{5}×1
Setting the expected value of Sam picking one to Sam picking three:
p_{1}×2  p_{2}×3  p_{3}×4  p_{4}×5  p_{5}×6 = p_{1}×4  p_{2}×5 + p_{3}×6  p_{4}×7  p_{5}×8
p_{1}×6 + p_{2}×2  p_{3}×10 + p_{4}×2 + p_{5}×2
Setting the expected value of Sam picking one to Sam picking four:
p_{1}×2  p_{2}×3  p_{3}×4  p_{4}×5  p_{5}×6 = p_{1}×5  p_{2}×6  p_{3}×7 + p_{4}×8  p_{5}×9
p_{1}×7 + p_{2}×3 + p_{3}×3  p_{4}×13 + p_{5}×3
Setting the expected value of Sam picking one to Sam picking five:
p_{1}×2  p_{2}×3  p_{3}×4  p_{4}×5  p_{5}×6 = p_{1}×6  p_{2}×7  p_{3}×8  p_{4}×9 + p_{5}×10
p_{1}×8 + p_{2}×4 + p_{3}×4 + p_{4}×4  p_{5}×16
It goes without saying that p_{1} + p_{2} + p_{3} + p_{4} + p_{5} = 1
Now we have five equations and five unknowns. Here is our matrix for these equations:
5  7  1  1  1  0 
6  2  10  2  2  0 
7  3  3  13  3  0 
8  4  4  4  16  0 
1  1  1  1  1  1 
A little matrix algebra gives us:
pr(1) = 77/548
pr(2) = 107/548
pr(3) = 117/548
pr(4) = 122/548
pr(5) = 125/548
Michael Shackleford, Jan. 15, 2018
  108.  Five people, a pile of coconuts and a monkey problem #2
   On a deserted island live five people and a monkey. One day everybody gathers coconuts and puts them together in a community pile, to be divided the next day. During the night one person, not trusting the other, decides to secretly take his share for himself. He divides the coconuts into five equal piles, with one coconut left over. He gives the extra coconut to the monkey, hides his pile, and puts the other four piles back into a single pile. The other four islanders then do the same thing, one at a time, each giving one leftover coconut to the monkey to make the piles divide equally. What is the smallest possible number of coconuts in the original pile?
Answer
Problem 108 Answer
Problem 108 Answer
The answer is 3121.
Michael Shackleford, A.S.A.
Solution
Problem 108 Solution
Problem 108 Solution
The original pile must be a number such that you can subtract one and multiply
by 4/5 and get an integer. Possibilities are 6, 11, 16, 21, 26 ...
After the first person is finished the remaining pile will have 4, 8, 12, 16, 20, ...
The remaining pile after the first person must be a number such that you can subtract
one then multiply by 4/5 and get an integer. Possibilities are 16, 36, 56, 76, 96, ...
After the second person is finished the remaining pile will have 12, 28, 44, 60, 76, ...
The remaining pile after the second person must be a number such that you can subtract
one then multiply by 4/5 and get an integer. Possibilities are 76, 156, 236, 316, 396, ...
After the third person is finished the remaining pile will have 60, 124, 188, 252, 316 ...
The remaining pile after the third person must be a number such that you can subtract
one then multiply by 4/5 and get an integer. Possibilities are 316, 636, 956, 1276, 1596, ...
After the fourth person is finished the remaining pile will have 252, 508, 764, 1020, 1276 ...
The remaining pile after the fourth person must be a number such that you can subtract
one then multiply by 4/5 and get an integer. The smallest possibilities is 1276
After the fifth person is finished the remaining pile will have 1020 ...
So the fifth person will get 127610201 = 255.
The fifth person will leave behind a pile of 1020.
The fourth person will have left behind a pile of 1020+255+1 = 1276.
The fourth person will get 1276/4 = 319.
The third person will have left behind a pile of 1276+319+1 = 1596.
The third person will get 1596/4 = 399.
The second person will have left behind a pile of 399+1596+1 = 1996.
The second person will get 1996/4 = 499.
The first person will have left behind a pile of 499+1996+1 = 2496.
The first person will get 2496/4 = 624.
The original pile must have had 624+2496+1 = 3121.
Michael Shackleford, A.S.A.
  85.  Five pirates and a 1000 coins problem
   Five pirates have come across a treasure of 1000 coins. According to pirate rules the pirate of highest rank must make a suggestion on how to divide the money. If a majority agree to his suggestion then it is to be followed by all the pirates. However, if the suggestion does not get a majority approval then the suggesting pirate is thrown overboard, after which time the remaining pirate of highest rank then makes a suggestion under the same rules. This process repeats, if necessary, until only the pirate of lowest rank is left, in which case he would get everything. Any pirate may suggest any distribution, rank does not guarantee getting more coins than anybody else. Assume that all pirates are infinitely greedy, infinitely logical, and infintely bloodthirsty, and that each pirate knows this to be true of every other pirate. The highest priority of each pirate is to get as much money for themselves as possible. The second highest priority is to throw overboard the other pirates. A pirate will vote to throw another one over even if they have no monetary gain by doing so, and even if it would cost them their own life, but would not if throwing them over would cost even 1 coin. How should the first pirate suggest dividing the money?
Answer
Problem 85 Answer
Problem 85 Answer
He should suggest giving himself 997 coins,
the pirate second in rank 0 coins,
the priate third in rank 1 coin,
the priate fourth in rank 0 coins,
and the pirate fifth in rank 2 coins.
Michael Shackleford, A.S.A.
Solution
Problem 85 Solution
Problem 85 Solution
Call the pirate of highest rank pirate 1, the pirate of
second highest rank pirate 2, and so on. Let
(x,y,z,...) denote giving pirate 5 x coins, pirate
4 y coins, and so on.
If only pirate 5 were left he would get everything.
If only pirate 4 and 5 were left then pirate 4 has no hope, regardless of what he suggests pirate 5 will vote to throw him overboard, and overboard he will go. Thus it doesn't make any difference what pirate 4 suggests.
If only pirates 3,4, and 5 were left then pirate 3 could count on pirate 4's support as long as
pirate 4 got at least 1 coin. This is because if pirate 3 goes overboard pirate 4 knows he will get nothing. So pirate 3 would suggest (0,1,999) to which he and pirate 4 would vote 'yes.' Note that according to the problem pirate 4 would vote 'no' to (0,0,1000), even though a 'yes' vote would save his own life his pleasure of throwing pirate 3 over would be worth his
own life.
If only pirates 2,3,4, and 5 were left then pirate 2 could count on pirate 3 voting 'no' unless pirate 3 got more than 999 coins, which would leave everybody else with 0, which obviously wouldn't pass. So pirate 2 should not count on pirate 3's support so he shouldn't waste any
coins on him. If pirate 2 goes overboard he knows the money will be distributed (0,1,999). So if he offers 2 coins to pirate 4 and 1 coin to pirate 5 then he will get their support since they will come out better voting 'yes.' Thus pirate 2's suggestion would be (1,2,0,997) to which pirates 2, 4 and 5 would vote 'yes.'
Thus if all five pirates were left pirate 1 could count on pirate 2 voting 'no' since pirate 2 would get 997 if pirate 1 went overboard. So pirate 1 should not waste any coins on pirate 2 since he will vote 'no' anyway and concentrate on winning 2 votes from the other 3. If pirate 1 goes overboard the money will be distributed (1,2,0,997). So it would cost 2 coins to get
pirate 5's support, 3 coins to get priate 4's support, and 1 coin to get pirate 3's support. Since he only needs to buy two votes the cheapest two are pirates 5 and 3. Thus pirate 1 would
suggest (2,0,1,0,997) to which pirate 1, 3, and 5 would vote 'yes'.
Thanks to
The Ultimate Puzzle Site which has also posted this problem, click
here to see their solution if this one was too confusing.
Michael Shackleford, A.S.A.
  162.  Five pumpkins problem
   Five pumpkins are weighed two at a time in all ten sets of two. The weights are recorded as 16, 18,19, 20, 21, 22, 23, 24, 26, and 27 pounds. All individual weights are also integers. How much does each pumpkin weigh?
Answer
Problem 162 Answer
Problem 162 Answer
The answer is 7, 9, 11, 12, and 15 pounds.
Thanks to Elizabeth Gomez for this problem.
Michael Shackleford, ASA  February 11, 2001
  98.  Five rectangles in a square problem
   You are to form five rectangles using each integer once from 1 to 10 as the lengths of the sides. How can you tile a square, without overlapping, with the five rectangles? There are four possible answers.
Answer
Problem 98 Answer
Problem 98 Answer
Here are the four possible answers, the last one appears in the
picture below.
 1x9, 2x8, 3x6, 4x7, 5x10 = 11x11 square
 1x6, 2x10, 3x9, 4x7, 5x8 = 11x11 square
 1x2, 3x7, 4x6, 5x10, 8x9 = 13x13 square
 1x2, 3x8, 4x5, 7x9, 6x10 = 13x13 square
. 












. 












. 












. 












. 












. 












. 












. 












. 












. 












. 












. 












. 












Go back to Problem 98
Go back to Shack's Math Problems
Michael Shackleford, A.S.A
Solution
Problem 98 Solution
Problem 98 Solution
This solution is hard to explain so if you don't understand my
explanation please forgive me.
First consider the minimum and maximum possible areas the rectangles
can cover, not neccessarily forming a square. The maximum area would
be (10*9)+(8*7)+(6*5)+(4*3)+(2*1)=190. The minimum area would
be (10*1)+(9*2)+(8*3)+(7*4)+(6*5)=110. The three possible squares between
these two bounds are 121 (11 by 11), 144 (12 by 12), and 169 (13 by 13).
Try to tile a 12 by 12 square. It makes common sense that one
rectangle should fit in each corner of the square with the fifth one
in the middle. There is no other possibility since the sides of the
rectangles are all different lengths. Each side of the square shall
consist of the sides of two different rectangles. Thus the sides
will consist of these sums: 10+2, 9+3, 8+4, 7+5. The middle
rectangle will be the leftover two numbers, 6 by 1.
It also stands to reason that each side of the square will consist of the
long side of one rectangle and the short side of another. To form a
middle rectangle of side 6 we must subtract the side of one rectangle from
the side of another, the possibilities being 104=6 and 82=6. 93=6 is
not possible since these two dimensions must form a side of the square.
In the 104 case we would have the 10+2 side of the square opposite of the
8+4 side. The other two opposite sides being 9+3 and 7+5, all odd numbers. Note that it
is impossible to subtract one odd number from another and obtain 1. Thus the
104 case is impossible. Under the 82 case we still have the 10+2 and the 8+4 on
opposite sides of the square. Again it will be impossible to form a 1 remainder
with odd numbered sides only the other way. Thus the 12 by 12 square is impossible
to tile.
The 13 by 13 square will have sides of lengths 10+3, 9+4, 8+5, and 7+6, with a 2 by 1
rectangle in the middle. Through a process of elimination as in the 12 by 12 case
we can't help but stumple upon the solutions.
The 11 by 11 square can have sides of lengths 10+1, 9+2, 8+3, 7+4, and 6+5. Four of
these will go along the sides and the fifth shall form the middle rectangle. Again
it is a process of trial and error to find the solutions.
This is a picture of just one solution:
. 












. 












. 












. 












. 












. 












. 












. 












. 












. 












. 












. 












. 












Go back to Problem 98
Go back to Shack's Math Problems
Michael Shackleford, A.S.A
  112.  Five socks problem
   In a drawer are two red socks and three blue socks. A sock is drawn at random from the drawer, with replacement, one million times. What is the range, with the expected outcome as the midpoint of the range, such that the probability is 95% that the number of red socks drawn falls within this range?
Answer
Problem 112 Answer
Problem 112 Answer
The range is 399,040 to 400,960
Michael Shackleford, A.S.A.
Solution
Problem 112 Answer
Problem 112 Solution
The mean number of reds is 1,000,000 * 0.4 = 400,000.
The standard deviation of the number of reds is sqr(1,000,000 * 0.4 * (1.4)) =~ 489.9
Let l denote the lower bound of the range, let u denote the upper bound, and let x denote the
number of reds drawn:
Pr (l <= x <= u) = .95
Pr (l400,000 <= x400,000 <= u400,000) = .95
Pr ((l.5400,000)/489.9 <= Z <= (u+.5400,000)/489.9) = .95 where Z denotes a random variable distributed
according to the standard normal distribution.
Next we want the probability that the number of reds will fall on either side of this range to
be .025 per side.
Pr( Z <= (u+.5400,000)/489.9) = .975
(u+.5400,000)/489.9 = 1.96
u399,999.5 = 960.2
u =~ 400,960
Pr( Z <= (l400,000.5)/489.9) = .025
(l400,000.5)/489.9 = 1.96
l = 399,040
So the range is 399,040 to 400,960
Michael Shackleford, A.S.A.
  229.  Flight around the world problem
   An island has a fleet of identical airplanes. Each plane can hold enough fuel to make it half way around the earth. The planes can refuel each other midflight. There is an unlimited supply of fuel on the island. Assuming no time or fuel lost in refueling or making a uturn, what is the fewest number of airplanes needed to fly around the earth and how can it be done such that all planes must return safely to the island?
Answer Problem 229 Answer
The around the world flight can be accomplished with as few as three planes.
Michael Shackleford, Dec. 3, 2017
Solution Problem 229 Solution
The around the world flight can be accomplished with only three planes.
The following table shows how it can be done. I'm sure there are variations of this that will also work. The time column is relative to the time required to fly around the earth nonstop. Fuel remaining is relative to a full tank. So a value of 1 would be full and 0 would be empty. The location is relative to a 360degree circle, starting at the 0 point.
Time  Fuel  Location 
Comment 
A  B  C  A  B  C 
0.0000  1.0000  1.0000  1.0000  0.0  0.0  0.0  A, B, and C depart clockwise 
0.0625  0.8750  0.8750  0.8750  22.5  22.5  22.5  C transfers 1/8 tank to A and B 
0.0625  1.0000  1.0000  0.6250  22.5  22.5  22.5  C turns back 
0.1250  0.8750  0.8750  0.5000  45.0  45.0  0.0  C refuels 
0.1250  0.8750  0.8750  1.0000  45.0  45.0  0.0  C leaves 
0.2500  0.6250  0.6250  0.7500  90.0  90.0  45.0  B refuels A 
0.2500  1.0000  0.2500  0.7500  90.0  90.0  45.0  B turns back 
0.3125  0.8750  0.1250  0.6250  112.5  67.5  67.5  C refuels B with a quarter tank 
0.3125  0.8750  0.3750  0.3750  112.5  67.5  67.5  C turns back 
0.5000  0.5000  0.0000  0.0000  180.0  0.0  0.0  A and B reach island on fumes, refuel 
0.5000  0.5000  1.0000  1.0000  180.0  0.0  0.0  A and B depart counterclockwise 
0.6875  0.1250  0.6250  0.6250  247.5  292.5  292.5  C transfers 1/4 tank to B 
0.6875  0.1250  0.8750  0.3750  247.5  292.5  292.5  C turns back 
0.7500  0.0000  0.7500  0.2500  270.0  270.0  315.0  B refuels A with 3/8 tank 
0.7500  0.3750  0.3750  0.2500  270.0  270.0  315.0  B turns back 
0.8750  0.1250  0.1250  0.0000  315.0  315.0  360.0  C refuels 
0.8750  0.1250  0.1250  1.0000  315.0  315.0  360.0  C turns back 
0.9375  0.0000  0.0000  0.8750  337.5  337.5  337.5  C refuels A and B with 7/24 tank each 
0.9375  0.2917  0.2917  0.2917  337.5  337.5  337.5  C turns back 
1.0000  0.1667  0.1667  0.1667  360.0  360.0  360.0  All arrive with 1/6 of a tank 
Notes:
* It was not required to transfer this much fuel. As long as all planes had at least 1/8 of a tank to get back all would have enough to make it back to the island. I chose 7/24 so that each plane would have the same amount of emergency fuel, in the interests of safety.
I found this problem in the book My Best Mathematical and Logic Puzzles by Martin Gardner, which I highly recommend. It is problem 19.
Michael Shackleford, Dec. 3, 2017
  55.  Flying bird in the wind problem
   A bird is flying towards its nest which is 100 feet away due east. Without considering the wind the bird can fly at a speed of 5 feet per second. There is a north blowing wind, however, at a speed of 2 feet per second. At all times the bird is flying in a direction such that he is directly facing the nest. The bird is not tacking into the wind as a boat would do but is continuously being blown off course and trying to set a new course. The graph above shows the path of the bird. How long does it take for the bird to reach the nest?
Answer
Problem 55 Answer
Problem 55 Answer
The answer is 500/21 seconds =~ 23.81 .
Michael Shackleford, A.S.A.
Solution
Problem 55 solution
Problem 55 Solution
I would like to thank Dr. ChuanMing Zou for his solution to this problem. The following
was taken from a fax he sent me.
First reword the problem such that the coordinates of the bird's initial position are (100,0)
and the nest is at (0,0). In polar coordinates the inital position of the bird
is (100,0^{o}). Let p stand for the distance to the origin in polar coordinates,
and let a stand for the angle.
Considering that the bird always flied towards the nest (p direction) and the wind blows
north (y direction) the following graph shows the change in the variables with respect to
a change of one infinitesimal unit of time in units of time. Note that as an angle
approaches zero its arc approaches a straight line.
(1) dp/dt = 5 + 2sin(a).
(2) da/dt = 2cos(a)/p.
(3) dx/dt = 5cos(a).
(4) dy/dt = 2  5sin(a).
Divide eq. (3) by eq. (2):
(5) dx/da = (dx/dt)/(da/dt) = 5pcos(a)/2cos(a) = 5p/2 = (5/2)*(x/cos(a)). (see graph above)
Rearrange:
(6) (2/x)dx = (5/cos(a))da.
Integrate both sides:
(7) ln(x^{2}) = ln(sec(a)+tan(a))^{5} + C you might need an
integral table for the integral of 1/cos(a).
Exp() both sides:
(8) x^{2}*(sec(a) + tan(a))^{5} = C.
Put in the intial condition of x=100, a=0 to solve for the constant:
(9) x^{2}*(sec(a) + tan(a)) = 10,000.
(sec(a) + tan(a))^{5} = (100/x)^{2}.
(sec(a) + tan(a)) = (100/x)^{2/5}.
Let b=(100/x)^{2/5}, then:
(10) sec(a) + tan(a) = b.
(11) (1+sin(a))/cos(a) = b (sec(a)=1/cos(a), tan(a)=sin(a)/cos(a)).
(12) (1+(1cos^{2}(a))^{1/2}) / cos(a) = b.
1+(1cos^{2}(a))^{1/2} = b*cos(a).
(1cos^{2}(a))^{1/2} = b*cos(a)  1.
1cos^{2}(a) = b^{2}*cos^{2}(a)  2*b*cos(a) + 1.
2*b*cos(a) = (1+b^{2})*cos^{2}(a).
2*b = (1+b^{2})*cos(a).
(13) cos(a) = 2*b / (1+b^{2}).
Substitute equation (13) into equation (3):
(14) dx/dt = 5*cos(a) = 10*b / (1+b^{2}).
Rearranging:
dt = (1+b^{2})/(10*b) dx.
10 dt = b+(1/b) dx
(15) 10 dt = (100/x)^{2/5} dx + (x/100)^{2/5} dx, remember b=(100/x)^{2/5}
Integrating:
(16) Constant 10t = (5/3)*100^{2/5}*x^{3/5} + (5/7)*(1/100)^{2/5}*x^{7/5}.
Put in the initial constants t=0 and x=100:
Const = (5/3)*100^{2/5}*100^{3/5} + (5/7)*(1/100)^{2/5}*100^{7/5}.
Const = (5/3)*100 + (5/7)*100 = 5000/21.
So the function of x and t is:
(17) 5000/21  10t = (5/3)*100^{2/5}*x^{3/5} + (5/7)*(1/100)^{2/5}*x^{7/5}.
The original question was how long does it take the bird to reach the nest, in other words what is
t when x=0. By substituting x=0 in the above equation t = Const/10 = 500/21.
Tristan Simbulan sent in the following solution which you may find better:
The initial position of the bird is ( 100, 0 ) and the nest is ( 0, 0 ) I am using the
velocity vector component parallel to y and x.
Let a be the angle between x axis and
bird's line of sight towards ( 0, 0 ).
(1)The wind velocity minus y component of the
bird's velocity = dy/dt = 2  5 sin a = 2  5 * y/sqrt(x^2 + y^2)
(2) Velocity x
component = dx/dt =  5 cos a =  5 * x / sqrt( x^2 + y^2 )
(3) (dy/dt)/(dx/dt) =
dy/dx = [ 5y 2 * sqrt(x^2 + y^2) ] / 5x 5xdy  5ydx + 2 * sqrt(x^2 + y^2) dx =
0 Let y = vx, dy = vdx + x dv , the result by substitution is dv/sqrt(1 + v^2) +
2/5 * dx/x = 0.
ln[ v + sqrt(1 + v^2) ] + 2/5 ln x = C since v = y/x, by substitution
ln [ y/x + sqrt(1 + y^2/x^2) ] + 2/5*ln x = C At y = 0 and x = 100 , then C =
ln 100^(2/5).
The equation of the curve is sqrt(x^2 + y^2) = x^(3/5) *100^(2/5) 
y.
Square both sides then y = [ 100^(4/5) * x^(3/5)  x^(7/5) ] / [ 2 * 100^(2/5)].
Solve (2) Velocity x component = dx/dt =  5 cos a =  5 * x / sqrt( x^2 + y^2 )
in terms of y.
By algebraic manipulation using fractional exponent and factoring
the result is 100^(4/5) * x^(2/5) dx + x^2/5 dx =  10^(9/5) dt 5/3 *100^(4/5) *
x^(3/5) + 5/7 * x^(7/5) =  10^(9/5) * t + C At x =100 and t = 0 , C = 100^(7/5) *
50/21.
At x = 0 the result is 0 =  10^(9/5) * t + C , substitute C = 100^(7/5) *
50/21 Solve: 10^(9/5)*t = 100^(7/5) * 50/21.
Therefore t = [100^(7/5)] / [10^(9/5)]
* 50/21
t = [ 10^(14/5) ] / 10^(9/5) * 50/21 = 10 *50/21 = 500/21.
Go back to Problem 55
Go back to Shack's Math Problems
Michael Shackleford, A.S.A.
  23.  Fork in the road problem
   You are traveling down a path and come to a fork in the road. A sign lays fallen at the path indicating that one path leads to a village where everyone tells the truth and the other to a village where everyone tells lies. The sign has been knocked down so you do not know which path leads to which village. Then someone from one of the villages (you don't know which one) comes down the path from which you came. You may ask him one question to determine which path goes to which village. What question do you ask?
Answer
Problem 23 Answer
Problem 23 Answer
Answer 1: Say, "Please point to the village
you are from." Regardless of where they are from they will point
to the truthful village.
Answer 2: You point to either path and say,
"Are you from this village?"
If the person answers 'yes' then you are pointing to the
truthful village, if they answer 'no' then you are pointing
to the lying village.
Answer 3: Ask "Which path would a person
from the other village say leads to the truthful (or untruthful) village?"
Regardless of whom you speak to the answer will filter through one lie, so if
you ask for the truthful village the person will point to the lying village,
and vise versa.
The Wizard of Odds, A.S.A.
Solution
Problem 23 Solution
Problem 23 Solution
You point to either path and say, "Are you from this village?"
If the person answers 'yes' then you are pointing to the
truthful village, if they answer 'no' then you are pointing
to the lying village.
This is not the only answer that works but is the simplist
I have heard so far.
I continue to receive a lot of email on this problem from
people who don't understand or disagree with my solution.
Before you question me on this answer these four questions:
 If you point to the truthful village what will a truthful person
say in answer to your question?
 If you point to the truthful village what will a untruthful person
say in answer to your question?
 If you point to the untruthful village what will a truthful person
say in answer to your question?
 If you point to the untruthful village what will a untruthful person
say in answer to your question?
The answers to these questions are: 1. "yes", 2. "yes", 3. "no", 4. "no".
Note that in the two cases where the answer is "yes" you are pointing to the
truthful village and in the two cases where the answer is "no" you are pointing
to the untruthful village. Thus you can tell from the reply to which village
you are pointing. You do not have to know to whom you are speaking.
Another possible solution which I have heard is to ask "Which path would a person
from the other village say leads to the truthful (or untruthful) village?"
Regardless of whom you speak to the answer will filter through one lie, so the
person will point to the lying village if you ask for the truthful village, and
vise versa.
Michael Shackleford, A.S.A.
  10.  Four businesses on a circle problem
   A town consists of only one street in the form of a circle. The town authorities give out four licenses for a particular kind of business. The inhabitants of the town live in equal density along the circle and will always go to the closest business for what they need. Business A gets to choose a location first, then business B, then C, and finally D. Each business desires to carve out as much business for themselves as possible but each knows the others all have the same motive. Assume that if a business is indifferent between locating in two different sections of the circle it will choose a section at random. Also assume that the business that goes last will choose a location in the middle of the largest (or one of the largest) sections. Where should business B choose relative to the location of A?
Answer
Problem 10 Answer
The answer is just barely less then 1/3 of the circumference of the circle
away from A.
Michael Shackleford, A.S.A., 10/20/1998
Solution
Problem 10 Solution
This is only a quick overview of the solution, the details are left up to you. Let the
circumference of the circle be 1 and that A chooses a location at point 0.
Business D will choose a location in the middle of
the largest section.
Business C will also choose the midpoint of the larger of the two gaps between the
first two businesses.
If business B chooses a point before 1/3 then C will choose a point halfway between
B and 1. Business D will choose randomly between the halfway point between A and C
or B and C. If x is the location of business B then the area which B will carve
out of the circle will be either (1+3x)/8 if D goes between B and C or (1+x)/4 if
D goes between A and C. The average of these is (3+5x)/16. The same logic
applies if B chooses a point after 2/3.
If business B chooses a point after 1/3 (but before 1/2) then C will choose a point halfway between
A and B going the long way and D will choose the halfway point between A and B the short way. This
will leave B exactly a 1/4 share of the business. The same logic applies if A chooses a point
between 1/2 and just before 2/3.
If B should choose a location at exactly 1/3 then C would choose at 2/3 and D would be
indifferent between 1/6, 1/2, and 5/6. B would have a 2/3 chance of having 1/4 of the
business share and 1/3 chance of having 1/3, the average being 5/18 =~ 0.27778 .
Thus B should try to maximize (3+5x)/16 without choosing x equal or greater to 1/3. The
optimal choice of location would be just a hair short of 1/3 (or just a hair after 2/3).
At this point B will have a 50/50 chance at having either 1/4 or 1/3 of the market share
for an average of 7/24=~ 0.29166667 of the market share.
Michael Shackleford, A.S.A., 10/20/1998
  111.  Four circles problem
   The three colored circles in the diagram above have radii of 1, 2, and 3, and each are tangent to the other two. A fourth interior circle is tangent to all three colored circles. What is the radius of the interior circle? For extra credit what is the radius of the exterior circle (not pictured) that is tangent to the three colored circles?
Answer
Problem 111 Answer
Problem 111 Answer
The radius of the interior solution is 6/23.
The radius of the exterior solution is 6.
Michael Shackleford, A.S.A.
Solution
Problem 111 Solution
Problem 111 Solution
The following solution and diagrams are courtesy of James Sutton.
The triangle marked by the centers of the three outer circles conveniently forms a right triangle with sides 3, 4, and 5 (3^{2} + 4^{2} = 5^{2}) as seen by the diagram below.
Next draw segments from the center of the interior circle and perpendicular to the outer triangle sides of length 3 and 4.
There are four right triangles in this picture, two of which are the same. Using the pythagorean formula we know:
(1): x^{2} + y^{2} = (1+r)^{2}
(2): x^{2} + (3y)^{2} = (r+2)^{2}
(3): y^{2} + (4x)^{2} = (r+3)^{2}
Combining equations (1) and (2) we get y=(3r)/3
Combining equations (1) and (3) we get x=(2r)/2
Substituting these values for x and y back into equation 1 we get:
((2r)/2)^{2} + ((3r)/3)^{2} = r^{2} + 2r + 1
This leads to...
23*r^{2} + 132r  36 = 0
Putting this through the quadratic formula we get r=6/23, 6
So the interior circle has radius 6/23 and the outer circle has radius 6.
Here is my old solution, as well as some comments from another reader.
The solution to this problem is rather hard to explain so I'm only going to go
over it briefly.
First arrange two of the circles so that their centers lay on the y axis. Then use
your trigonomtry to determine the coordinates of the center of the third circle (2.4,.2)
if the large circle is on top, the middle circle on bottom, and the third circle is to the
right. See problem 92 for help in how to find such coordinates.
Second, invert all three circles such that the third circle is mapped onto itself. Note
that the top and bottom circles are mapped to a horizontal lines that are tanjent to the
third circle.
Note: To invert any figure take all points (t,r) and map them onto the point
(t,c/r) where t is the angle in polor coordinates and r is the distance from the
center. In this case choose the constant c carefully such that the third circle is
mapped onto itself (okay, c=4.8 in this case).
The image of the center circle will be tanjent to both horizontal lines and the
third circle. Going to the right of the third circle will lead us to the interior
circle.
Consider the line that goes through (0,0) and the center of the circle in the above step. The
distance from the center of the points where this line intersect the circle are sqr(19.4)1 and
sqr(19.4)+1. Invert these two points and their distance from the center are
4.8/(sqr(19.4)1) and 4.8/(sqr(19.4)+1). Both these points lay on a diameter of the interior
circle so take their difference and divide by 2 to get the radius. The answer simplifies
to 6/23.
To find the exterior solution shift the third circle to the left rather than the right. The
radius of the exterior solution is exactly 6!
Note: The general case if the third circle has coordinates (x,y) and radius r
then the radii of the solution circles are (x^{2} + y^{2}  r^{2}) /
(4xr +/ (x^{2}+y^{2}+3*r^{2})). The + will determine the interior
solution, the  the exterior solution.
Here is another answer from Giulio Zambon
(Giulio.Zambon@aar.alcatelalsthom.fr) suggesting another
solution and some justified complaints on my own:
I would like to propose a solution, also using some trigonometry. First of
all, some definitions:
The centres of the four circles:
'A' for radius 3.
'B' for radius 2
'C' for radius 1
'Z' for the middle circle.
Two angles:
'a' between CA and CZ
'b' between CB and CZ
Our unknown:
'x' is the radius of the middle circle
Then, we make some obvious statements:
CA = 1 + 3 = 4
CB = 1 + 2 = 3
AB = 2 + 3 = 5
AZ = 3 + x
BZ = 2 + x
CZ = 1 + x
Followed by a trivial consequence:
(1) a + b = PI/2 (because CA^2 + CB^2 = AB^2)
Now we establish two equations using trygonometry:
(2) BZ^2 = CZ^2 + CB^2  2*CZ*CB*cos(b)
(3) AZ^2 = CZ^2 + CA^2  2*CZ*CA*cos(a)
By replacing the values we know and introducing 'x', (2) and (3) become:
(2) (2 + x)^2 = (1 + x)^2 + 9  2*3*(1 + x)*cos(b)
(3) (3 + x)^2 = (1 + x)^2 + 16  2*4*(1 + x)*cos(a)
By expanding the squares and simplifying (I also omit some '*' to make the
formulae more readable) we obtain:
(2) 6 + 2x + 6(1 + x)cos(b) = 0
(3) 8 + 4x + 8(1 + x)cos(a) = 0
By adding '+2 2' to the left side of (2) and '+4 4' to the left side of
(3), and then grouping '(1 + x)' terms, we obtain:
(2) (1 + x)(1 + 3cos(b)) = 4
(3) (1 + x)(1 + 2cos(a)) = 3
>From (1), we can express cos(a) in terms of cos(b):
(1) cos(a) = cos(PI/2  b) = sin(b) = sqrt(1  (cos(b)^2))
By substituting cos(a) in (3):
(3) (1 + x)(1 + 2sqrt(1  (cos(b)^2))) = 3
Now we can calculate cos(b) from (2) and substitute it into (3):
(2) cos(b) = (4/(1 + x)  1)/3 ==>
(3) (1 + x)(1 + 2sqrt(1  ((4/(1 + x)  1)/3)^2)) = 3
For better reading, we can define y = 1/(1 + x), with which (3) becomes (we
can multiply both sides by (1 + x) because x > 0; therefore, (1 + x) > 0 too):
1 + (2/3)sqrt(9  (4y  1)^2) = 3y
A bit of simple algebra brings us to:
4sqrt(2 + y  2y^2) = 9y  3
We will verify later that '(2 + y  2y^2) >= 0'. For the time being, we
square both sides and we get (after some simplification):
145y^2  86y 23 = 0
This resolves into: (43 + 72)/145. By taking the positive solution, we
obtain:
y = 23/29 ==> x = 29/23  1 = 6/23.
To check that '2 + y  2y^2 >= 0', we calculate:
2 + y  2y^2 = 2 + 23/29  2(23/29)^2 = 1.5+ > 0
I haven't worked on the "external" circle (yet), and my solution relies on
the fact that the centres of the original circles form a right triangle.
You do show a more general solution, but I confess (please, do not take
offense) that I am a bit skeptical that such a simple equation would apply.
In fact, I did not manage to understand it. Perhaps (time permitting), you
could add an explanatory figure.
You say that the two centres A and B are along the Yaxis. What you don't
say is that the Xaxis is the tangent between the two bigger circles. This
implies the two following equations (C must be below the Xaxis, because
the biggest circle is above it. Therefore, we can write the formulae in
such a way that y calculates to a negative number):
(1) BC^2  (2 + y)^2 = x^2
(2) AC^2  (3  y)^2 = x^2
This system provides x = 2.4 and y = 0.2, which is what you say in your
solution.
So far so good, but then I really get lost...
Ciao, Giulio
Michael Shackleford, A.S.A.
  159.  Four couples on a deserted island problem
   Four men and four women are shipwrecked on a deserted island. Eventually each person falls in love with one person and is loved by one person. You are given the following information: Who loves who?
Answer
Problem 159 Answer
Problem 159 Answer
Ellen loves Chad who loves Mary who loves David who loves Gloria who loves Arthur who loves Helen who loves Bruce who loves Ellen.
Here is an email I received from Dave Stigant before adding the last two clues.
I think there's another solution with 2 cycles instead of 1:
Bruce loves Mary loves Bruce (By my reading this fulfills the requirement that Bruce loves the woman (Mary) who loves the man (Bruce) who loves Mary)
Arthur loves Hazel loves Chad loves Ellen loves David loves Gloria
Incidently, assuming everybody is heterosexual there are no other 2 cycle solutions and no 3 cycle solutions. Arthur cannot love Ellen: either Bruce loves Mary (in which case you must have Hazel loves x loves Gloria) or not (in which case either H loves x loves G or G loves B). Chad and David must be in the same cycle, so the only way to have 3 cycles is A loves E and B loves M and the only way to have 2 cycles is if either A loves E or B loves M.
If you allow for homosexuality, we know that Chad, Arthur and Bruce must be straight. We know that Hazel is straight. We know that at least 2 women are straight (A loves girl, Gx, who loves the man who loves E... and B loves the girl, Gy, who loves the man who loves M. Gx and Gy must be straight but not the same since each is loved by a different man) Also, the number of gay men must be the same as the number of gay women (the number of straight men + the number of gay women = the number of women = number of gay men + the number of straight women = the number of men = 4). So if anybody is gay it must be David and one woman other than Hazel.
This is not an exhaustive list, but here are a few possible solutions:

Chad loves Mary loves David loves Arthur loves Hazel loves Bruce loves Ellen loves Gloria loves Chad
 Chad loves Mary loves David loves Arthur loves Hazel loves Bruce loves Ellen loves Chad and then Gloria loves herself (narcissism is not explicitly ruled out either)
 Chad loves Ellen loves David loves Arthur loves Hazel loves Chad and Bruce loves Gloria loves Mary loves Bruce.
Cheers,
Dave Stigant
This problem appears in Games for the Superintelligent by James F. Fixx (page 54).
Michael Shackleford, ASA  November 29, 2000
  29.  Four dogs and a square problem
   Four dogs occupy the four corners of a square with side of length a. At the same time each dog starts walking at the same speed directly toward the dog on his left. Eventually all four dogs will converge at the center of the square. What path does each dog follow and what is the distance each dog walks until he reaches the center?
Answer
Problem 29 Answer
Problem 29 Answer
Each dog travels in an eliptical spiral of the form r=(a/2 ^{1/2})*e ^{x},
where x is the angle relative to the starting angle.
The total distance covered is a.
Michael Shackleford, A.S.A.
Solution
Problem 29 Solution
Problem 29 Solution
Hard Solution
It is intuitive that at all times the dogs will form the corners of a square, ever decreasing in size and rotating about the center.
Let C be the center of the square at point (0,0). Let the dogs be at initial points (.5,.5), (.5,.5), (.5,.5), and (.5,.5). Let dog 1 bet at (.5,.5) who is chasing dog 2 at (.5,.5). Consider the line from C to dog 1. Next consider the line from dog 1 to dog 2. The angle between these lines is 135 degrees.
Theorem: Let C be a fixed point, D a point on a curve, r the distance from C to D, x the polar angle formed by r, and y the angle formed by the angle between CD and the tangent line to the curve at point D. Then tan(y)=r/(dr/dx).
So the line of sight from the center to the initial point of dog 1 would be be at a 270 degree angle. The line of sight from dog 1 to dog 2 would be a vertical line of 90 degrees. The angle between these lines is 270 degrees. The tangent of 270 is 1. So we have r/(dr/dx) = 1. Thus r(x)=c*e^{x}.
At the moment the dogs start walking x equals 0. At this moment the distance from the center to any of the dogs is 2^{1/2}. So r(x)=2^{1/2}*e^{x}.
The arc length is given by the formula:
Integral from 0 to infinity of ((f(x))^{2}+(f^{'}(x))^{2})^{1/2} =
Integral from 0 to infinity of ((2^{1/2}*e^{x})^{2} + (2^{1/2}*e^{x})^{2})^{1/2} =
Integral from 0 to infinity of (2^{1}*e^{2x} + 2^{1}*e^{2x})^{1/2} =
Integral from 0 to infinity of e^{x} =
e^{x} from 0 to infinity =
=0  (e^{0}) =
0 (1) = 1
The path the dogs take is called a logarithmic spiral. It is an interesting paradox that the dogs will make an infinite number of circles, yet the total distance is constant. Every revolution the size of the square will to e^{2*pi}/2^{1/2} = 0.03055663 times it's size before the revolution.
Easy Solution
David Wilson suggested the more simple solution which follows...
Suppose dog A is pursuing dog B who is pursuing dog C. During the entire pursuit,
the dogs remain at the corners of a square, and angle ABC is a constant 90 degrees.
That is, B's path towards C is always perpendicular to A's path towards B, and B
has zero velocity in the direction along A's path. Since B neither approaches nor
recedes from A during the walk, A simply covers the intial separation of 1.
Antonio Molins went further to state that as dog A is heading direcctly towards dog B, dog B is moving towards from dog A at rate equal to cosine of any of the interior angles of the shape made by the dogs. He also provided the following graphics.
For example consider a pentagon. Remember that for a ngon of n sides each angle will be 180*(1(2/n)). So a pentagon will have angles of 108°. Let's assume each dog walks at a rate of 1 unit per minute, where the initial shape has sides of 1 unit. As pointed out by David Wilson we can forget about the spiral paths but instead think of the dogs walking in straight lines. In the case of the penagon for every one unit dog A draw towards dog B's initial position, dog B will draw cos(180°)=0.309 units towards dog A. The question is, how long will it take for them to meet? Let the answer to that question be t. In t seconds dog A will cover t units, and dog B will cover .309*t units. The total distance traveled by A equals the distance traveled by B plus 1 (the initial separation). So solving for t: t=0.309+1 > t=1/(1.309) > t=1.4472.
Thus the general formula is 1/(1+cos(t)), where t is an interior angle of the shape.
Number of sides = 3
Lengh of initial side = 1
Each angle = 180*(1(2/3)) = 60°
Length of path = 1/(1+cos(60)) = 1/(1+0.5) = 2/3
Number of sides = 4
Lengh of initial side = 1
Each angle = 180*(1(2/4)) = 90°
Length of path = 1/(1+cos(90)) = 1/(1+0) = 1
Number of sides = 5
Lengh of initial side = 1
Each angle = 180*(1(2/5)) = 108°
Length of path = 1/(1+cos(108)) = 1/(1.309) = 1.4472
Number of sides = 6
Lengh of initial side = 1
Each angle = 180*(1(2/6)) = 120°
Length of path = 1/(1+cos(120)) = 1/(1.5) = 2
Number of sides = 10
Lengh of initial side = 1
Each angle = 180*(1(2/10)) = 144°
Length of path = 1/(1+cos(144)) = 1/(1.809) = 5.2361
Number of sides = 15
Lengh of initial side = 1
Each angle = 180*(1(2/15)) = 156°
Length of path = 1/(1+cos(156)) = 1/(1.9135) = 11.5668
Michael Shackleford, A.S.A.
  151.  Four mathematicians problem
   Four mathematicians have the following conversation: Alice: I am insane. Bob: I am pure. Charlie: I am applied. Dorothy: I am sane. Alice: Charlie is pure. Bob: Dorothy is insane. Charlie: Bob is applied. Dorothy: Charlie is sane. You are also given that:  Pure mathematicians tell the truth about their beliefs.
 Applied mathematicians lie about their beliefs.
 Sane mathematicians beliefs are correct.
 Insane mathematicians beliefs are incorrect.
Describe the four mathematicians.
Answer
Problem 151 Answer
Problem 151 Answer
 Alice: applied and insane.
 Bob: pure and sane.
 Charlie: pure and insane.
 Dorothy: pure and insane.
I'd like to thank Mathematics and Informatics Quartly
9:4 Dec 1999, page 162 for this problem.
Michael Shackleford, ASA, Januray 25, 2000
Solution
Problem 151 Solution
Problem 151 Solution
There are four possible kinds of mathematicians:
 Pure and sane
 Pure and insane
 Applied and sane
 Applied and insane
The first step is to determine what each kind of mathematician would say about
themselves:
 Pure and sane: "I am pure and sane."
 Pure and insane: "I am applied and sane."
 Applied and sane: "I am pure and insane."
 Applied and insane: "I am applied and insane."
Note that the applied and insane mathematician is lying about a lie so is
actually speaking the truth.
Next consider the first statement by Alice, "I am insane." Based on the four
groups there are two that will say they are insane:
 The applied and sane.
 The applied and insane.
What these two groups have in common is that both are applied, thus we can determine
that Alice is applied.
By the same logic we can determine that Bob is sane, Charlie is insane, and Dorothy
is pure.
Now we know one of the two characteristics of each mathematician. Next jump to the
last statement by Dorothy, "Charlie is sane." We already know that Dorothy is pure
so she actually believes that Charlie is sane. However we also know that Charlie is
infact insane so Dorothy's belief must be incorrect. Since she is not deliberately telling a
lie her belief is incorrect, making her insane. So Dorothy is pure and insane.
Let review what we know thus far:
 Alice: applied.
 Bob: sane.
 Charlie: insane.
 Dorothy: pure and insane.
Next look at what Bob says about Dorothy, "Dorothy is insane." This is a true statement and since
Bob is already sane we can conclude that he is also pure. Now we have:
 Alice: applied.
 Bob: sane and pure.
 Charlie: insane.
 Dorothy: pure and insane.
Next consider what Charlie says about Bob, "Bob is applied." This is not true since Bob is pure.
We already know Charlie is insane thus he must also be pure. If he were applied he would be lying about
a lie, thus telling the truth. Now we have:
 Alice: applied.
 Bob: sane and pure.
 Charlie: pure and insane.
 Dorothy: pure and insane.
Finally consider Alice's second statement, "Charlie is pure." Charlie is pure thus Alice is making a
true statement. However Alice is applied, thus she thinks she is telling a lie. So Alice is lying about
an incorrect belief, making a true statement. Since Alice's belief is incorrect she must be insane. Thus:
 Alice: applied and insane.
 Bob: pure and sane.
 Charlie: pure and insane.
 Dorothy: pure and insane.
I'd like to thank Mathematics and Informatics Quartly
9:4 Dec 1999, page 162 for this problem.
Michael Shackleford, ASA, Januray 25, 2000
  95.  Four people, a bridge, and a flashlight problem
   Four people need to cross a bridge at night. The bridge is only strong enough to hold at most two people at once. Because it is night a flashlight must be used for all crossings. It takes person A ten minutes to cross, person B five minutes, person C two minutes, and person D one minute. If two cross at the same time they must walk at the slower man's pace. How can you get everyone across in 17 minutes?
Answer
Problem 95 Answer
Problem 95 Answer
First, C and D cross together and C returns (4 minutes consumed).
Second, A and B cross together and D returns (11 minutes consumed).
Third, C and D cross together (2 minutes consumed).
Thanks to Antonio Cordoba for this problem. He says that Microsoft
applicants are tested with this one.
Michael Shackleford, A.S.A.
Solution
Problem 95 Solution
Problem 95 Solution
Michael Shackleford, A.S.A.
  148.  Four people, bridge, and flashlight problem
   It takes Jimmy 1 minute to cross a bridge. It takes Ronald 2 minutes. It takes Bill 5 minutes. It takes George 10 minutes. At most two may cross at a time. However, it is night and each group must take a flashlight. The group has one flashlight only. What is the least amount of time for all to cross?
Answer
Problem 148 Answer
Problem 148 Answer
Answer 1:
 Ronald and Jimmy cross in 2 minutes, total time 2 minutes.
 Ronald crosses back in 2 minutes, total time 4 minutes.
 Bill and George cross in 10 minutes, total time 14 minutes.
 Jimmy cross back in 1 minute, total time 15 minutes.
 Ronald and Jimmy cross in 2 minutes, total time 17 minutes.
Answer 2:
 Ronald and Jimmy cross in 2 minutes, total time 2 minutes.
 Jimmy crosses back in 1 minutes, total time 3 minutes.
 Bill and George cross in 10 minutes, total time 13 minutes.
 Ronald cross back in 2 minutes, total time 15 minutes.
 Ronald and Jimmy cross in 2 minutes, total time 17 minutes.
Thanks to Glenn Yoshimura for this problem. He says it has been asked
to prospective employees at Microsoft, with five minutes given to
solve it.
Michael Shackleford, ASA, August 20 1999
  14.  Four pints of beer problem
   A bartender has a three pint glass and a five pint glass. A customer walks in and orders four pints of beer. Without a measuring cup but with an unlimited supply of beer how does he get four pints in either glass?
Answer
Problem 14 Answer
Problem 14 Answer
Follow the following steps:
3 5
pint pint
glass glass
 
0 5
3 2
0 2
2 0
2 5
3 4
Or, here is another way that takes
more steps but wastes less beer:
3 5
pint pint
glass glass
 
3 0
0 3
3 3
1 5
1 0
0 1
3 1
0 4
In other words, fill the 3 pint glass and pour it into the 5 pint glass. Then
fill it again and from it top off the 5 pint glass, leaving 1 pint in the
3 pint glass. Then empty the 5 pint glass and move the 1 pint to the 5 pint
glass. Then fill the 3 pint glass and pour that into the 5 pint glass.
A problem like this appeared in the movie Die Hard III.
Michael Shackleford, A.S.A.
Solution
Problem 14 Solution
Problem 14 Solution
Follow the following steps:
3 5
pint pint
glass glass
 
0 5
3 2
0 2
2 0
2 5
3 4
At this point there are four pints in the five pint glass.
A problem like this appeared in the movie Die Hard III.
Michael Shackleford, A.S.A.
  68.  Four weights and a scale problem
   Using a balance scale and four weights you must be able to balance any integer load from 1 to 40. How much should each of the four weights weigh?
Answer
Problem 68 Answer
The answer is 1, 3, 9, and 27.
Thanks to Nick Hobson for this one.
Michael Shackleford, A.S.A.
Solution
Problem 68 Solution
Here is how to balance any amount between 1 and 40 using weights
of 1, 3, 9, and 27, letting W stand for the weight:
Weight
1 W=1
2 W+1=3
3 W=3
4 W=1+3
5 9=1+3+W
6 9=3+W
7 9+1=3+W
8 9=1+W
9 9=W
10 9+1=W
11 9+3=W+1
12 9+3=W
13 9+3+1=W
14 27=W+9+3+1
15 27=W+9+3
16 27+1=W+9+3
17 27=W+9+1
18 27=W+9
19 27+1=W+9
20 27+3=W+9+1
21 27+3=W+9
22 27+3+1=W+9
23 27=W+3+1
24 27=W+3
25 27+1=W+3
26 27=W+1
27 W=27
28 W=1+27
29 W+1=3+27
30 W=3+27
31 W=1+3+27
32 9+27=1+3+W
33 9+27=3+W
34 27+9+1=3+W
35 27+9=1+W
36 27+9=W
37 27+9+1=W
38 27+9+3=W+1
39 27+9+3=W
40 27+9+3+1=W
Nick Hobson sent in the following proof that 1, 3, 9, 27 is a unique
solution:
Hi Mike,
Here is my proof of the uniqueness of the solution to the weights puzzle. The
answer "drops out" of the proof, though I must admit I first guessed the answer,
and then came up with this proof!
First of all, think of each of the counterweights {a,b,c,d} as having a
coefficient which can be one of {1,0,+1}. Three coefficients varying over four
weights give 3 to the power 4 = 81 possible combinations.
When the coefficients are all 0, the result is clearly 0. Otherwise, if a given
set of coefficients gives +n, you can flip all the signs to get n. Therefore,
if there is a solution to the problem, all 81 integer weights from 40 to +40
must be generated.
We now prove the weights must sum to 40 and must all be integers. Clearly, the
sum must be >= 40, or 40 itself could not be generated. If the sum was > 40 you
could generate a result outside the solution set, and hence must miss a number
within the set. Similarly for a noninteger weight.
Now, add one to each of the coefficients, giving {0,1,2}. The equivalent
problem is now: what weights can generate all integers from 0 thru 80?
Thinking in base 3, the powers of 3 {1,3,9,27} stand out as one obvious
solution.
Finally, we can prove uniqueness. Note that we now have no negative
coefficients. Therefore, to generate 1, we need a weight of 1. Using this we
can generate 2. But for 3, we need a weight of 3. Now we can generate up to 8
(22 in base 3); so next we need a weight of 9. Then 27. Hence the solution is
unique. Clearly this can be generalized to n weights.
Regards,
Nick
Michael Shackleford, A.S.A.
  73.  Free gift in the cereal box problem #1
   There is a free gift in my breakfast cereal. The manufacturers say that the gift comes in four different colors, and encourage one to collect all four (thus eating lots of cereal). Assuming there is an equal chance of getting any one of the colors, what is the expected number of boxes I must consume to get all four?
Answer
Problem 73 Answer
Problem 73 Answer
The answer is 25/3 =~ 8.333 .
Michael Shackleford, A.S.A.
Solution
Problem 73 Solution
Problem 73 Solution
First I'll show that if the probability of an event happening is p then
the mean number of trials to obtain a sucess is 1/p.
Number Probability
of of
Trials Success
 
1 p
2 pq
3 pq^{2}
4 pq^{3}
. .
. .
. .
Since there must eventually be a sucess the sum of probabilities is:
p + pq + pq ^{2} + pq ^{3} + ... = 1.
The mean number of trials (m) is:
m = p + 2pq + 3pq^{2} + 4pq^{3} + ...
mq = pq + 2pq^{2} + 3pq^{3} + ...
mqm= p + pq + pq^{2} + pq^{3} + ...
m(1q) = 1.
m = 1/(1q) = 1/p.
Second, the answer to the problem can be express as the sum of the following:
 Number of trials to get a first toy
 Number of trials to get a second toy once you have one toy
 Number of trials to get a third toy once you have two toys
 Number of trials to get the final toy once you have three toys
The number of trials to get one toy is obviously one.
Once you have one the probability of getting a different toy in the next box is 3/4,
thus the expected number of trials is 1/(3/4) = 4/3 to get the second toy.
By the same logic the number of trials to get the third is 1/(1/2)=2 and 1/(1/4)=4
for the final toy.
Summing these yields 1 + 4/3 + 2 + 4 = 25/3.
I'd like to thank Michael Brasher for providing this solution. Below is my old
solution which is much less elegant, I recommend you ignore it.
The probability of it taking i boxes to get all four prizes is
the probability that the first (i1) will contain three different
kinds of the toy, times (1/4).
The probability that in j boxes there will be three or less kinds of
toys is 4*(3/4)^{j} where j>3.
The probability that in j boxes there will be exactly three kinds of
toys is 4*(3/4)^{j}  12*(1/2)^{j} + 12*(1/4)^{j}.
This is rather hard to explain why but basically there are 4 different sets of
3 toys, 6 different sets of 2 toys, and 4 different sets of 1 toy, use this
information to not double count combinations of 2 prizes or 1 prize.
Thus the answer is 1/4*sum for j=3 to infinity of
j*(4*(3/4)^{j}  12*(1/2)^{j} + 12*(1/4)^{j}).
Theorem: The sum for j=1 to infinity of i*(1/n)^{i} = n/(n1)^{2}.
From there is is just simple math to find that the answer is 25/3.
Michael Shackleford, A.S.A.
  74.  Free gift in the cereal box problem #2
   What is the answer for problem 73, only instead of 4 different prizes there are n different prizes?
Answer
Problem 74 Answer
Problem 74 Answer
The answer is n*(1 + 1/2 + 1/3 + ... + 1/n).
I got this answer from a Usenet discussion on probability, I can't
prove it myself at this time.
Michael Shackleford, A.S.A.
Solution
Problem 73 Solution
First I'll show that if the probability of an event happening is p then
the mean number of trials to obtain a sucess is 1/p.
Number Probability
of of
Trials Success
 
1 p
2 pq
3 pq^{2}
4 pq^{3}
. .
. .
. .
Since there must eventually be a sucess the sum of probabilities is:
p + pq + pq^{2} + pq^{3} + ... = 1.
The mean number of trials (m) is:
m = p + 2pq + 3pq^{2} + 4pq^{3} + ...
mq = pq + 2pq^{2} + 3pq^{3} + ...
mqm= p + pq + pq^{2} + pq^{3} + ...
m(1q) = 1.
m = 1/(1q) = 1/p.
Second, the answer to the problem can be express as the sum of the following:
 Number of trials to get a first toy
 Number of trials to get a second toy once you have one toy
 Number of trials to get a third toy once you have two toys
 .
 .
 .
 Number of trials to get the final toy once you have all but one toys
The number of trials to get one toy is obviously one.
Once you have one the probability of getting a different toy in the next box is (n1)/n,
thus the expected number of trials is 1/((n1)/n) = n/(n1) to get the second toy.
By the same logic the number of trials to get the third is n/(n2), and n/(n3) to
get the fourth, ... and n to get the final toy.
Summing these yields n * (1/n + 1/(n1) + 1/(n2) + ... + 1 ).
I'd like to thank Michael Brasher for providing this solution.
Nick Hobson sent in the following method to approximate the solution:
you can use Euler's approximation to get a good idea of how the
expected number grows with increasing n.
1 + (1/2) + ... + (1/n) ~= log n + (1/2n) + Y, where Y ~= 0.57721, and is known
as Euler's constant.
So, E(n) ~= n * (log n + Y) + (1/2).
Even for n = 4 this approximation is within 3% of the true value. For larger n
the percentage error is much smaller.
Here's elegant way of proving that if the probability of an event happening is p
then the mean number of trials to obtain a success is 1/p.
Consider the situation after the first trial.
Either the event has happened: contribution 1 * p, or it has not: contribution
(1 + m) * q.
Therefore, m = p + (1 + m)*(1  p), from which m = 1/p.
Michael Shackleford, A.S.A.
  134.  Free trade problem
   Assume two countries, call them the United States and Mexico, both have dual product economies. The table below shows how many days it takes for one person in each country to produce a unit of food or a unit of clothing. Country  Food  Clothing  United States  1  2  Mexico  3  4  Although the United States is more efficient at producing both items is it possible to have mutually beneficial free trade between the two countries? Assume no tariffs or transport costs.
Answer
Problem 134 Answer
Problem 134 Answer
The answer is yes.
Michael Shackleford, A.S.A., March 9, 1999
Solution
Problem 134 Solution
Problem 134 Solution
"The benefit of international trade  a more efficient employment of the
productive forces of the world."  J.S. Mill
Although the United States can make either product more efficiently it has a
comparative advantage in food, and Mexico has a comparative advantage in clothing.
Without free trade 2 units of food will be worth 1 unit of clothing in the United States.
In Mexico 4 units of food will be worth 3 units of clothing. So the exchange rate of
food to clothing in the U.S. will be 2:1, or 6:3, in the US and 4:3 in Mexico.
Next lets assume the traders set a compromise exchange rate of 5:3, or 5 units of
food for 3 units of clothing. Before free trade
if someone in the U.S. had 60 units of food they could exchange that for 30 units of
clothing. With free trade they could exchange the 60 units of food for 36 units of
clothing. Before free trade if someone in Mexico had 60 units of clothing they could exchange
that for 80 units of food. With free trade they could exchange the 60 units of clothing
for 100 units of clothing. The trader will be taking in food from the U.S. and clothing
from Mexico, allowing for supply and demand for both products and no shortages or surpluses
of either.
The net effect will be that the price of clothing will drop in the U.S. from 2 days labor to
5/3 of a day, and the price of food will drop in Mexico from 3 days labor to 2.4 days.
Yes, there will be a temporary loss of clothing jobs in the U.S. and farming jobs in
Mexico but with a movement in the labor force into more farming in the U.S. and more
clothing in Mexico everybody will be able to enjoy a higher standard of living.
For more information please read the chapter 'International Trade and the
Theory of Comparative Advantage' in Economics by Paul J. Samuelson which is
seems to invariably be the textbook of choice for freshmen in economics.
Michael Shackleford, A.S.A., March 9, 1999
  28.  Furniture factory problem
   A factory that produces tables and chairs is equipped with 10 saws, 6 lathes, and 18 sanding machines. It takes a chair 10 minutes on a saw, 5 minutes on a lathe, and 5 minutes of sanding to be completed. It takes a table 5 minutes on a saw, 5 minutes on a lathe, and 20 minutes of sanding to be completed. A chair sells for $10 and a table sells for $20. How many tables and chairs should the factory produce per hour to yield the highest revenue, and what is that revenue?
Answer
Problem 28 Answer
Problem 28 Answer
The maximum profit per hour, of $1200, comes when 24 chairs and
48 tables are produced per hour.
Michael Shackleford, A.S.A.
Solution
Problem 28 Solution
Problem 28 Solution
The 10 saws can produce 600 minutes of work per hour (10 saws * 60 minutes).
The 6 lathes can produce 360 minutes of work per hour (6 lathes * 60 minutes).
The 18 sanding machines can produce 1080 minutes of work per hour (18 sanding machines * 60 minutes).
Let c be the number of chairs produced per hour and t the number of tables produced per hour.
The number of saws limit the combination of chairs and tables to 600=10c+5t.
The number of lathes limit the combination of chairs and tables to 360=5c+5t.
The number of sanding machines limit the combination of chairs and tables to 1080=5c+20t.
Next graph these three lines. It should be expected that the answer will lie on the intersection
of two of these lines or to make all chairs or all tables. The intersection of the saw and sanding
machine line occurs outside of how many chairs the lathe can make so this combination is not
a viable answer. The saw and lathe lines cross at 48 chairs and 24 tables. The lathe and sanding
machine lines cross at 24 chairs and 48 tables.
Next determine the revenue at all points of intersection.
Chairs 
Tables 
Revenue 
60 
0 
600 
48 
24 
960 
24 
48 
1200 
0 
54 
1080 
So the optimal answer is to make 24 chairs and 48 tables for revenue of $1200 per hour.
To check it will take 24*10 + 48*5 = 480 minutes of saw time. There are 600 minutes available so
the saws will be idle 20% of the time.
It will take 24*5 + 48*5 = 360 minutes of lathe time which is exactly what we have.
It will take 24*5 + 48*20 = 1080 minutes of sanding machine time which is exactly what we have.
I would like to thank Brain Storm for this problem.
Michael Shackleford, A.S.A.
  75.  Gift exchange problem
   Suppose there are n people in an office. At Christmas they have a random gift exchange in which every name is writen on scraps of paper, mixed around in a hat, and then everyone draws a name at random to determine who they are to get a gift for. What is the probability nobody draws their own name?
Answer
Problem 75 Answer
Problem 75 Answer
The answer is 1/2!  1/3! + 1/4!  1/5! ... +/ 1/n!.
Here are some approxmiate probabilities for number of people n:
n Prob(no matches)
 
2 50.000000%
3 33.333333%
4 37.500000%
5 36.666667%
6 36.805556%
7 36.785714%
8 36.788194%
9 36.787919%
10 36.787946%
11 36.787944%
12 36.787944%
13 36.787944%
14 36.787944%
15 36.787944%
Note that the probabilities quickly converge to 1/e.
Michael Shackleford, A.S.A.
Solution
Problem 75 Solution
Problem 75 Solution
Lets take the case of 12 people for example.
There are 12! ways of drawing names.
Next determine the number of these 12! ways that have no matches.
Subtract from 12! the number of ways that one person matches, which would be
12*11! (12 people times 11! ways to arrange the other 11 names).
However this would be double counting those combinations in which two people
match, which is (12:2)*10! = 12!/2!. Note that (x:y) means x choose y or x!/(y!*(xy)!).
So add this group back in.
Yet we must take out from those we put in those combinations in which three people match,
which is (12:3)*9! = 12!/3!.
So keep repeating this process and the numerator or number of combinations in which
there are no matches is 12!  12! + 12!/2!  12!/3! + 12!/4!  12/5! ... + 12!/12!.
Divide this by the total number of combinatios, 12!:
1  1 + 1/2!  1/3! + 1/4!  1/5! ... +1/12!.
This same method works for any n.
Note that it approaches 1/e.
Michael Shackleford, ASA
  114.  Gold mining problem
   A gold mining company is testing locations for its next mine. From location A eight samples were taken of units of gold per ton of ore. The results were 1.23, 1.42, 1.41, 1.62, 1.55, 1.51, 1.60, and 1.76 . From location B six samples were taken with the following results 1.76, 1.41, 1.87, 1.49, 1.67, and 1.81 . It is assumed that the amount of gold in a sample in either location have a standard normal distribution with a fixed, yet unknown, mean and variance, and that the variance in both locations is equal. Test the hypothesis that the mean gold content of both locations is equal. Use a 10% level of significance, in other words assume that if the two means were the same the test would pass 90% of the time.
Answer
Problem 114 Answer
Problem 114 Answer
The samples pass the test of equal means.
Michael Shackleford, A.S.A.
Solution
Problem 114 Solution
Problem 114 Solution
The mean of the sample from location A is 1.5125 and the mean from location B is 1.6683 .
Let S_{A}^{2} = The sum for i=1 to 8 of (s_{i}1.5125)^{2}
= .10875, where S_{i} is the ith sample from location A.
Let S_{B}^{2} = The sum for i=1 to 6 of (s_{i}1.5125)^{2}
= .167673, where S_{i} is the ith sample from location B.
Let U be the statistic for testing the the hypothesis that the mean of location A does
not equal the mean of location B.
U = (sqr(8+62)*(1.51251.6683)) / (sqr(1/8 + 1/6)+sqr(.10875+.167673)) = 1.693354 .
The t value, with 12 degrees of freedom, at the .95 level of significance is 1.782 .
Thus if 1.782 <= U <= 1.782 the test will pass, which it does.
Michael Shackleford, A.S.A.
  220.  Good Will Hunting problem
   Draw all homeomorphically irreducible trees of size n=10.
If this sounds familiar, it was the math problem posed in the movie Good Will Hunting, that Will solved. It was actually not so hard. Here is my attempt to put it in plain simple English:
Using only straight lines, draw all figures where the sum of intersections and dead ends equals 10. You may not have any closed loops. You may also not have two equivalent figures. Any intersection must have at least three paths leading from it. I'll give you a hint  there are ten figures. In the movie, Will got only eight of them.
Answer Problem 220 Answer
Here is the answer, including two patterns Will didn't get.
Michael Shackleford, April 5, 2017
Solution
Problem 220 Solution
This solution is nothing very scientific but my attempt at a common sense answer.
 First, there is the case of just one intersection:
 Second, let's consider all the cases of two intersection points.
Let's start with the case of two nodes on one side. Considering both intersection points, and the two nodes on one side, there are six nodes left for the other side:
 Next, move a node from the side with six to the side with two, for a 35 split:
 Next, move a node from the side with five to the side with three, for a 44 split:
 Third, let's consider all the cases of three intersection points.
Let's start with the case of two nodes off one side. In the middle we can have just one node, since it has two intersection lines braching off of it. Consider the three intersections and three nodes, there are four nodes left for the other end:
 Next, move a node from the side with four to the other end with only two:
 Next, move a node from either side to the middle:
 Next, move a node from the other side (that still has three nodes) to the middle:
 Fourth, let's consider all the cases of four intersection points in a row.
Considering the two ends must have at least two nodes and the middle two must have at least one each, and the four intersection points themselves, we've already spoken for all ten points:
 Fifth, let's consider all the cases of four intersection points where there is one in the middle and three other branching off.
We've already spoken for four intersection points and each must have at least two nodes branching off, so we've already spoken for all 4+6=10 nodes:
Here is the full answer:
Michael Shackleford, ASA — April 5, 2017
  137.  Grave digging problem
   A small town in Alaska is approaching winter. Because the soil will soon freeze they need to dig enough graves in the town cemetary now in anticipation of the number of deaths until the ground thaws in spring. The town's population is 1000 and it is assumed that each person has a 1% chance of dying during the winter. What is the least number of graves should they did so that the probability of having enough is at least 90%? What about 95% and 99%?
Answer
Problem 136 Answer
Problem 136 Answer
Posted June 24, 1999
At 90% the answer is 14, at 95% the number is 15, at 99% the number is 18.
Michael Shackleford, ASA
Solution
Problem 137 Solution
Problem 137 Solution
Posted June 24, 1999
Binomial Distribution Solution
This is a basic binomial distibution problem. Given n people and a
probability of death of p the probability of x deaths is
n!/(x!*(nx)!)*p ^{x}*(1p) ^{(nx)}. In this
particular case the formula is 1000!/(x!*(1000x)!)*.01 ^{x}*.99 ^{(1000x)}.
The table below shows the probability of specific numbers of death using this formula:
Problem 137 
Number of Deaths (x) 
Probability total=x 
Probability total<=x 
0  0.00004317  0.00004317 
1  0.00043607  0.00047924 
2  0.00220019  0.00267943 
3  0.00739322  0.01007265 
4  0.01861375  0.02868640 
5  0.03745311  0.06613951 
6  0.06273711  0.12887663 
7  0.08998657  0.21886319 
8  0.11282407  0.33168726 
9  0.12561333  0.45730059 
10  0.12574021  0.58304080 
11  0.11430928  0.69735009 
12  0.09516152  0.79251160 
13  0.07305328  0.86556489 
14  0.05202279  0.91758768 
15  0.03454173  0.95212941 
16  0.02147955  0.97360896 
17  0.01255845  0.98616742 
18  0.00692759  0.99309501 
19  0.00361663  0.99671164 
20  0.00179188  0.99850352 
21  0.00084466  0.99934817 
22  0.00037967  0.99972784 
23  0.00016307  0.99989092 
24  0.00006705  0.99995797 
25  0.00002644  0.99998441 
26  0.00001002  0.99999443 
27  0.00000365  0.99999808 
28  0.00000128  0.99999936 
29  0.00000043  0.99999979 
30  0.00000014  0.99999994 
31  0.00000004  0.99999998 
32  0.00000001  0.99999999 
33  0.00000000  1.00000000 
34  0.00000000  1.00000000 
The table shows that if 14 graves were dug then the probability of not running out
would be 0.91758768 (the smallest number greater or equal to 90%). So they must
dig 14 graves to have a 90% of not running out. To have a 95% chance they must dig
15 and to have a 99% chance they must dig 18.
Normal Distribution Solution
We can expect that the number of deaths to approximate a normal distribution curve.
The mean is 1000*.01=10. The standard deviation is (1000*(.01)*(.99)) ^{1/2} =~
3.1464 . Let g be the number of graves such that the probability of running out is
less than 90%. Let d be the number of deaths.
Pr(d <= g+.5) = 0.90 (The 0.5 is added because the number of deaths must be an integer)
Pr(d9.5 <= g10) = 0.90
Pr((d9.5)/3.1464 <= (g10)/3.1464) = 0.90
Pr(Z <= (g9.5)/3.1464) = 0.90
(g9.5)/3.1464 = 1.28
g =~ 13.527
The number of graves must be an integer, thus we round up to 14.
Michael Shackleford, ASA
  25.  Guess the age problem
   Person x and y have the following conversation:
x: I forgot how old your three kids are. y: The product of their ages is 36. x: I still don't know their ages. y: The sum of their ages is the same as your house number. x: I still don't know their ages. y: The oldest one has red hair. x: Now I know their ages!
How old are they?
Answer Problem 25 AnswerProblem 25 AnswerTheir ages are 2, 2, and 9. Michael Shackleford, A.S.A. Solution Problem 25 SolutionProblem 25 SolutionFrom the statement that the product of their ages is 36 the possibilities of the three individual ages are: 1,1,36 1,2,18 1,3,12 1,4,9 1,6,6 2,2,9 2,3,6 3,3,4 From the statement that the sum equals the house number it is possible to eliminate all but two possibilities. The sums of the rest are unique and would allow for an immediate answer. For example if the house number were 16 the ages must be 1, 3, and 12. The two remaining possibilities are 2, 2, and 9; or 1, 6, and 6. After the clue that the oldest has red hair you can eliminate 1, 6, and 6 because the oldest two have the same age thus there is no oldest son. The only remaining posibility is 2, 2, and 9. Michael Shackleford, A.S.A.   235.  Hanging rope problem
   A 100meter rope is suspended from the top of two 50meter poles. The lowest point of the rope is 10 meters from the ground. How far apart are the poles?
For some helpful equations, please see these hints.
Answer
Problem 235 Answer
The answer is 45×ln(3) = 49.4376 meters.
Michael Shackleford
Solution   211.  Hat and river problem
   Joe puts his canoe in the river and starts paddling upstream. After a mile his hat falls in the river. Ten minutes later he realizes his hat is missing and immediately paddles downstream to retrieve it. He catches up to it at the same place he launched his canoe in the first place. What is the speed of the river current?
Answer Problem 211 Answer
Question
Joe puts his canoe in the river and starts paddling upstream. After a mile his hat falls in the river. Ten minutes later he realizes his hat is missing and immediately paddles downstream to retrieve it. He catches up to it at the same place he launched his canoe in the first place.
What is the speed of the river current?
Answer
Three miles per hour.
Michael Shackleford, ASA — Feb. 7, 2012
Solution Problem 211 Solution
Question
Joe puts his canoe in the river and starts paddling upstream. After a mile his hat falls in the river. Ten minutes later he realizes his hat is missing and immediately paddles downstream to retrieve it. He catches up to it at the same place he launched his canoe in the first place.
What is the speed of the river current?
Easy Solution
If Joe paddles upstream and them downstream the same amount of time in both directions then his effort paddling in both directions an equal amount of time would cancel out. In other words, he would end up in the same place if he floated the whole time.
We know he paddled upstream for ten minutes. So, after paddling downstream for another ten minutes he would end up in the same spot as he would had he floated for 20 minutes. We also know from the problem that in this same 20 minutes the hat floated one mile. So if the hat travels a mile in 20 minutes then it would travel three miles in an hour, for a speed of 3 m.ph..
Hard Solution
Let:
p=paddling speed in still water.
c=current of the water.
d=distance Joe paddled upstream since his hat fell out of the canoe.
Recall that distance = rate * time. Consider the unknown distance Joe went upstream since his hat fell out:
(1) d=(pc)*(1/6)
This is because his net speed is the paddling speed less the current speed. We know he paddled without his hat upstream for 1/6 of an hour.
We're also given that it took the same time for the hat to float downriver a mile as it took for Joe to travel d miles upstream and 1+d miles downstream. Let's set up an equation, balancing for time.
(2) Time for hat to float one mile downstream = Time for Joe to travel d miles upstream + Time for Joe to travel 1+d times downstream.
Time for hat to float one mile downstream = one mile/rate of current = 1/c.
Time for Joe to travel d miles upstream = 1/6 (in hours), as provided in the question.
Time for Joe to travel 1+d times downstream = distance Joe traveled downstream / rate Joe traveled downstream = [1 + (1/6)*(pc) ]/ (p+c).
The distance is the one mile the hat floated in the river + the (1/6)*(pc) from equation (1). Joe’s speed going downstream is p+c, which is the sum of his paddling speed and current speed.
Now we’re ready to solve for equation (2)
1/c = (1/6) + [1 + (1/6)*(pc) ]/ (p+c)
1/c = (1/6) + (6+pc)/(6p+6c)
6/c = 1 + (6+pc)/(p+c) Multiplying both sides by 6.
6/c = (p+c)/(p+c) + (6+pc)/(p+c) Finding a common denominator
6/c = (6+2p)/(p+c)
6p + 6c = 6c + 2pc
6p = 2pc
3=c
So the current is 3 miles per hour.
Michael Shackleford, ASA — June 16, 2012
  156.  Heatseeking missle problem
   A plane flies at 10 km./minute in a straight direction and level altitude. When it is directly 5 km. overhead you shoot a heatseeking missile at it, straight up. The missile always points directly at the plane. The missile travels at a constant speed of 11 km./minute. How long does it take to hit the plane?
Answer
Problem 156 Answer
Problem 156 Answer
The answer is 55/21 minutes.
Michael Shackleford
Solution   123.  Height of a pentagon problem
   What is the area of an equalateral pentagon of side 1? Answer many not be expressed in trigonometric functions. Hint: solve problem problem 47 first.
Answer
Problem 123 Answer
Problem 123 Answer
The answer is sqr(25+10*sqr(5))/4 =~ 1.720477.
Michael Shackleford, A.S.A., November 23, 2000
Solution
Problem 123 Solution
Problem 123 Solution
Solution #1
Consider the diagram from problem 47.
The area of the entire pentagon is x * ( a + 1/2)  two isocoles triangles of base 1 with the other two sides of length a.
We established from problem 47 that the height of these isocoles triangles is x. So, each one has area 1/2 * 1 * x, so two of them would have area x.
Thus, the area of the penagon is x * ( a + 1/2)  x =
x * ( a  1/2) =
sqrt(5 + 2*sqrt(5))/2 * [(1 + sqrt(5))/2  1/2] =
sqrt(5 + 2*sqrt(5))/2 * sqrt(5)/2 =
sqrt(25 + 10*sqrt(5))/4 =~ 1.7205.
My thanks to Bob H. for this solution.
Solution #2
I'd like to thank James Sutton for suggestion the following solution and for the diagram. My old solution, which is not as elegant, follows his.
Let x be the height of the pentagon and r be the distance from the center to any corner. The height of the pentagon was solved in problem 47 as (2*5^{1/2}+5)^{1/2}/2. The area of the pentagon equals 10 times the area of the triangle shown in the following diagram.
We already know r so let's solve for x in terms of r using the pythagorean formula on the triangle in the diagram above.
(1/2)^{2} + (xr)^{2} = r^{2}
1/4 + x^{2} 2xr + r^{2} = r^{2}
1/4 + x^{2} 2xr = 0
r = 1/(8x) + x/2
So xr = x/2  1/(8x)
The area of each triangle is 1/2*base*height = x/8  1/(32x)
The area of the pentagon is 10 of these triangles = 10x/8  10/(32x) = 5x/4  5/(16x) = (4x^{2}  1)/(32x).
We know from problem 47 that x = (2*5^{1/2}+5)^{1/2}/2. Substituting this value for x in the above equation results in an area of (25+10*5^{1/2})^{1/2}/4 =~ 1.7205.
Solution #3
From the solution to problem 47 we know the height of this pentagon is sqr(5+2*sqr(5))/2 and distance from any two nonadjacent corners is (1+sqr(5))/2 (the golden mean).
The next step is hard to explain without a picture, however I will try. Consider three consecutive corners of the pentagon and label them a, b, and c. Next draw a line from a to c. Call d the point in the middle of segment ac. Now consider the triangle abd. This triangle has angles 36, 54, and 90. We also know the hypotenuse is 1 and longer base is half the distance between any two nonadjacent corners, or (1+sqr(5))/4. From this we can deduce that cos(36)=(1+sqr(5))/4.
Next extend a segment from all five corners of the pentagon and all five midpoints of each side. This will divide the pentagon into 10 equal triangles. Each of these triangles conveniently also has angles of 36, 54, and 90. Next lets define the following:
r=distance from any corner of pentagon to center of pentagon
y=distance from the midpoint of any side to the center of pentagon
r+y is the height of the pentagon, which from problem 47 is sqr(5+2*sqr(5))/2.
The ratio y/r = cos(36) = (1+sqr(5))/4.
Now we have two equations and two unknowns. At this point we grind out the algebra to find that:
r=2*sqr(5+2*sqr(5))/(5+sqr(5))=~0.850651
y=sqr(25+10*sqr(5))/10=~0.688191
The pentagon is composed of 10 right triangles, each of area 1/2*1/2*y. The area of the pentagon is thus (10/4)*y = sqr(25+10*sqr(5))/4 =~ 1.720477.
Thanks to Mark Hoyle for some help with this problem.
Michael Shackleford, A.S.A., October 9, 2001
  47.  Height of pentagon problem
   What is the height of a pentagon with side length of 1? Answer may not be expressed in trigonometric functions.
Answer
Problem 47 Answer
Problem 47 Answer
sqr(5+2*sqr(5))/2 =~ 1.538842
Michael Shackleford, ASA
Solution
Problem 47 Solution
Problem 47 Solution
I would like to thank Douglas Lerner for the following solution and
illustrations. My solution, which is much less elegant, appears afterward.
Let x be the height of the pentagon and a be the length of the
extensions of the sides, as shown in the following diagram:
Using the Pythagorean formula we know
x^{2} + (a+1/2)^{2} = (a+1)^{2}, or
(1) x^{2} = a + 3/4.
Now consider the added two chords in the next diagram:
The isosceles triangle formed by these two chords and one side of the pentagon is
congruent to the triangle formed by two extensions of length a and the same side
of the pentagon.
Thus the distance between two nonadjacent corners of the pentagon is also a. Consider
the chord going from either corner of the base to the top of the pentagon. We can use
the Pythagorean formula, with this chord as the hypotenuse to obtain:
x^{2} + (1/2)^{2} = a^{2}.
Substituting for x^{2} from above we can solve for a:
a + 3/4 = a^{2}  (1/2)^{2}, or
a^{2}  a  1 = 0, or
a=(1+sqr(5))/2.
Now that we know a we can solve for x:
x^{2} = ((1+sqr(5))/2)^{2}  1/4, or
x^{2} = (5+2*sqr(5))/4, or
x = sqr(5+2*sqr(5))/2 =~ 1.538842
Old Solution
Here is my old solution which used the following hints:
 sin(5x)=5*sin(x)20*sin^{3}(x)+16sin^{5}(x).
 tan(x)=sin(x)/(1sin^{2}(x))^{1/2}.
 The interior angles of an ngon are each 180*(n2)/n.
With some simple trig we can express the height as 1/2*[1/sin(36) + 1/tan(36)].
From the formula in the hint we can express tan(36) as sin(36)/sqr(1sin^{2}(36)).
After a few simple steps we can express the height as (1) 1/2*[(1+sqr(1sin^{2}(36))/2*sin(36)].
Next lets work on expressing sin(36).
From the formula in the hint we know sin(180) = 16*sin^{5}(36)  20*sin^{3}(36) + 5*sin(36)
= 16*sin^{4}(36)  20*sin^{2}(36) + 5
= 16x^{4}  20*x^{2} + 5 (where x=sin(36)).
= 16y^{2}  20*y + 5 (where y=x^{2}).
y = (5sqr(5))/8.
x = sqr((5sqr(5))/8) = sin(36).
Now put this expression for sin(36) into equation (1):
1/2 * [ (1+sqr(1(5sqr(5))/8)) / sqr((5sqr(5))/8) ] =
1/2 * [ (1+sqr(3+sqr(5))/8)) / sqr((5sqr(5))/8) ] =
1/2 * [ sqr(8/(5sqr(5))) + sqr((3+sqr(5))/(5sqr(5))) ] =
1/2 * [ (sqr(8) + sqr((3+sqr(5)))/(5sqr(5))) ] =~ 1.5389
Michael Shackleford, ASA
  78.  House painting problem
   If it takes Bill 20 hours to paint a house and it takes George 30 hours, how long will it take if they work together yet independently?
Answer
Problem 78 Answer
Problem 78 Answer
The answer is 12 hours.
Thanks to Marilyn Vos Savant for this one.
Michael Shackleford, A.S.A.
Solution
Problem 78 Solution
Problem 78 Solution
Bill can paint at a rate of 1/20th of a house per hour.
George can paint at a rate of 1/30th of a house per hour.
Combine their efforts and they can paint 1/20 + 1/30 = 1/12th of a house
in an hour.
So if their rate is 1/12th of a house per hour it will take 12 hours to
paint an entire house.
Thanks to Marilyn Vos Savant for this one.
Michael Shackleford, A.S.A.
  87.  How far can the plane fly problem
   An airplane weighs 10,000 pounds, has a 1000 gallon tank, burns fuel at a rate of 5 gallons per minute, and has a speed in miles per hour of 6,000,000 divided by the sum of weight of the plane and remaining fuel. Each gallon of fuel weighs 10 pounds. With a full tank how far can the plane fly?
Answer
Problem 87 Answer
Problem 87 Answer
The answer is 2000*ln(2) =~ 1386.3 miles .
Michael Shackleford, A.S.A.
Solution
Problem 87 Solution
Problem 87 Solution
The weight of the plane as a function of time, in pounds per hour of
flying time, is f(w)=20,0003,000*t
The plane will run out of fuel after 10/3 hours.
The distance the plane can cover is the integral from 0 to 10/3 of (6,000,000/(20,0003,000t)) dt.
= 6,000,000 * 1/3000 * ln(20,0003,000t) from 10/3 to 0.
= 2,000 * ln(1/2)
= 2000 * ln(2) =~ 1386.3 miles .
Michael Shackleford, A.S.A.
  171.  How many ways to get from A to C problem
   A, B, and C are three downs, each connected by a network of roads. There are 82 ways to get from A to B, including those routes that pass through C. There are 62 ways to get from B to C, including those routes that pass through A. The number of ways to get from A to C, including those ways passing through B, is less than 300. How many ways are there to get from A to C?
Answer
Problem 171 Answer
Problem 171 Answer
The answer is 46.
Michael Shackleford, A.S.A.
Solution
Problem 171 Solution
Problem 171 Solution
This solution was submitted by Shawn Stackhouse. My original solution follows his, however I feel Shawn's is better.
Let's define the number of direct routes from A to B as x.
Let's define the number of direct routes from B to C as y.
Let's define the number of direct routes from A to C as z.
From the information given we know:
x + yz = 82
y + xz = 62
Adding them we get:
x + xz + y + yz = 144
(x + y) * (z + 1) = 144
and subtract them to get:
x  xz  y + yz = 20
(y  x) * (z  1) = 20
The factors of 20 are {1, 2, 4, 5, 10, 20}. So (z  1) is one of {1, 2, 4, 5, 10, 20}.
then (z + 1) is one of {3, 4, 6, 7, 12, 22}.
The factors of 144 are: {1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 36, 48, 72, 144}. So (z + 1) must be one of these factors.
Common possible (z + 1) values are {1, 2, 3, 4, 6, 12}.
Given (x + y) * (z + 1) = 144, possible values of (x + y) are: {144, 64, 48, 36, 24, 12}.
Possible values of (z  1) are: {1, 0, 1, 2, 4, 10}.
Concurrently since (y  x) * (z  1) = 20, then possible values of (y  x) are: {20, 10, 5, 2, 20}.
This leads to five possible solutions. Note that x+y=64 is not listed because then z would be 1 and (yx)*(z1) could not equal 20 for any values of x and y.
 If x+y=144 then z=0, yx=20, y=62, x=82
 If x+y=48 then z=2, yx=20, y=34, x=14.
 If x+y=36 then z=3, yx=10, y=23, x=13.
 If x+y=24 then z=5, yx=5, y=14.5, x=9.5.
 if x+y=12 then z=11, yx=2, y=7, x=5
The fourth solutions is not possible because there can not be half a path. The problem limits us to z+xy<300, so that eliminates the first three. Thus the fifth solution is the correct one. If x=5, y=7, and z=11 then the final answer is z+xy=11+5*7=46.
Alternative Solution
Here is my original solution.
Let's define the distance from A to B as x.
Let's define the distance from B to C as y.
Let's define the distance from A to C as z.
The information given tells us that:
x+yz=82
y+xz=62
z+xy<300
Combining the first two equations and solving for z we get:
z=(82x)/y, and z=(62y)/x
Equating the two we get x/y = (62y)/(82x)
Unfortunately trial and error seems the only solution from here. The way I did it was to create a spreadsheet with values of x along one axis and values of y along the other. In the body of the spreadsheet I had the following formula (62y)/(82x)  (x/y). Then I simply looked for a zero.
If your ranges for x and y are small then the values x=5 and y=7 should pop out, which lead to the correct solution. However other values that work are (x,y)=(13,23), (14,28), (14,34), (12,42), (5,55).
The problem tell us that z+xy<300. Multiplying x and y easily eliminates all possible values except (5,7) and (13,23). If x=23 and y=23 then z is easily solved as 3. In this case the total paths from A to C would be 3+13*23=302, which is too many.
So x=5 and y=7. Simple algebra tells us that z=11. So 11+5*7 = 46.
Michael Shackleford, A.S.A.
  135.  How much is that puppy problem
   A man wishes to sell a puppy for $11. A customer wants to buy it but only has foreign currency. The exchange rate for the foreign currency is 11 round coins = $15, 11 square coins = $16, 11 triangular coins = $17. How many of each coinage should the customer pay?
Answer
Problem 135 Answer
Problem 135 Answer
Posted May 27, 1999
7 circular coins and 1 square coin.
I would like to thank Recreations in the Theory of Numbers by Albert H.
Beiler for this problem (#42 on page 299).
Michael Shackleford, ASA
Solution
Problem 135 Solution
Problem 135 Solution
Posted May 27, 1999
Let x be the number of circular coins, y be the number of square coins,
and z be the number of triangular coins. Thus:
x*(15/11) + y*(16/11) + z*(17/11) = 11. Multiplying by 11:
15x + 16y + 17z = 121.
From here we have to use trial and error or educated guessing. Personally I
divided 121 by 15 to get 8 plus a remainder of 1. It was then obvious that
x=7, y=1, and z=0 would solve the equation.
Michael Shackleford, ASA
  170.  How old am I problem
   I am as old as John will be when I am twice as old as John was when my age was half the sum of our present ages. John is as old as I was when John was half the age he will be 10 years from now. How old am I?
Answer
Problem 170 Answer
Problem 170 Answer
I am 40.
Michael Shackleford, A.S.A.
Solution
Problem 170 Solution
Problem 170 Solution
90% of this problem is working through the wording into the form of an equation. Let's let i be my age and j be the age of John. The first piece of information tells us...
i = 2*((i+j)/2 + (ji)) + (ji)
= 2*(3ji)/2 + (ji)
= 4j  2i
= (4/3)*j
The second piece of information tells us...
j = (j+10)/2 + (ij)
= i  j/2 + 5
= (2i+10)/3
Combining the two equations we get...
i = (4/3)*(2i+10)/3
= (4/9)*(2i+10)
= (8/9)*i + 40/9
= 40
So I am 40 (and John is 30).
Michael Shackleford, A.S.A.
  169.  Hungry spider problem
   A spider eats 3 flies a day. Until the spider fills his quota a fly has a 50% chance of survival if he attempts to pass the web. Assuming 5 flies have already made the attempt to pass, what is the probability that the 6th fly will survive the attempt?
Answer
Problem 169 Answer
Problem 169 Answer
The answer is 0.75.
Michael Shackleford, ASA  August 17, 2001
Solution
Problem 169 Solution
Problem 169 Solution
Using the binomial distribution, the probability that n flies out of 5 have been eaten is combin(5,n)*(1/2) ^{5}, where combin(5,2)=5!/(n!*(5n)!).
The probability that 0 flies were eaten is combin(5,0)*(1/2)^{5}=1/32.
The probability that 1 fly was eaten is combin(5,1)*(1/2)^{5}=5/32.
The probability that 2 flies were eaten is combin(5,2)*(1/2)^{5}=10/32.
So the probability that the spider is still hungry is 1/32 + 5/32 + 10/32 = 16/32 = 1/2. The probability the spider is full is 11/2=1/2. Thus the probability of a successful attempt to pass is (1/2)*1 + (1/2)*0.5 = 0.75 .
Michael Shackleford, ASA  August 10, 2001
  63.  Impossible to draw figure problem
   It is impossible to draw this figure without taking the pen off the paper, redrawing any lines, or other trickery. Explain why it is impossible.
Answer
Problem 63 Answer
Problem 63 Answer
Notice that the two points on the bottom are both
the junction of three different line segments. The
two points above these two are the junction of five
line segments. As you draw the path each point you
enter you must also exit, except the beginning and
ending points. Thus line segments are added to
any given point two at a time. So for any figure
you draw, without
lifting the pen from the paper, must have an even
number of line segments leaving each point, except
the beginning and ending point. Given the beginning
and ending point, you are allowed two grace points
that may have an odd number of segments. However the
figure in this problem has four points with an odd
number of segments, thus it is impossible.
Michael Shackleford, A.S.A.
  188.  Infinite fraction problem
   This problem has been removed.
Answer
Problem 188 Answer
Problem 188 Answer
(1+5 ^{1/2})/2 =~ 1.61803
Michael Shackleford, A.S.A.
Solution
Problem 188 Answer
Problem 188 Answer
This could be rephrased as x=1+(1/x).
x1 = 1/x
x^{2}x1 = 0
x=(1+5^{1/2})/2
Michael Shackleford, A.S.A.
  133.  Inheritance problem
   A man has $1,000,000 he wishes to divide up in his will. He wants to give each person named in his will an amount of money, in dollars, which is a power of 7 ($7^{0}=$1, $7^{1}=$7, $7^{2}=$49, $7^{3}=$343, ...). He does not want to give more than six people the same amount. How can he divide the money?
Answer
Problem 133 Answer  Math Problems.info
Problem 133 Answer
He should give:
 1 person $1
 1 person $7
 3 people $49
 3 people $343
 3 people $2401
 3 people $16807
 1 person $117649
 1 person $823543.
I would like to thank Recreations in the Theory of Numbers by Albert H.
Beiler for this problem (#52 on page 300).
Michael Shackleford, A.S.A., May 27, 1999
Solution
Problem 133 Answer
Problem 133 Solution
Lets call a the number of people he leaves $1 to, b the number
of people he leaves $7 to, c the number
of people he leaves $49 to, etc. The solution equation is thus:
1,000,000 = a + 7b + 49c + 343d + 2401e + 16807f + 117469g + 822283h.
Next divide 1,000,000 by 7. If 1,000,000 is divisible by 7 we could divide the
money without giving anyone only $1. However 1,000,000/7 = 142857 plus a remainder
of 1. By giving one person $1 (a=1) we now have:
999,999 = 7b + 49c + 343d + 2401e + 16807f + 117469g + 822283h.
Next divide by 7:
142,857 = b + 7c + 49d + 343e + 2401f + 16807g + 117469h.
Dividing 142,857 by 7 we get 20,408 plus a remainder of 1. So let b=1, divide by 7,
and we get:
20,408 = c + 7d + 49e + 343f + 2401g + 16807h .
Dividing 20,408 by 7 we get 2,915 plus a remainder of 3. So let c=3, divide by 7,
and we get:
2915 = d + 7e + 49f + 343g + 2401h.
Dividing 2915 by 7 we get 416 plus a remainder of 3. So let d=3, divide by 7,
and we get:
416 = e + 7f + 49g + 343h.
Dividing 416 by 7 we get 59 plus a remainder of 3. So let e=3, divide by 7,
and we get:
59 = f + 7g + 49h.
Dividing 59 by 7 we get 8 plus a remainder of 3. So let f=3, divide by 7,
and we get:
8 = g + 7h.
Obviously g=1 and h=1.
I would like to thank Recreations in the Theory of Numbers by Albert H.
Beiler for this problem (#52 on page 300).
Michael Shackleford, A.S.A., May 27, 1999
  153.  Insurance company problem
   An insurance company issues policies to two classes of people, as shown in the table below. Problem 153  Class  Probability of Death  Death Benefit  Number in Class  A  .01  200  500  B  .05  100  300 
The company charges each customer the product of expected claim amount and a constant k. How much should k be so that the probability of total claims exceeding total revenue is 5%?
Answer
Problem 153 Answer
Problem 153 Answer
The answer is 1.38
This is a Society of Actuaries sample problem
for course 100.
Michael Shackleford, ASA  March 17, 2000
Solution
Problem 153 Solution
Problem 153 Solution
Let a be an individual in class A and b be an individual in class B.
E(a) = 200*.01=2
E(a^{2}) = 200^{2}*.01=400
Var(a)=E(a^{2})  (E(a))^{2} = 400  4 = 396
E(b) = 100*.05=5
E(b^{2}) = 100^{2}*.05=500
Var(b)=E(b^{2})  (E(b))^{2} = 500  25 = 475
The variance of total claims is 500*396+300*475 = 340,500
The standard deviation of total claims is 340500^{1/2} =~ 583.52
Expected total claims is 500*2 + 300*5 = 2500
Pr(total claims <= total revenue) = .95
Pr(C<=R) = .95 (where C=total claims, R=total revenue)
Pr(C<=R) = .95
Pr(Cm<=Rm) = .95 (where m is the expected total claims)
Pr(C2500<=R2500) = .95
Pr((C2500)/s<=(R2500)/s) = .95 (where s is the standard deviation)
Pr((C2500)/583.52<=(R2500)/583.25) = .95
We can now use the central limit theorem because (C2500)/583.52 has a normal
distribution with mean of 0 and stanrdard deviation of 1.
(R2500)/583.25 = 1.645
R = 3459.89
So the insurance company needs 3459.89 in revenue. The expected costs of
claims is 2500. So k = 3459.89/2500 = 1.38
This is a Society of Actuaries sample problem
for course 100.
Michael Shackleford, ASA  March 17, 2000
  67.  Insurance in blackjack problem
   In blackjack if the dealer is showing an ace you make "take insurance" or bet that his/her other card is a ten. A winning bet pays 2:1 (win $2 for every $1 bet, keeping original bet). What is the house advantage in this bet? How many nontens would you have to remove from a single deck game to make it a fair bet? Assume you know nothing about the cards that have already been played (including the dealer's ace and your own).
Answer
Problem 67 Answer
The house advantage is aproximately 7.69%.
Removing 4 nontens from a single deck game would make it a fair bet.
Michael Shackleford, A.S.A.
Solution
Problem 67 Solution
The house advantage is what $1 minus what you would expect to have after
$1 bet. The probability of winning this bet is 16/52 (16 tens out of 52
cards) =~ 30.8%. If you win you triple your money so you $1 bet on
insurance on average will yield $3 * .308 = 92.3 cents. Thus the house
advantage is 1.00  .923 = 7.69%.
Let x be the number of nontens you would have to remove to make this a fair
bet. Since it pays 2:1 you should have a 1 in 3 chance of winning for it
to be fair.
1/3 = 16/(52x)
52x = 48
x=4.
Thus you would have to remove four nontens from a
single deck game to make this a fair bet.
Michael Shackleford, A.S.A.
  91.  Keno problem
   This problem has been removed.
Answer
Problem 91 Answer
Problem 91 Answer
The answer is (x:y)*(80x:20y)/(80:20), where (a:b)=a!/(b!*(ab)!), where a!=1*2*...*a.
Click here for specific answers given x and y.
Michael Shackleford, A.S.A.
  205.  Keno with Replacement
   Imagine a keno game where the numbers are drawn WITH replacement. In other words, 20 balls are drawn from an urn with balls numbered 1 to 80, one at time. As each ball is drawn the number is noted, and then the ball is returned to the urn. The player picks four different numbers from 1 to 80. For each number picked it is considered a match if the casino draws that number at least once. What is the probability for each possible outcome of 0 to 4 matches?
Answer Problem 205 AnswerQuestion Imagine a keno game where the numbers are drawn WITH replacement. In other words, 20 balls are drawn from an urn with balls numbered 1 to 80, one at time. As each ball is drawn the number is noted, and then the ball is returned to the urn. The player picks four different numbers from 1 to 80. For each number picked it is considered a match if the casino draws that number at least once. What is the probability for each possible outcome of 0 to 4 matches? Answer Problem 205 Answer  Matches  Probability  4  0.001906  3  0.031326  2  0.179819  1  0.428464  0  0.358486  Total  1  Acknowledgement: My thanks to Eliot J. for this help with this problem. Michael Shackleford, ASA — June 2, 2009 Solution Problem 205 SolutionQuestion Imagine a keno game where the numbers are drawn WITH replacement. In other words, 20 balls are drawn from an urn with balls numbered 1 to 80, one at time. As each ball is drawn the number is noted, and then the ball is returned to the urn. The player picks four different numbers from 1 to 80. For each number picked it is considered a match if the casino draws that number at least once. What is the probability for each possible outcome of 0 to 4 matches? Solution First determine the probability that the casino will drawn any of the four balls x times, including duplicates. The probability of m such draws is combin(80,m)×(4/80)^{m}×(76/80)^{20m}, where combin(80,x)=80!/(x!×(80x)!). The following table shows the probability of 0 to 20 total matches. Table 1 — Total Match Probabilities  Total Matches  Probability  20  9.53674E27  19  3.62396E24  18  6.54125E22  17  7.45703E20  16  6.02155E18  15  3.6611E16  14  1.73902E14  13  6.60829E13  12  2.04031E11  11  5.16878E10  10  1.08028E08  9  1.86593E07  8  2.65895E06  7  3.10893E05  6  0.000295348  5  0.002244646  4  0.013327586  3  0.059582148  2  0.188676801  1  0.377353603  0  0.358485922  Total  1  The hard part is determining the probability of d distinct matches given m matching draws. Let's let f(m,d)=probability of d distinct matches, given m total matches. It is obvious that f(0,0)=1 and f(1,1)=1. If d=2 then the probability the second matching ball drawn is the same as the first is 1/4. So f(2,1)=0.25, and f(2,2)=0.75. If d=3 then the probability that the second and third matching balls drawn are the same as the first is (1/4)×(1/4)=1/16. So f(3,1)=0.0625. The probability of three different numbers is (3/4)×(2/4)=6/16=3/8=0.375. The only other possibility is d=2, so f(3,2)=10.06250.375=0.5625. If d=4 the probability the probability of only one distinct number drawn is f(4,1)=(1/4)^{3}=1/64. The probability of four distinct numbers is (3/4)×(2/4)×(1/4)=6/64=3/32. Things get a little harder with three distinct mathces. There are 4!/(2!×1!×1!) = 12 ways you can pick one number for the one that is called twice, and two numbers called once. Once you do that there are the same 4!/(2!×1!×1!) = 12 ways you can choose the order of those numbers called. So out of 4^{4}=256 ways to pick 4 numbers out of 4, with replacement, 12×12=144 of them result in three distinct numbers. So f(4,3)=144/256=0.5625. By ommission, f(4,2)=1f(4,1)f(4,3)f(4,4)=1(1/64)(144/256)(3/32)=0.328125. The fun part is calculating the probability of 1 to 4 distinct matches given 5 to 20 total matches. It is fairly obvious that there are 4 (one for each pick) ways out of 4^{m} matches that all the number can be the same. So the f(m,1)=4/4^{m} = 1/4^{m1}. For two distinct matches there are combin(4,2)=6 ways to picks two numbers out of the four picks. It would be incorrect to say that f(m,2)=6×(2/4)^{m}, because it would include, multiple times, the probability of drawing only one number. There are four possible numbers that can be drawn only once, and each of them is triple counted. For example, of the six sets of two numbers: (1,2), (1,3), (1,4), (2,3), (2,4), and (3,4) any given single number appears three times. So, the probability of picking exactly two distinct numbers is 6×(2/4)^{m}  4×3×(1/4)^{m}. Let's look at the case of m=10, for example. f(10,2)=6×(2/4)^{10}  12×(1/4)^{10} = 0.005847931. Things get even messier with three distinct matches. The following equations show that the probability of exactly three matches is the probability of 13 matches, less the probability of exactly 2 matches, less the probability of exactly one match. So f(m,3)=4×(3/4)^{m}  12×(2/4)^{m} + 12×(1/4)^{m}. For example, for 10 matches, the probability of exactly 3 distinct numbers is 4×(3/4)^{10}  12×(2/4)^{10} + 12×(1/4)^{10} = 0.213546753. The probability of exactly four matches can be found by omission, subtracting the probability of 1 to 3 matches from 1. The following table shows the conditional probability of 0 to 4 matches for 0 to 20 total matches. Table 2 — Conditional Probability of 0 to 4 Distinct Matches by 0 to 20 Total Matches  Total Matches  Distinct Matches  4  3  2  1  0  20  0.987320874287  0.012673403675  0.000005722035  0.000000000004  0  19  0.983098313736  0.016890242201  0.000011444048  0.000000000015  0  18  0.977472047671  0.022505064262  0.000022888009  0.000000000058  0  17  0.969977988862  0.029976235237  0.000045775669  0.000000000233  0  16  0.960001168773  0.039907280356  0.00009154994  0.000000000931  0  15  0.946729257703  0.053087644279  0.000183094293  0.000000003725  0  14  0.929094403982  0.070539414883  0.000366166234  0.000000014901  0  13  0.905703306198  0.093564391136  0.000732243061  0.000000059605  0  12  0.874759197235  0.123776435852  0.001464128494  0.000000238419  0  11  0.833988189697  0.163084030151  0.002926826477  0.000000953674  0  10  0.780601501465  0.21354675293  0.005847930908  0.000003814697  0  9  0.711364746094  0.276947021484  0.011672973633  0.000015258789  0  8  0.622924804688  0.353759765625  0.023254394531  0.000061035156  0  7  0.5126953125  0.44091796875  0.046142578125  0.000244140625  0  6  0.380859375  0.52734375  0.0908203125  0.0009765625  0  5  0.234375  0.5859375  0.17578125  0.00390625  0  4  0.09375  0.5625  0.328125  0.015625  0  3  0  0.375  0.5625  0.0625  0  2  0  0  0.75  0.25  0  1  0  0  0  1  0  0  0  0  0  0  1  To get the total probability of 0 to 4 distinct matches, take the dot product of m total matches by the conditional probability of d distinct matches. The following table shows the results. Table 3 — Total Probability of 0 to 4 Distinct Matches by 0 to 20 Total Matches  Total Matches  Distinct Matches  4  3  2  1  0  20  0  0  0  0  0  19  0  0  0  0  0  18  0  0  0  0  0  17  0  0  0  0  0  16  0  0  0  0  0  15  0  0  0  0  0  14  0  0  0  0  0  13  0.000000000001  0  0  0  0  12  0.000000000018  0.000000000003  0  0  0  11  0.000000000431  0.000000000084  0.000000000002  0  0  10  0.000000008433  0.000000002307  0.000000000063  0  0  9  0.000000132736  0.000000051676  0.000000002178  0.000000000003  0  8  0.000001656327  0.00000094063  0.000000061832  0.000000000162  0  7  0.000015939328  0.000013707822  0.00000143454  0.00000000759  0  6  0.000112486115  0.000155750006  0.000026823612  0.000000288426  0  5  0.000526088909  0.001315222272  0.000394566681  0.000008768148  0  4  0.001249461158  0.007496766948  0.004373114053  0.000208243526  0  3  0  0.022343305413  0.03351495812  0.003723884236  0  2  0  0  0.141507600951  0.047169200317  0  1  0  0  0  0.377353602535  0  0  0  0  0  0  0.358485922409  Total  0.001905773455  0.031325747161  0.179818562032  0.428463994944  0.358485922409  The bottom row shows the probabilities for 0 to 4 distinct matches. Acknowledgement: My thanks to Eliot J. for this help with this problem. Michael Shackleford, ASA — July 3, 2009   40.  Land and sea race problem
   You are in a race in which the starting line is at a certain point on a straight beach. The finish line is in the water. One way to arrive at the finish line is to run 4 kilometers down the beach, make a 90 degree turn and swim 1 kilometer. However, you may cut into the water at any point. You speed on land is 6 k.p.h and you speed in water is 2 k.p.h.. At what point, measured from the starting line, should you cut into the water?
Answer
Problem 40 Answer
Problem 40 Answer
The answer is 4  8 ^{1/2} kilometers.
Michael Shackleford, A.S.A.
Solution
Problem 40 Solution
Problem 40 Solution
Let y be the solution. Let x=4y. The time to
reach the finish line is:
t=(4x)/6 + (1+x^{2})^{1/2}/2.
Set the derivative equal to 0:
dt/dx= 1/6 + 1/2 * 1/2 * 2x * (1+x^{2})^{1/2} = 0.
The solution is x=1/8^{1/2}.
y=4x.
Michael Shackleford, A.S.A.
  163.  Large tank with 10 outlets problem
   A large tank has a steadily flowing intake and 10 outlet valves, the latter being all of the same size. With 10 outlets open, it takes two and one half hours to empty the tank; with 6 outlets open it takes five and one half hours to empty the tank. After the tank is empty and with all 10 outlets closed, how long will it take to fill the tank?
Answer
Problem 163 Answer
Problem 163 Answer
The answer is 6.875 hours.
Thanks to the Baltimore Actuaries Club and Alan Goldberg for this problem.
Michael Shackleford, ASA  June 22, 2001
Solution
Problem 163 Solution
Problem 163 Solution
Let x be the input rate in gallons per hour into the tank.
Let y be the output rate in gallons per hour from the tank for each value.
Let w be the number of gallons in the tank.
We know that it takes 2.5 hours to empty the tank with 10 valves open. Over the 2.5 hours the sum of the water in the tank initially and the water going into the tank will equal the amount of water leaving the tank. Let's set that up as an equation.
w + 2.5x = 2.5*10y
We also know that it takes 5.5 hours to empty the tank with 6 valves open. Let's set that up as an equation.
w + 5.5x = 5.5*6y
I don't like decimals so let's multiply both equations by 2:
2w+5x=50y
2w+11x=66y
To solve the problem we need to know the relationship between w and x, so lets solve for y in the first equation and subsitute in the the second:
y=(2w+5x)/50.
Substituting this in the second equation...
2w+11x=66*(2w+5x)/50
100w+550x=132w+330x
220x=32w
w=220x/32
w=6.875x
So 6.875 times the input rate equals the capacity of the tank. Thus it would take 6.875 hours to fill the tank.
Andrew N wrote it to add his two cents on this problem, arguing in real life the rate at which the water leaves will depend on the water pressure in the tank. Here are his comments.
Hi Michael,
First off, I'd like to say that you've done a fine job with your math problems website. The problems are interesting, and the solutions are quite thorough and easy to understand for the most part. This is no easy feat. I also love your gambling website, the Wizard of Odds. You taught me how to play blackjack! Thanks for that.
Now, problem 163 about emptying the water tanks has a physical wrinkle which in missing from the problem statement: if water is constantly entering the tank, there is no way the tank can be emptied. This is because the flow rate through the outlets is related to the height of the water in the tank. I can prove this assertion as follows.
Setting up Bernoulli's equation for one outlet, where point 1 is the top of the tank and point 2 is on the free jet at the outlet, we have:
P1/gamma + v1^2/(2g) + h1 = P2/gamma + v2^2/(2g) + h2
where:
P1, P2 are pressures
v1, v2 are flow velocities
h1, h2 are heights
gamma is the specific gravity of water = density * gravity
g is the gravitational constant
We set h2 equal to 0 as a reference height. Now, assuming both water surfaces are exposed to the atmosphere, and the height of the tank is not several miles high so atmospheric pressure is constant, we can cancel P1/gamma and P2/gamma. Also, assuming that the water at the top of the tank is not moving quickly compared to the jet, we can assume v1 = 0. After simplifying, we are left with:
h1 = v2^2/(2g)
Solving for v2 we get:
v2 = (2g*h1)^0.5
By the continuity equation, the flow rate out from the outlet is equal to the flow velocity v2 times the outlet area A, so:
Qout = A*v2 = A*(2g*h1)^0.5
Assuming a cylindrical tank, the volume of water in the tank V is the height of the water h1 times the crosssectional area R, so:
Qout = A*[2g*(V/R)]^0.5 = k*(V^0.5) for simplicity's sake.
Now that we have an expression for the flow rate out of the outlet as a function of the volume in the tank, we can set up a differential equation to solve for the volume in the tank as a function of time.
change in volume = flow in  flow out
dV/dt = Qin  n*Qout (< n outlets, all assumed to flow at the same rate)
dV/dt = Qin  n*k*(V^0.5)
This is a seperable differential equation. Seperating both parts we get:
dV/(Qin  n*k*(V^0.5)) = dt
Integrating both sides, we get:
(1/nk)^2 * [2*(V0^0.5  V^0.5) + 2*Qin*LN[(Qin  nk*V0^0.5)/(Qin  nk*V^0.5)]] = t
The interesting thing here is that if Qin = nk*V^0.5, or V = (Qin/nk)^2, then the LN part of this equation will be undefined. Values of V less than (Qin/nk)^2 give negative values for t, which are clearly not possible. The physical explanation for this is that as the volume in the tank is reduced, there is not enough driving force (hydrostatic pressure in this case) to evacuate the fluid in the tank at a faster rate than it's coming in. In effect, it is not possible to empty the tank so long as water in flowing in. I used a cylindrical tank as the basis for this proof for simplicity, but this principle applies to any shaped tank.
Just my 2 cents.
Thanks,
Andrew N
Michael Shackleford, ASA  June 22, 2001
  172.  Late train problem
   Two trains, A and B, leave their respective starting points at the same time and travel in opposite directions. They travel at constant speeds, and pass at point M. One travels at twice the speed of the other. If one of the trains leaves five minutes late they pass at a point 2 miles from point M. What is the speed of the slow train, in miles per hour?
Answer
Problem 172 Answer
Problem 172 Answer
The answer is 36 miles per hour.
Michael Shackleford, A.S.A.
Solution
Problem 172 Solution
Problem 172 Solution
As long as both trains are moving they will always meet at a point 1/3 of the way between their two positions. Lets say the initial positions for the slow train is 0 and the fast train is 3x. Thus if they both left on time they would meet at point x. Let's assume that in 5 minutes the slow train can travel u units of distance and the fast train can cover 2u units.
Assume first the slow train leaves late. At the moment the slow train does leave the fast train will be at point 3x2u. The meeting point will be 1/3 of the way between 0 and 3x2u = (3x2u)/3 = x2u/3. We also know the distance from this point to x is 2 miles. In other words x(x2u/3)=2 ... 2u/3=2... u=3.
It is not necessary but we can also calculate u if the fast train leaves late. In this case when the fast train does leave from point 3x the slow train will already be at point u. The meeting point will be u+(3xu)/3 = x+2u/3. Again the distance between this point and x is 2 miles. In other words (x+2u/3)x=2 ... 2u/3=2 ... u=3.
Now we just plug what we know into the d=rt formula (distance=rate * time). The distance is one unit, which equals 3 miles. The time is 5 minutes, or 1/12 of an hour. So we have 3=r*(1/12) ... r=36 m.p.h..
Michael Shackleford, A.S.A.
  245.  Leaping frog
   There is a frog at the bottom of 20 steps. With each jump, the frog can jump up one or two steps. The frog eventually lands on exactly on the 20th step. What is the number of possible permutations of jumps, in order, the frog can take. For example, 111112222211111.
Answer
Problem 245 Answer
Problem 245 Answer
The answer is 10,946.
Michael Shackleford
Solution
Problem 245 Solution
Problem 245 Solution
Let a jump of one step be a "small jump" and a jump of two steps be a "big jump."
 There is only one way to get to the first step, with a single small jump.
 There are two ways to get to the second step, either two small jumps or one big jump.
 There are two ways to arrive at the third step, either with a big jump from step 1 or a small jump from step 2. There is one way to arrive at step 1 and two ways to arrive at step 2. Thus, the number of ways to get to step 3 is 1 + 2 = 3.
 There are two ways to arrive at the step 4, either with a big jump from step 2 or a small jump from step 3. There are 2 ways to arrive at step 2 and 3 ways to arrive at step 3. Thus, the number of ways to get to step 4 is 2 + 3 = 5.
 There are two ways to arrive at the step 5, either with a big jump from step 3 or a small jump from step 4. There are 3 ways to arrive at step 3 and 5 ways to arrive at step 4. Thus, the number of ways to get to step 4 is 3 + 5 = 8.
 Keep following this logic and you will find that there are 10,946 ways to arrive at the 20th step.
Here is the full Fibbonachi sequence, up to 20:
 1
 2
 3
 5
 8
 13
 21
 34
 55
 89
 144
 233
 377
 610
 987
 1597
 2584
 4181
 6765
 10946
Michael Shackleford
  21.  License plates and a scale problem
   You are the foreman in charge of license plate production. Your equipment can produce two different gauges of steel, one weighing about 90% of the other. Normally you use the heavier gauge but one week you leave that responsibility to someone else who sets the gauge randomly every day. After a week you realize that some days the license plates may have been made of the lighter gauge steel. Every days license plates are kept in separate piles. You may use a digital scale once which will give an exact weight of whatever you decide to weigh. Assume that the heavy gauge license plates weigh 1 unit each and the light ones 0.9 units each. You may use the scale only once. How many license plates from each pile should you put on the scale?
Answer
Problem 21 Answer
You should weigh 1 license plate from the first day, 2 from the second, 4
from the third, 8 from the fourth, 16 from the fifth, 32 from the sixth, and
64 for the seventh. In other words 2^{(x1)} for day x.
Michael Shackleford, A.S.A.
  35.  Light bulb problem #1
   You have ten light bulbs. Five have an average life of 100 hours, and the other five have a average life of 200 hours. These light bulbs have a memoryless property in that their current age (measured in how long they have already been on) has no bearing on their future life expectancy. Assuming they are all already on what is the expected number of hours before the first one burns out?
Hint: The density function for this kind of light bulb with average life of n hours is f(x)=1/n * e^{x/n}.
Answer
Problem 35 Answer
Problem 35 Answer
The answer is 40/3 hours.
Michael Shackleford, A.S.A.
Solution
Problem 35 Solution
Problem 35 Solution
Warning: This is just a quick explanation which assumes some knowledge of
the exponential distribution and integral calculus.
The density function of the shorter lived life bulbs is f(x)=1/100 * e^{x/100}.
The density function of the shorter lived life bulbs is f(x)=1/200 * e^{x/200}.
The survival is 1 minus the integral of the density function. S(x)=e^{x/100}, and
S(x)=e^{x/200}.
The survival function for the entire group is (e^{x/100})^{5} *
(e^{x/200})^{5}.
The density function for the group is the derivitive of 1 minus the survival
function = e^{3x/40}.
The mean for the group is the integral from 0 to infinity of x times the density
function for the group = 40/3 hours.
Michael Shackleford, A.S.A.
  36.  Light bulb problem #2
   In problem 35 what is the probability a light bulb with 100 hour life expectancy burns out first?
Answer
Problem 36 Answer
Problem 36 Answer
The answer is 2/3.
Michael Shackleford, A.S.A.
Solution
Problem 36 Solution
Problem 36 Solution
Warning: This is just a quick explanation which assumes some knowledge of
the exponential distribution and integral calculus.
The density function of the shorter lived life bulbs is f(x)=1/100 * e^{x/100}.
The density function of the longer lived life bulbs is f(x)=1/200 * e^{x/200}.
The probability of any given 100 hour bulb burning out first is the integral from 0 to infinity of (1/100)*e^{x/100}*(e^{x/100}>^{4}*(e^{x/200})^{5} =
1/100 * integral e^{15x/200} =
(1/100)*(200/15)=2/15.
The probability that ANY of the 100 hour bulbs burn out first is five times this answer, or 10/15=2/3.
Michael Shackleford, A.S.A.
  70.  Lottery problem
   In the "Match 5" game of the Maryland state lottery the player chooses any five numbers from 1 to 39. At a designated time there will be a drawing in which five numbers from 1 to 39 will be drawn at random, plus a "bonus ball" from the remaining 34. The following is the payout schedule for a $1 ticket plus an example of a winning ticket given that the numbers drawn were 12345, with a bonus number of 6:  Match all 5 numbers (bonus ball n/a): Win $50,000 (12345)
 Match 4 numbers plus bonus ball: Win $600 (12346)
 Match 4 numbers w/o bonus ball: Win $400 (12347)
 Match 3 numbers plus bonus ball: Win $30 (12367)
 Match 3 numbers w/o bonus ball: Win $15 (12378)
 Match 2 numbers plus bonus ball: Win $2 (12678)
What is the expected return of this game?
Answer
Problem 70 Answer
Problem 70 Answer
The answer is 7,434,440 / 19,575,738 =~ 37.98 cents.
Note: In all fairness to the state I should mention that you
can buy three tickets for the price of two, increasing the expected return to 56.97 cents
(but still a waste of money).
Michael Shackleford, A.S.A.
Solution
Problem 70 Solution
Problem 70 Solution
The number of ways the six balls can be drawn are (39:5)*34 = (39!/(34!*5!) * 34 = 19,575,738.
Lets assume the chosen numbers are 12345. Of the 19,575,738 ways to arrange the six
drawn balls how many will fit the drawn numbers to make a winner? Take the ratio of that
number of 19,575,738 and you have the probability.
There are 34 ways to draw the 6 numbers to match 12345 (you must match all of the first
5, the bonus ball can be any of the remaining 34).
The number of ways to match four plus the bonus ball are 170 (the bonus ball can be any
of the 5 you got right, and the last ball can be any of the remaining 34).
The number of ways to match four w/o the bonus ball are 5610 (5 different ways to arrange
the 4 correct numbers * the 5th number can be any of the remaining 34 * the bonus ball
can be any of the remaining 33).
Keep following the above logic for all possible winning combinations:
Ways Probability Expected
Type of win to win of winning Payoff return
    
Pick 5 34 .0000017368 $50000 .0868
Pick 4 + BB 170 .0000086842 $600 .0052
Pick 4 w/o BB 5610 .0002865792 $400 .1146
Pick 3 + BB 11220 .0005731585 $30 .0172
Pick 3 w/o BB 179520 .0091705355 $15 .1376
Pick 2 + BB 179520 .0091705355 $2 .0183
Total 376074 .3798
Note: In all fairness to the state I should mention that you
can buy three tickets for the price of two, increasing the expected return to 56.97 cents
(but still a waste of money).
Michael Shackleford, A.S.A.
  187.  Marching soldiers problem
   A column of soldiers one mile long is marching forward at a constant rate. The soldier at the front of the column has to deliver a message to the soldier at the rear. He breaks rank and begins marching toward the rear at a constant rate while the column continues forward. The soldier reaches the rear, delivers the message and immediately turns to march forward at a constant rate. When he reaches the front of the column and drops back in rank, the column has moved one mile. Question How far did the soldier delivering the message march?
Answer
Problem 187 Answer
Problem 187 Answer
The answer is 1+2 ^{1/2} =~ 2.4142 miles.
Michael Shackleford, A.S.A.
Solution
Problem 187 Solution
Problem 187 Solution
Let r _{1} = rate of the column = 1 mile/hr.
Let r_{2} = rate of the messanger.
d = distance covered by messanger.
t = time to move one mile = 1 hour.
The meeting point of the messanger and the person at the rear is a distance of r_{2}/(1+r_{2}) from the starting point of the messanger. This is just the ratio of the messanger's rate divided by the sum of the two rates.
The messanger will have to cover the distance to the meeting point, turn around and cover it again, and then march one more mile. So...
d = 2*r_{2}/(1+r_{2}).
distance = rate * time, so...
2*r_{2}/(1+r_{2}) = r_{2}*1
2*r_{2}/(1+r_{2}) = r_{2} 1
2*r_{2} = r_{2}^{2} 1
r_{2}^{2}  2*r_{2}  1 = 0
Use the quadratic equation to solve for r_{2}...
r_{2} = (2 + 8^{1/2})/2 = 1+2^{1/2} =~ 2.4142 miles/hr
Since he is marching for one hour is distance is 2.4142 miles.
I would like to thank Michael Whitfield for suggesting this problem.
Michael Shackleford, A.S.A.
  208.  Maximum volume of a cylinder problem
   A cylinder with an open top, has a surface area of S. What is the maximum volume it can hold?
Answer Problem 208 Answer
Question
A cylinder with an open top, has a surface area of S. What is the maximum volume it can hold?
The answer is [S^{2}×(3×pi/S)^{1/2}  pi×(S/(3×pi))^{1/2}]/(6*pi)
You may also wish to know that:
height=(Spi×r^{2})/(2×pi×r)
radius=(S/(3×pi))^{1/2}
Michael Shackleford, ASA — Feb. 7, 2012
Solution Problem 208 Solution
Question
A cylinder with an open top, has a surface area of S. What is the maximum volume it can hold?
Let V be the volume of the cylinder.
Let S be the surface area of the can.
Let h be the height of the can.
Let r be the radius of the can.
V = pi×r ^{2}×h
Let's express V as a function of r only.
S = pi×r ^{2} + 2×pi×r×h
Solving for h...
h = (Spi×r ^{2})/(2×pi×r)
So...
V = ((Spi×r ^{2})/(2×pi×r)) × pi×r ^{2}
V = r×(Spi×r ^{2})/2
Next, take the derivative of V and set equal to 0...
V = (1/2)×(S×pi×r ^{2}) = 0
S = 3×pi×r ^{2}
r ^{2} = S/(3×pi)
r = (S/(3×pi)) ^{1/2}
If we plug r in the equation for h, and then plug r and h into the equation for V we get...
V = [S ^{2}×(3×pi/S) ^{1/2}  pi×(S/(3×pi)) ^{1/2}]/(6×pi)
In the case of a surface area of 1, for example:
r = 0.325735
h = 0.325735
v = 0.108578
Michael Shackleford, ASA — Feb. 7, 2012
  242.  Maximum volume of cone
   You are in charge of making coneshaped paper cups. Your goal is to maximize the ratio of volume to surface area. What is the maximum that ratio can be? Assume the length from the tip of the cup to any point on the edge is 1.
Answer
Problem 242 Answer
Problem 242 Answer
The answer is 1/6.
Michael Shackleford
Solution
Problem 241 Solution
Problem 242 Solution
Please see my problem 242 solution (PDF  97K)
Michael Shackleford
  51.  Maximum volume of cone problem
   You have a tortilla with radius 1 and wish to form a cone. You may cut out any wedge you like from the tortilla. The point of the wedge must be at the center of the circle. After cutting out the wedge you then attach the two straight edges remaining to form a cone. What is the maximum ratio of the volume of the cone to the remaining surface area?
Answer
Problem 51 Answer
Problem 51 Answer
The answer is 1/6.
Michael Shackleford, A.S.A.
Solution   53.  Mean distance between two points in a square problem
   What is the mean distance between two random points in a unit square? Hint
Answer
Problem 53 Answer
Problem 53 Answer
The answer is [2 ^{1/2} + 2 + 5*ln(1+2 ^{1/2})] / 15 =~ 0.521405433 .
Michael Shackleford, A.S.A.
Solution
Problem 53 Solution
Problem 53 Solution
Here is my solution inspired by Presh Talwalker. Here is also a video solution:
VERY HARD Puzzle: What Is The Distance Between Two Random Points In A Square?.
To appreciate his solution, you'll need to know this integral:
Integral of sec^3(x) = (sec(s) + tan(x))/2 + ln(sec(x) + tan(x))/2
Here is another solution that doesn't rely on knowing that integral, submitted by Igor Volkov:
First, let I(x,y) stand for "the integral from x to y of ...", and sqr stand for "the square root of...".
To find the answer we must solve I(0,1) I(0,1) I(0,1) I(0,1) sqr((xa)^{2}+(yb)^{2}) db da dy dx.
Even the first integration of this is very messy, requiring a big table
of integrals, and I think at best a small book of integrals would only
take you through the second one. So we need to be more creative to
make the problem more managable. First we introduce a change of
variables:
Or:
(1): Now we have I(0,1) I(t,t1) I(0,1) I(s,s1) sqr(u^{2} + v^{2}) * J du ds dv dt
Where J = the Jacobian =
 db/du db/ds db/dv du/dt   0 0 1 1 
   
 da/du da/ds da/dv da/dt   1 1 0 0 
  =   = 1
 dy/du dy/ds dy/dv dy/dt   0 0 0 1 
   
 dx/du dx/ds dx/dv dx/dt   0 1 0 0 
Now lets solve the following inner two integrals of equation (1):
(2): I(0,1) I(s,s1) sqr(u^{2} + v^{2}) du ds.
The area of integration is bounded by the quadrilateral
with vertices (0,0), (1,1), (0,1), and (1,0), with s being
the horizontal axis and u being the vertical.
Changing the order of integral we get:
I(1,0) I(0,u+1) sqr(u^{2} + v^{2}) ds du +
I(0,1) I(u,1) sqr(u^{2} + v^{2}) ds du =
I(1,0) sqr(u^{2} + v^{2}) * (1+u) du +
I(0,1) sqr(u^{2} + v^{2}) * (1u) du =
I(1,0) sqr(p^{2} + v^{2}) * (1p) dp +
I(0,1) sqr(u^{2} + v^{2}) * (1u) du
(letting p=u in the first integral) =
I(0,1) sqr(p^{2} + v^{2}) * (1p) dp +
I(0,1) sqr(u^{2} + v^{2}) * (1u) du (Notice the two terms are equal)=
2* I(0,1) sqr(u^{2} + v^{2}) * (1u) du.
Now put this into equation (1), and moving the constant terms to the left:
2* I(0,1) I(t,t1) I(0,1) sqr(u^{2} + v^{2}) * (1u) du dv dt =
2* I(0,1) I(0,1) I(t,t1) sqr(u^{2} + v^{2}) * (1u) dv dt du (changing the order of integration)=
2* I(0,1) (1u) I(0,1) I(t,t1) sqr(u^{2} + v^{2}) dv dt du =
Note that the inner two integral are of the same form as equation (2), thus:
4* I(0,1) I(0,1) sqr(u^{2} + v^{2}) * (1u) * (1v) dv du =
4* I(0,1) I(0,1) sqr(u^{2} + v^{2}) * (1u) * (1v) du dv (changing the order of integration)
Next we make another substitution:
This time the Jacobian shall equal:
 du/dr du/dw   sin(w) r*cos(w) 
  =   = r
 dv/dr dv/dw   cos(w) r*sin(w) 
Thus the integral as a function of w and r is:
8 * I(0,pi/4) I(0,1/cos(w)) r * sqr(r*sin^{2}(w) + r*cos^{2}(w)) * (1r*cos(w)) * (1r*sin(w)) dr dw.
Note that we halved the area of integration, going from 0 to pi/4 as opposed to pi/2, and
multiplying by 8 instead of the 4. This is permissible because of the symetry of
the two regions.
= 8 * I(0,pi/4) I(0,1/cos(w)) r^{2} * (1r*cos(w)) * (1r*sin(w)) dr dw.
= 8 * I(0,pi/4) I(0,1/cos(w)) r^{2}  r^{3}*(cos(w)+sin(w)) +
r^{4}*(cos(w)+sin(w)) dr dw.
= 8 * I(0,pi/4) r^{3}/3  r^{4}*(cos(w)+sin(w))/4 +
r^{5}*(cos(w)+sin(w))/5 ( from 0 to 1/cos(w) ) dw.
= 8 * I(0,pi/4) (3*cos^{3}(w))^{1}  (cos(w)+sin(w))/(4*cos^{4})
+ (cos(w)+sin(w))/(5*cos^{5}(w)) dw.
= 8 * I(0,pi/4) (3*cos^{3}(w))^{1} 
(4*cos^{3}(w))^{1} 
sin(w)*(4*cos^{4}(w))^{1} +
sin(w)*(5*cos^{4}(w))^{1} dw.
= 8 * I(0,pi/4) (12*cos^{3}(w))^{1} 
sin(w)*(20*cos^{4}(w))^{1} dw.
(3) = 2/15 * I(0,pi/4) 5/cos^{3}(w) 
3*sin(w)/cos^{4}(w) dw.
Next we must dust off our table of integrals to help us with the integral of cos^{3}(w).
I dx/cos^{n}(x) = (1/(n1)) * sin(x)/cos^{n1}(x) + ((n2)/(n1)) * I dx/cos^{n2}(x).
In our case n=3 so,
I dx/cos^{3}(x) = (1/2) * sin(x)/cos^{2}(x) + (1/2) * I sec(x) dx.
Now lets look up the integral of sec(x):
I sec(x) dx = (1/2) * ln((1+sin(x))/(1sin(x))).
Now we are ready to integrate cos^{3}(w):
I(0,pi/4) cos^{3}(w) dw = sin(w)/(2*cos^{2}(w)) (from 0 to pi/4) + (1/2)*I(0,pi/4) sec(x) dx.
= (1/2)*(2^{1/2}/2^{1}  0) + (1/4)*ln((2+sqr(2))/(2sqr(2))).
= (1/2)*sqr(2) + (1/4)*ln(((2+sqr(2))/2+sqr(2))/((2sqr(2))*(2+sqr(2))).
= 1/sqr(2) + (1/4)*ln((4+4*sqr(2)+2)/(42)).
= 1/sqr(2) + (1/4)*ln(3+2*sqr(2)).
= 1/sqr(2) + (1/2)*ln(3+2*sqr(2))^{1/2}.
= 1/sqr(2) + (1/2)*ln(1+sqr(2)).
Now lets integrate sin(w)/cos^{4}(w).
let m=cos(w), thus dm=sin(w) dw.
I(0,pi/4) sin(w)/cos^{4}(w).
= I(1,1/sqr(2)) 1*m^{4} dm.
= 1/(3*m^{3}) from 0 to 1/sqr(2).
= (1/3) * (2*sqr(2)  1).
Not it is time to plus these integrals into equation (3):
(2/15) * [ 5*(1/sqr(2) + (1/2)*ln(1+sqr(2))) + 3*((1/3) * (2*sqr(2)  1)) ]
= (2/15) * [ 5*sqr(2)/2 + (5/2)*ln(1+sqr(2))  4*sqr(2)/2 + 1 ]
= (2/15) * [ sqr(2)/2 + (5/2)*ln(1+sqr(2)) + 1 ]
= (sqr(2) + 2 + 5*ln(1+sqr(2))) / 15.
Congradulations and thanks to Igor Volkov for this solution.
Michael Shackleford, A.S.A.
  38.  Meeting at a restaurant problem
   Two people arrive in a restaurant independently. Each arrives a random time between 5pm and 6pm, distributed uniformaly (no moment in this range is any more likely for arrival than another). What is the probability they arrived within 10 minutes of each other?
Answer
Problem 38 Answer
Problem 38 Answer
The answer is 11/36 =~ 30.6%
Michael Shackleford, A.S.A.
Solution
Problem 38 Answer
Problem 38 Answer
Draw a 6 by 6 square. Plot the arrival time of one person vertically. Plot
the arrival time of the other person horizontally. The times they arrive
within 10 minutes of each other can be represented by a diagonal stripe
across the square. The area of the entire square is 36. The area of the portions
outside the stripe is 25. Thus the area of the stripe is 11. Considering the
uniformity of the arrival times the probability of arriving within ten
minutes is the area of the stripe divided by the area of the square = 11/36.
Michael Shackleford, A.S.A.
  81.  Military attack problem
   Country x is planning to attack country y, and country y is anticipating the attack. Country x can either attack by land or by sea. Country y can either prepare for a land defense or a sea defense. Both countries must choose either an all land or all sea strategy, they may not divide their forces. The following are the probabilities of a successful invasion according to both strategies used.  If x launches an attack by sea and y prepares a defense by sea the probability of a successful invasion is 80%.
 If x launches an attack by sea and y prepares a defense by land the probability of a successful invasion is 100%.
 If x launches an attack by land and y prepares a defense by land the probability of a successful invasion is 60%.
 If x launches an attack by land and y prepares a defense by sea the probability of a successful invasion is 100%.
What should the strategy of country x be, assuming the goal is to maximize the probability of a successful invasion? Assume the goal of country y to be to minimize the probability of a successful invasion. What is the final probability of a successful invasion assuming both utilize an optimal strategy?
Answer
Problem 81 Answer
Problem 81 Answer
X should attack by sea with probability 2/3 and by land with probability 1/3. The probability of a successful invastion is 13/15.
Michael Shackleford, A.S.A.
Solution
Problem 81 solution
Problem 81 Solution
For purposes of this solution I shall denote the country names with capital letters.
Both X and Y should choose their strategy randomly. Let x be the probability that country X attacks by sea. Let y be the probability that country Y defends by sea.
The probability of a successful invasion is :
f(x,y)=.8xy + x(1y) + (1x)y + .6(1x)(1y) =
.8xy + x  xy + y  xy + .6  .6x  .6y + .6xy =
.6xy + .4x + .4y + .6 .
Y is obviously going to try to minimize the probability of a successful attack. Taking the derivative of f(x,y) with respect to y yields:
.6x + .4 = 0.
x=2/3.
Thus X should attack by sea with probability 2/3 and by land with probability 1/3. Y should also defend by sea with probability 2/3 and by land with probability 1/3. The probability of a successful invation is .6*(2/3)*(2/3) + .4*(2/3) + .4*(2/3) + .6 = 13/15.
Michael Shackleford, A.S.A.
  166.  Miss America problem
   Fifty contestants for the Miss America pageant were arranged in ten rows and five columns on the Boardwalk in Atlantic City. One viewer of this display noticed that of the tallest women in each column that Miss Maryland was the shortest of the five. Another viewer noticed that of the shortest women in each row that Miss Nevada was the tallest of the ten. There are no ties, all fifty women have different heights. Who is taller, Miss Maryland or Miss Nevada?
Answer
Problem 166 Answer
Problem 166 Answer
Miss Maryland is taller.
Michael Shackleford, ASA  July 19, 2001
Solution
Problem 166 Solution
Problem 166 Solution
There are three possible scenarios:
 If Miss Maryland and Miss Nevada are in the same column then Miss Maryland would be taller, because she is the tallest in that column.
 If Miss Maryland and Miss Nevada are in the same row then Miss Maryland would be taller, because Miss Nevada is the shortest in that row.
 If Miss Maryland and Miss Nevada share neither a row nor column then there must be some other contestant who is both in Miss Maryland's column and Miss Nevada's row. Call this woman Miss Texas. To illustrate her position consider the following diagram.
xxxxx
xmxxx
xxxxx
xxxxx
xxxxx
xxxxx
xtxxn
xxxxx
xxxxx
xxxxx
In this digram the m stands for Miss Maryland, the n for Miss Nevada, and the t for Miss Texas.
Miss Maryland must be taller than miss Texas, because miss Maryland is the tallest in her column. Miss Texas must be taller than Miss Nevada, because Miss Nevada is the shortest in her row. Thus if M>T>N then M>N, or Miss Maryland is taller than Miss Nevada.
Michael Shackleford, ASA  July 19, 2001
  186.  Monty Hall problem
   On a game show there are three doors. Behind one door is a new car and behind the other two are goats. Every time the game is played the contestant first picks a door. Then the host will open one of the other two doors and always reveals a goat. Then the host gives the player the option to switch to the other unopened door. Should the player switch?
Answer
Problem 186 Answer
Problem 186 Answer
The answer is YES.
Michael Shackleford, A.S.A.
Solution
Problem 186 Solution
Problem 186 Solution
The key to this problem is that the host is predestined to open a door with a goat. He knows which door has the car so regardless of which door the player picks he always can reveal a goat behind another door.
Let's assume that the prize is behind door 1. Following are what would happen if the player had a strategy of not switching.
 Player picks door 1 > player wins
 Player picks door 2 > player loses
 Player picks door 3 > player loses
Following are what would happen if the player had a strategy of switching.
 Player picks door 1 > Host reveals goat behind door 2 or 3 > player switches to other door > player loses
 Player picks door 2 > Host reveals goat behind door 3 > player switches to door 1 > player wins
 Player picks door 3 > Host reveals goat behind door 2 > player switches to door 1 > player wins
So by not switching the player has 1/3 chance of winning. By switching the player has a 2/3 chance of winning. So the player should definitely switch.
The plain simple English reason the probability of winning increases to 66.67% by switching, lays in the fact that the host always reveals a goat. If the host didn't know which door had the car, then the probability of having a win would go up to 50% after revealing one of the goats. Such is the case on the show "Deal or no Deal." On that show, the host does not know where the million dollar case is. So, as cases are eliminated that do not have the million dollars, the probability increases that every remaining case has it, equally. When there are only two cases left and two prizes, each case has a 50% chance of each prize.
Links
Monty Hall problem at Wikipedia.
Michael Shackleford, A.S.A.
  90.  Mortgage payment problem
   Given a $100,000 mortgage, a payment period of 30 years, monthly payments (the first due at the end of the first month), and an interest rate of 7.5% compounded monthly, what will be the monthly payments?
Answer
Problem 90 Answer
Problem 90 Answer
The answer is approximately $699.21
Michael Shackleford, A.S.A.
Solution
Problem 90 Solution
Problem 90 Solution
Since there are monthly payments there will be 12*30=360 total payments. The
present value of these 360 payments must be $100,000. Lets call the montly payment
p, then the present value of all payments can be expressed below where i = the
interest rate = .075/12.
p * [ 1/(1+i) + 1/(1+i)^{2} + 1/(1+i)^{3} + ... + 1/(1+i)^{360} ] =
p * [ (1  1/(1+i)^{360}) * ( 1/(1+i) + 1/(1+i)^{2} + 1/(1+i)^{3} + ... ) ].
Now recall that that the sum for n=1 to infinity of x^{n} = x/1x, where x is less than 1, so
we can further simplify:
p * [ (1  1/(1+i)^{360}) * (1/(1+i))/(1(1/(1+i))) =
p * [ (1  1/(1+i)^{360}) / i = $100,000.
Solving for p (remember that i=.075/12):
p = $100,000 * i / (1  1/(1+i)^{360}) =~ $699.21
For your own information here are what the monthly payments would be
under various other interest rates:
0% $277.78
1% $321.64
2% $369.62
3% $421.60
4% $477.42
5% $536.82
6% $599.55
7% $665.30
8% $733.76
9% $804.62
10% $877.57
11% $952.32
12% $1028.61
13% $1106.20
14% $1184.87
15% $1264.44
16% $1344.76
17% $1425.68
18% $1507.09
19% $1588.89
20% $1671.02
25% $2084.58
30% $2500.34
35% $2916.76
40% $3333.36
45% $3750.01
50% $4166.67
55% $4583.33
60% $5000.00
65% $5416.67
70% $5833.33
75% $6250.00
80% $6666.67
85% $7083.33
90% $7500.00
95% $7916.67
Michael Shackleford, A.S.A.
  122.  Mouse and a cube of cheese problem
   A cubic piece of cheese has been subdivided into 27 subcubes (so that it looks like a Rubik's Cube). A mouse starts to eat a corner subcube. After eating any given subcube it goes on to another adjacent subcube. Is it possible for the mouse to eat all 27 subcubes and finish with the center cube?
Answer
Problem 122 Answer
The answer is no.
Michael Shackleford, A.S.A.
Solution
Problem 122 Solution
Consider the subcubes to form a three dimensional checkerboard
composed of subcubes of swiss and cheddar cheese. Assume the corners and
centers of each face are swiss cheese and the rest of the subcubes,
including the center, are cheddar cheese. Assume the mouse
eats two subcubes an hour. At the beginning of every hour the mouse will be
starting on a swiss subcube and will end the hour with a subcube
of cheddar cheese. After 13 hours the mouse will have just finished a
cheddar subcube. However the center subcube is also cheddar. Since
two adjacent cubes must be of different kinds of cheese the mouse can
not eat the center cube last.
Thanks to Nick Hobson for suggesting this one.
Michael Shackleford, A.S.A.
  20.  Nine minute egg problem
   You are a cook in a remote area with no clocks or other way of keeping time other than a 4 minute hourglass and a 7 minute hourglass. You do have a stove however with water in a pot already boiling. Somebody asks you for a 9 minute egg, and you know this person is a perfectionist and will be able to tell if you undercook or overcook the eggs by even a few seconds. What is the least amount of time it will take to prepare the egg?
Answer
Problem 20 Answer
Problem 20 Answer
The answer is 9 minutes.
Thanks to The Grey Labyrinth for this one.
Michael Shackleford, A.S.A.
Solution
Problem 20 Solution
Problem 20 Solution
Problem 20 Solution 
Time 
Action 
Time remaining before action 
Time remaining after action 
4 Minute 
7 Minute 
4 Minute 
7 Minute 
0 
Turn over both hourglasses 
0 
0 
4 
7 
4 
Turn over 4 minute hourglass 
0 
3 
4 
3 
7 
Turn over 7 minute hourglass 
1 
0 
1 
7 
8 
Turn over 7 minute hourglass 
0 
6 
0 
1 
9 
Take egg off stove 
0 
0 
0 
0 
You can also try this problem witha a 15 minute egg and hourglasses of 7 and 11 minutes.
Thanks to The Grey Labyrinth for this one.
Michael Shackleford, A.S.A.
  71.  Nine pearls and a scale problem
   You have nine pearls, eight are real and one is fake. All the real ones weigh the same and the fake weighs less than the real ones. Using a balance scale twice how can you weed out the fake one?
Answer
Problem 71 Answer
Problem 71 Answer
Weigh any three pearls against any other three. If one side
is higher than the other the fake pearl is on that side. If the
sides stay equal the fake pearl is in the three you didn't use.
For the second weighing take any two of the bad three and
weigh one against the other. If one side goes up that side
has the fake pearl. If the two sides stay the same the unused
pearl of the three is fake.
Michael Shackleford, A.S.A.
  203.  OneCard Poker
   The game of onecard poker has a threecard deck, an ace, deuce, and trey. The ace is lowest and the trey is highest. Each of two players ante $1 into the pot. Then, each player gets one card. The order of betting is predetermined, with player 1 to act first. Player 1 may either bet $1 or check. If player 1 bets, player 2 may either call or fold. If player 1 checks, player 2 may either bet $1 or check. If player 1 checks, and player 2 bets, then player 1 may either call or fold. If both players check, or one bets and one calls, then the higher card wins the pot. Assuming both players are perfect logicians, what is the optimal strategy for each player?
Answer
Problem 203 Answer
Problem 203 Answer
Question
The game of onecard poker has a threecard deck, an ace, deuce, and trey. The ace is lowest and the trey is highest. Each of two players ante $1 into the pot. Then, each player gets one card. The order of betting is predetermined, with player 1 to act first. Player 1 may either bet $1 or check. If player 1 bets, player 2 may either call or fold. If player 1 checks, player 2 may either bet $1 or check. If player 1 checks, and player 2 bets, then player 1 may either call or fold. If both players check, or one bets and one calls, then the higher card wins the pot. Assuming both players are perfect logicians, what is the optimal strategy for each player?
Answer
The obvious plays are neither player should call with the ace, always call with the trey, and never bet with a deuce.
For the more difficult decisions, let's define the following probabilities:
 p_{1} = Probability player 1 bets with an ace (bluffing).
 p_{2} = Probability player 1 calls with a deuce after player 2 bets (hoping player 2 is bluffing).
 p_{3} = Probability player 1 bets with a trey (hoping player 2 will call).
 q_{1} = Probability player 2 bets with an ace, after player 1 checks (bluffing).
 q_{2} = Probability player 2 calls with a deuce, after player 1 bets (hoping player 1 is bluffing).
Player 2 has probabilities q_{1}=1/3 and q_{2}=1/3.
Player 1 has more flexibility. He may use any probabilites as long as these too equations are satisfied:
3p_{2}p_{3}=1
3p_{1}=p_{3}
For example, p_{1}=1/5, p_{2}=8/15, p_{3}=3/5.
Acknowledgement: The idea of this problem came from the article Game Theory and Poker, by Jason Swanson (April 3, 2005). The paper works out the same solution. I kept the variables names the same for my own benefit, and yours.
Michael Shackleford — Apr. 28, 2009
Solution
Problem 203 Solution
Problem 203 Answer
Question
The game of onecard poker has a threecard deck, an ace, deuce, and trey. The ace is lowest and the trey is highest. Each of two players ante $1 into the pot. Then, each player gets one card. The order of betting is predetermined, with player 1 to act first. Player 1 may either bet $1 or check. If player 1 bets, player 2 may either call or fold. If player 1 checks, player 2 may either bet $1 or check. If player 1 checks, and player 2 bets, then player 1 may either call or fold. If both players check, or one bets and one calls, then the higher card wins the pot. Assuming both players are perfect logicians, what is the optimal strategy for each player?
Solution
The obvious plays are neither player should call with the ace, always call with the trey, and never bet with a deuce. By raising with a deuce, you will win $1 against an ace, because your opponent will fold, and lose $2 against a trey, because he will raise. A superior strategy with a deuce is to always check, and then to always call if the other player bets. By following the check and call strategy, you'll win $2 instead of $1 against an ace, and stay equal against a trey at a $2 loss. If your opponent checks with the ace, you'll still win $1, and be no worse off. You can only gain by checking and calling with a deuce.
There are five other, more difficult, decision points to be faced by both players, as follows:
 p_{1}
 Probability player 1 bets with an ace (bluffing).
 p_{2}
 Probability player 1 calls with a deuce after player 2 bets (hoping player 2 is bluffing).
 p_{3}
 Probability player 1 bets with a trey (hoping player 2 will call).
 q_{1}
 Probability player 2 bets with an ace, after player 1 checks (bluffing).
 q_{2}
 Probability player 2 calls with a deuce, after player 1 bets (hoping player 1 is bluffing).
The key to game theory problems like this is to put your opponent to an indifference point at nonobvious decisions. If the odds favor one way or the other, the person making the decision will always go with the odds, and gain an advantage. To do that, let's express player 1's expected value.
EV1 =
pr(player 1 has A)×pr(player 2 has 2)×pr(player 1 checks)×pr(player 2 checks)×1 +
pr(player 1 has A)×pr(player 2 has 2)×pr(player 1 bets)×pr(player 2 folds)×+1 +
pr(player 1 has A)×pr(player 2 has 2)×pr(player 1 bets)×pr(player 2 calls)×2 +
pr(player 1 has A)×pr(player 2 has 3)×pr(player 1 checks)×pr(player 2 bets)×pr(player 1 folds)×1 +
pr(player 1 has A)×pr(player 2 has 3)×pr(player 1 bets)×pr(player 2 calls)×2 +
pr(player 1 has 2)×pr(player 2 has A)×pr(player 1 checks)×pr(player 2 checks)×+1 +
pr(player 1 has 2)×pr(player 2 has A)×pr(player 1 checks)×pr(player 2 bets)×pr(player 1 folds)×1 +
pr(player 1 has 2)×pr(player 2 has A)×pr(player 1 checks)×pr(player 2 bets)×pr(player 1 calls)×+2 +
pr(player 1 has 2)×pr(player 2 has 3)×pr(player 1 checks)×pr(player 2 bets)×pr(player 1 folds)×1 +
pr(player 1 has 2)×pr(player 2 has 3)×pr(player 1 checks)×pr(player 2 bets)×pr(player 1 calls)×2 +
pr(player 1 has 3)×pr(player 2 has A)×pr(player 1 checks)×pr(player 2 checks)×+1 +
pr(player 1 has 3)×pr(player 2 has A)×pr(player 1 checks)×pr(player 2 bets)×pr(player 1 calls)×+2 +
pr(player 1 has 3)×pr(player 2 has A)×pr(player 1 bets)×pr(player 2 folds)×+1
pr(player 1 has 3)×pr(player 2 has 2)×pr(player 1 checks)×pr(player 2 checks)×+1 +
pr(player 1 has 3)×pr(player 2 has 2)×pr(player 1 bets)×pr(player 2 folds)×+1
pr(player 1 has 3)×pr(player 2 has 2)×pr(player 1 bets)×pr(player 2 calls)×+2
EV1 =
(1/3)×(1/2)×(1p_{1})×1×1 +
(1/3)×(1/2)×p_{1}×(1q_{2})×1 +
(1/3)×(1/2)×p_{1}×q_{2}×2 +
(1/3)×(1/2)×(1p_{1})×1×1×1 +
(1/3)×(1/2)×p_{1}×1×2 +
(1/3)×(1/2)×1×(1q_{1})×1 +
(1/3)×(1/2)×1×q_{1}×(1p_{2})×1 +
(1/3)×(1/2)×1×q_{1}×p_{2}×2 +
(1/3)×(1/2)×1×1×(1p_{2})×1 +
(1/3)×(1/2)×1×1×p_{2}×2 +
(1/3)×(1/2)×(1p_{3})×(1q_{1})×1 +
(1/3)×(1/2)×(1p_{3})×q_{1}×1×2 +
(1/3)×(1/2)×p_{3}×1×1 +
(1/3)×(1/2)×(1p_{3})×1×1 +
(1/3)×(1/2)×p_{3}×(1q_{2})×1 +
(1/3)×(1/2)×p_{3}×q_{2}×2
EV1 = (1/6)×(
1+p_{1}
+p_{1}p_{1}q_{2}
2p_{1}q_{2}
1+p_{1}
2p_{1}
+1q_{1}
q_{1}+q_{1}p_{2}
+2q_{1}p_{2}
1+p_{2}
2p_{2}
1q_{1}p_{3}+p_{3}q_{1}
+2q_{1}2p_{3}q_{1}
+p_{3}
+1p_{3}
+p_{3}p_{3}q_{2}
+2p_{3}q_{2})
= (1/6)×(p_{1}q_{1}p_{2}3p_{1}q_{2}+3q_{1}p_{2}p_{3}q_{1}+p_{3}q_{2}) Equation 1
= (1/6)×[q_{1}×(1+3p_{2}p_{3}) + q_{2}×(3p_{1}+p_{3}) + p_{1}p_{2}]
What values for q_{1} and q_{2} will make EV1 equal to a constant, causing player 1 to be indifferent to all values of p_{1}, p_{2}, and p_{3}?
The answer is q_{1}=1/3, and q_{2}=1/3, so...
EV1 = (1/6)×[(1/3)×(1+3p_{2}p_{3}) + (1/3)×(3p_{1}+p_{3}) + p_{1}p_{2}]
= (1/6)×[1/3 +p_{2} p_{3}/3  p_{1} +p_{3}/3 +p_{1} p_{2}]
= (1/6)×(1/3)
= 1/18
So, player 2 should bluff with an ace after a check, and call with a deuce, each with probability 1/3. If player 2 does that, it won't make any difference what player 1 does at his three difficult decision points, his overall expected value will be 1/18.
Next, what should player's 1 optimal strategy be, so that player 2 can't exploit a bad strategy? Let's go back to equation 1. The expected value of player 2 will be the opposite of the expected value of player 1. So...
EV2 = (1/6)×(p_{1}q_{1}p_{2}3p_{1}q_{2}+3q_{1}p_{2}p_{3}q_{1}+p_{3}q_{2}) =
EV2 = (1/6)×[p_{1}×(13q_{2})+p_{2}×(3q_{1}1)+p_{3}×(q_{2}q_{1})q_{1}]
Next, determine values for p_{1}, p_{2}, and p_{3} so that the q_{1} and q_{2} terms cancel out, leaving a constant. This will result in a strategy where player 2 is indifferent at the difficult decision points.
If 3p_{2}p_{3}=1, then the q_{1} terms will cancel out.
If 3p_{1}=p_{3}, then then q_{2} terms will cancel out.
So, two equations must be satisfied. Examples of values that work are p_{1}=1/5, p_{2}=8/15, p_{3}=3/5.
Acknowledgement: The idea of this problem came from the article Game Theory and Poker, by Jason Swanson (April 3, 2005). The paper works out the same solution. I kept the variables names the same for my own benefit, and yours.
Michael Shackleford, ASA — Apr. 28, 2009
  160.  Paint mixing problem
   Your paint inventory consists of 60 gallons of blue, 40 gallons of red, and 30 gallons of yellow. To make purple paint you mix equals parts of blue and red. To make orange paint you mix equal parts of red and yellow. To make green paint you mix equal parts of blue and yellow. Purple paint sells for $6 a gallon, orange for $20, and green for $9. There is a fixed disposal charge for every unused gallon. How much of each of purple, orange, and green paint should you mix to maximize profits if the disposal cost is (a) $4 per gallon, and (b) $6 per gallon.
Answer
Problem 160 Answer
Problem 160 Answer
(a) Purple=20, Orange=60, Green=0, revenue=$1120
(b) Purple=70, Orange=10, Green=50, revenue=$1070
Thanks to Contingencies Magazine for this problem, which appears in the November/December 2000 issue.
Michael Shackleford, ASA  December 1, 2000
  105.  Parking cars problem
   There is a street of length 4. The street is initially empty. Cars then come along to fill the street until there is no space left that is large enough to park a car in. Every car is length 1. Drivers will choose a location to park at random among all possible locations left. No consideration, whether good or bad, is given to other cars. What is the expected number of cars that will be able to park?
Answer
Problem 105 Answer
Problem 105 Answer
11/3  4/3*ln(2) =~ 2.7425
Michael Shackleford, A.S.A.
Solution
Problem 105 Solution
Problem 105 Solution
Let f(x) denote the answer.
Let I(a,b) denote the integral from a to b.
It is obvious that f(x)=0 for x<1.
It is obvious that f(x)=1 for 1<=x<2.
If 2<=x<=3 then f(x)=1+2*(x2)/(x1). This can be found by using simple geometry which is
left for you to do.
Now lets solve x=4. The first car will divide the rest of the street into a small and a
large portion, lets call them s and l. s is equally likely to range from 0 to 3/2. l
is equally likely to range from 3/2 to 3. The probability that the smaller portion
is less than 1 is 2/3. Likewise the probability that smaller portion is greater
than 1 is 1/3. If the smaller portion is greater than 1 then exactly 3 cars will be
able to park. If the smaller portion is less than one the number than will be able
to park is 1+f(l).
f(x) = 1/3*3 + 2/3*(1+f(l), where l goes from 2 to 3)
= 1 + 2/3*(1 + I(2,3) 1+2*(x2)/(x1) dx)
= 1 + 2/3 + 2/3 + 4/3*I(2,3) (x2)/(x1) dx, next let y=x1
= 1 + 4/3 * 4/3 * I(1,2) (y1)/y dy
= 7/3 + 4/3 * I(1,2) 1  1/y dy
= 7/3 + 4/3 * (yln(y) for y from 1 to 2)
= 7/3 + 4/3 * (2ln(2)1+ln(1))
= 7/3 + 4/3  4/3*ln(2)
= 11/3  4/3*ln(2)
=~ 2.7425
Michael Shackleford, A.S.A.
  89.  Pass line and odds problem
   This problem has been removed.
Answer
Problem 89 answer
Problem 89 answer
The player's advantage is 1/165 per 1 unit bet, or approximately 0.61%.
Michael Shackleford, A.S.A.
Solution
Problem 89 solution
Problem 89 Solution
There are two solutions presented here:
Solution 1
The house advantage in just the pass line bet is 7/495, see the
solution to problem 82 for an explanation of
this step.
There is a 1/3 chance that the come out roll will be a 2,3,7,11, or 12, click
here if this step needs explanation.
There is a 2/3 chance that double odds will be taken. Given that $1 is bet
on the pass line the expected money bet per round is 1/3*$1 + 2/3*$3 = $7/3.
Since $1 is always bet on the pass line 1/(7/3)=3/7 of money bet, overall, is on the pass
line, and the other 4/7 is on the odds.
Thus the house advantage is 3/7*(7/495) + 4/7*0 = 3/495 = 1/165. The players
advantage is just the opposite of this, 1/165.
Solution 2
The player's advantage is the sum over all possible outcomes of the product of
the probability of the outcome and the net profit or loss, divided by the sum over
all possible outcomes of the expected amount bet.
Lets say the original wager on the pass line is 5. By summing the following
possible outcomes you will have the expected gain or loss.
 Roll 2 on come out roll: 1/36 * 5 = 5/36 = 4950/35640
 Roll 3 on come out roll: 2/36 * 5 = 10/36 = 9900/35640
 Roll 4 on come out roll, then win: 3/36 * 3/9 * (5+20) = 225/324 = 24750/35640
 Roll 4 on come out roll, then lose: 3/36 * 6/9 * (510) = 270/324 = 29700/35640
 Roll 5 on come out roll, then win: 4/36 * 4/10 * (5+15) = 320/360 = 31680/35640
 Roll 5 on come out roll, then lose: 4/36 * 6/10 * (510) = 360/360 = 35640/35640
 Roll 6 on come out roll, then win: 5/36 * 5/11 * (5+12) = 425/396 = 38250/35640
 Roll 6 on come out roll, then lose: 5/36 * 6/11 * (510) = 450/396 = 40500/35640
 Roll 7 on come out roll: 6/36 * 5 = 30/36 = 29700/35640
 Roll 8 on come out roll, then win: 5/36 * 5/11 * (5+12) = 425/396 = 38250/35640
 Roll 8 on come out roll, then lose: 5/36 * 6/11 * (510) = 450/396 = 40500/35640
 Roll 9 on come out roll, then win: 4/36 * 4/10 * (5+15) = 320/360 = 31680/35640
 Roll 9 on come out roll, then lose: 4/36 * 6/10 * (510) = 360/360 = 35640/35640
 Roll 10 on come out roll, then win: 3/36 * 3/9 * (5+20) = 225/324 = 24750/35640
 Roll 10 on come out roll, then lose: 3/36 * 6/9 * (510) = 270/324 = 29700/35640
 Roll 11 on come out roll: 2/36 * 5 = 10/36 = 5/18 = 9900/35640
 Roll 12 on come out roll: 1/36 * 5 = 5/36 = 4950/35640
The sum of all these expected gains/losses is 2520/35640.
Now lets sum the expected amount wagered:
If the first roll is a 2,3,7,11, or 12 the wager is 5.
If the first roll is anything else the wager is 15.
Thus the expected wager is (1+2+6+2+1)/36 * 5 + (3+4+5+5+4+3)/36 * 15 = 60/36 + 360/36 = 420/36 = 35/3.
Thus the player's advantage is (2520/35640) / (35/3) = 7560/1,247,400 = 378/62370 = 1/165.
In other words the house has an advantage of approximately 0.61%.
Michael Shackleford, A.S.A.
  82.  Pass line bet problem
   What is the probability of winning a bet on the pass line in craps? If you aren't familiar with the bet click here.
Answer
Problem 82 Answer
Problem 82 Answer
The answer is 244/495 =~ 49.3%
Michael Shackleford, A.S.A.
Solution
Problem 82 solution
Problem 82 Solution
For those of you coming from The Wizard of Odds
this is the derivation of the house advantage on the pass line in craps.
First review the probability of throwing any given number on any given throw:
 2: 1/36
 3: 2/36
 4: 3/36
 5: 4/36
 6: 5/36
 7: 6/36
 8: 5/36
 9: 4/36
 10: 3/36
 11: 2/36
 12: 1/36
If you don't understand why the above is true please review my
page of dice
probabilities basics.
From the above the probalility of rolling 7 is 6/36 = 1/6 and the
probability of rolling an 11 is 2/36 = 1/18. These are the numbers
that win on the first throw.
Next lets assume the point thrown on the first roll is a 4, what is the probability
of throwing it again before a 7?
Let pr(x) stand for the probability
of event x happening on any given roll. The answer is:
pr(4) +
pr(anything other than 4 and 7) * pr(4) +
pr(anything other than 4 and 7)^{2} * pr(4) +
pr(anything other than 4 and 7)^{3} * pr(4) +
pr(anything other than 4 and 7)^{4} * pr(4) +
+ ...
Pr(4) = 3/36 = 1/12, pr(anything other than 4 and 7) = 13/366/36 = 27/36 = 3/4.
pr(rolling a 4 before a 7)
= 1/12 + (3/4 * 1/12) + ((3/4)^{2} * 1/12) + ((3/4)^{3} * 1/12) + ...
= 1/12 * sum for i = 0 to infinity of (3/4)^{i}
= 1/12 * (1/(13/4))
= 1/12 * 4
= 1/3.
The probability of rolling a 10 is the same as the probability of rolling a 4
so pr(rolling a 10 before a 7) also equals 1/3.
Next assume the point thrown is a 5.
Pr(5) = 4/36 = 1/9, pr(anything other than 5 and 7) = 14/366/36 = 26/36 = 13/18.
pr(rolling a 5 before a 7)
= 1/9 + (13/18 * 1/9) + ((13/18)^{2} * 1/9) + ((13/18)^{3} * 1/9) + ...
= 1/9 * sum for i = 0 to infinity of (13/18)^{i}
= 1/9 * (1/(113/18))
= 1/9 * 18/5
= 2/5.
The probability of rolling a 9 is the same as the probability of rolling a 5
so pr(rolling a 9 before a 7) also equals 2/5.
Next assume the point thrown is a 6.
Pr(6) = 5/36, pr(anything other than 6 and 7) = 15/366/36 = 25/36.
pr(rolling a 6 before a 7)
= 5/36 + (25/36 * 5/36) + ((25/36)^{2} * 5/36) + ((25/36)^{3} * 5/36) + ...
= 5/36 * sum for i = 0 to infinity of (25/36)^{i}
= 5/36 * (1/(125/36))
= 5/36 * 36/11
= 5/11.
The probability of rolling an 8 is the same as the probability of rolling a 6
so pr(rolling a 8 before a 7) also equals 5/11.
Now we're ready to add all this together...
The probability of winning the pass line bet is:
pr(7) + pr(11) +
pr(4)*pr(4 before a 7) +
pr(5)*pr(5 before a 7) +
pr(6)*pr(6 before a 7) +
pr(8)*pr(8 before a 7) +
pr(9)*pr(9 before a 7) +
pr(10)*pr(10 before a 7)
= 6/36 + 1/18 + (1/12 * 1/3) + (1/9 * 2/5) + (5/36 * 5/11) +
(5/36 * 5/11) + (1/9 * 2/5) + (1/12 * 1/3)
= 6/36 + 1/18 + 1/36 + 2/45 + 25/396 + 25/396 + 2/45 + 1/36
= 330/1980 + 110/1980 + 55/1980 + 88/1980 + 125/1980 + 125/1980 + 88/1980 + 55/1980
= 976/1980 = 244/495.
The expected return on this wager is the product of the probability of
winning and the ratio of what you keep (including your original wager)
to the original bet if you do win (in this case 2). Thus
the expected return is 244/495 * 2 = 488/495 =~ 98.59%.
The house advantage is what the casino gets to keep, on average, which is 1 minus
the expected return, which equals 1(488/495) = 7/495 =~ 1.41%
In the appendix to this problem I show the house advantage on
the don't pass bet.
Michael Shackleford, A.S.A., 9/15/1998
  177.  Picking apples problem
   Three men and a boy are gathered in an orchard to shake apples off the trees and gather them. You are given:  The boy can shake 300 apples off the trees per hour.
 Each man can gather 100 apples off the ground per hour.
 A man can shake the apples off the trees 25% faster than two men and the boy can gather them.
 The group experiments with having two men and the boy gather apples for an entire hour, and divide the time of third man between shaking and gathering. They realize that the most they can accomplish with this strategy is still gathering 300 apples per hour.
At the end of a day of work the group is paid $500. How should the money be fairly divided between the three men and the boy, according to their individual contributions?
Answer
Problem 177 Answer
Problem 177 Answer
Each man should get $500*(1085 ^{1/2})/3 = $130.08.
The boy should get $2000/(9+85^{1/2}) = $109.77
Michael Shackleford, A.S.A.
Solution
Problem 177 Solution
Problem 177 Solution
First let's define some terms:
m = man's shaking speed per hour
b = boy's gathering speed per hour
r = ratio of the time man is shaking, if they decide to have a man as shaker.
Let's look at the number of apples gathered if one of the men is the shaker for a ratio r of the time.
r×(200+b) + (1r)×(300+b) = 300
200r + rb + 300 + b 300r  rb = 300
b  100r = 0
100r = b
(1) r = b/100
Next, let's look at the fact that a man can shake 25% faster than two men and the boy can gather them. Assume he does this for ratio r of the hous. Then the number of apples shaken is:
r×(5/4)×(200+b) = 300
(b/100)×(5/4)×(200+b) = 300 Substituting equation (1) for r
b×(5/4)×(200+b) = 30,000
b×(200+b) = 24,000
b^{2}2 + 200b  24,000=0
Using the quadradic formula:
b = [200 +/ sqr(40,000+96,000)]/2 = 84.3909
We're given that m = 1.25×(200+b). We know b, so we can solve for m:
m = 1.25×(200+84.3909)
m = 355.4886
So, the man can shake 355.4886/300 = 18.5% faster than the boy.
The man can also gather 100/84.3909 = 18.5% faster than the boy.
It stands to reason he should get paid 18.5% more as well.
Dividing up the $500, let:
m' = man's pay
b' = boy's pay
500 = 3m' + b'
500 = 3×1.185×b' + b'
500 = 4.555 b'
b' = $109.77
That leaves $500$109.77 = $390.23 between the three men. So each one should get $390.23/3 = $130.08.
Michael Shackleford, A.S.A.
  115.  Poissoned win problem
   The police commissioner hired a mathematician to help at a crime scene. At the scene were between 100 and 200 glasses of wine. Exactly one glass was poisoned. The police lab could test any sampling for poison. A group of glasses could be tested simultaneously by mixing a sample from each glass. The police commissioner desires only to minimize the maximum possible tests required to determine which exact glass was poisoned. The mathematician started by asking a detective to select a single glass at random for testing. "Wouldn't that waste a test?", the detective asked. "No, besides I'm in a gambling mood.", the mathematician replied. How many glasses were there?
Answer
Problem 115 Answer
Problem 115 Answer
The answer is 129 glasses.
Michael Shackleford, A.S.A.
Solution
Problem 115 Solution
Problem 115 Solution
If the goal is to minimize the maximum number of tests then that maximum for n glasses is log _{2}n (log to the base 2 of n), rounded up.
If the number of glasses is a power of 2 (32, 64, 128, 256, etc.) then the correct procedure on the first step would be to test exactly half the glasses. Then test half of the glasses in the positive half next. Keep repeating this procedure. Each step will narrow down the field of poisoned glasses by a factor of 2.
If the number number of glasses is not a power of 2 then there are lots of ways you can test them and still achieve the same maximum number of tests. One way would be to find the closest power of 2 less than n, set this many glasses aside, and then test the others. The only n for which this remainder is 1, between 100 and 200, is 129. In the case of 129 the first test would likely be negative. Then proceed with the procedure for testing 128. Between the first test and the log_{2}128 the maximum number of tests is 8.
Credit for this problem belongs to Stephen Barr (Mathematical brain benders; ISBN 048624260
9 published by mamillan NY 1969), previously published in 'Scientific American.'
Michael Shackleford, A.S.A.
  161.  Poker problem
   Suppose you have two fivecard poker hands dealt from separate decks. You are told hand A contains at least one ace. You are told hand B contains the ace of spades. Which hand is more likely to contain at least one more ace?
Answer
Problem 161 Answer
Problem 161 Answer
Hand B.
Michael Shackleford, ASA  December 26, 2000
Solution
Problem 161 Solution
Problem 161 Solution
The following table shows the probability of 0 to 5 aces in a totally random hand.
Ace Probabilities  Random Hand

Aces 
Formula 
Combinations 
Probability 
0  Combin(48,5)  1712304  0.658842 
1  combin(4,1)*combin(48,4) 
778320  0.299474 
2  combin(4,2)*combin(48,3)  103776  0.03993 
3  combin(4,3)*combin(48,2)  4512  0.001736 
4  combin(4,4)*combin(48,1)  48  0.000018 
Total   2598960  1 
Take the sum for 1 to 4 aces we see the probability of at least one ace is 0.341158. The probability of two or more aces is 0.041684.
The probability of there being at least one more ace given there is at least one can be restated per Bayes' theorem as probability(two more aces given at least one ace) = probability (two or more aces)/probability(at least one ace) = 0.041684/ 0.341158 = 0.122185.
For those rusty on Bayes' Theorem is states the probability of A given B equals the probability of A and B divided by the probability of B, or Pr(A given B) = Pr(A and B)/Pr(B).
The next table shows the combinations and probability for each number of other aces given that the ace of spades was removed from the deck.
Ace Probabilities  Ace Removed Hand

Aces 
Formula 
Combinations 
Probability 
0  combin(3,0)*combin(48,4)  194580  0.778631 
1  combin(3,1)*combin(48,3) 
51888  0.207635 
2  combin(3,2)*combin(48,2)  3384  0.013541 
3  combin(3,3)*combin(48,1)  48  0.000192 
Total   249900  1 
This shows the probability of at least one more ace is 0.221369.
However suppose you want to solve it the same way as hand A, using Bayes' Theorem. The probability of at least one additional ace, given the hand contains the ace of spades, could be rewritten as probability(at least two aces given ace of spades is in hand). According to Bayes' theorem this equals Probability(hand contains ace of spades and at least one more ace) / Pr(hand contains ace of spades). We can break up the numerator as Probability(2 aces including ace of spades) + Probability(3 aces including ace of spades) + Probability(4 aces). Using the first table this equals 0.039930*(2/4) + 0.001736*(3/4) + 0.000018 = 0.021285. The probability of the ace of spades is 5/52 = 0.096154. So the probability of at least two aces given the aces of spades is 0.021285/0.096154 = 0.221369.
So the probability of two or more aces given at least one ace is 12.22% and given the ace of spades is 22.14%.
Q: Okay, I believe your numbers but it still doesn't make sense to me. I should think the probabilities would be equal. What difference does the suit make of the one ace you are given?
To make another example suppose you go to Jiffy Lube and they offer two deals for the same price. Deal A is they will check four parts and replace only the first defective one found. Deal B is they will check only one problem and fix it if found. Wouldn't you rather take deal A? Your car came in with the same number of expected bad parts but the probability of finding a problem is greater under deal A, and thus you will leave with a small number of expected defective parts under that plan. Likewise a test for any ace will probably turn up the only ace, while a test for the ace of spades does not check for the other three suits, leaving them more likely to be aces.
Michael Shackleford, ASA  December 26, 2000
  213.  Poker theory game #1
    Players X and Y must each ante $1 into the pot.
 X and Y are each given a random number from a uniform distribution from 0 to 1.
 X may bet $1 or check.
 If X checks Y must check and the higher number will win the pot.
 If X bets, then Y may call or fold. If Y calls then the higher number will win the pot.
What is the optimal strategy for both players? Also, what is the expected win/loss of player X?
Answer Problem 213 Answer
Answer
Player X Strategy:
 0.7 to 1.0: Bet
 0.1 to 0.7: Check
 0.0 to 0.1: Bet (bluff)
Player Y Strategy after a bet:
 0.4 to 1.0: Call
 0.0 to 0.4: Fold
The expected value of X is $0.10.
Michael Shackleford, ASA — Sept. 6, 2012
Solution Problem 213 Solution
Solution
My method of solving this problem comes from The Mathematics of Poker by Bill Chen and Jerrod Ankenman.
First, we have to come up with a structure of the strategy for both players, without knowing yet the specifics. In this case, X should obviously raise above a certain point. Call it x2.
Second, X should also bluff below a certain point, call it x1.
With anything above x1 and below x2, X should check.
Y is the last to act, so there is no opportunity to bluff. He should simply fold below a certain point, and raise above it. Call that point y1.
The crux of their solution is to find indifference points where each player perceives the same expected value between two options. Let’s look at this problem as an example.
Next, we must decide which number will be greater, x1 or y1. It seems obvious to me that X will bluff only with a pretty low probability. Meanwhile, Y fold after a raise fairly often. So, based on common sense, I assume that x1
Now we're ready to get our hands dirty with some math. Let's start by finding an equation for x1. We do this by examining the possible outcomes for all possible values of Y, and how much X will win whether he bluffs or checks with exactly x1 points.
To be more specific,
 If Y has 1 to y1, then X will win 2 (or lose 2) if he bluffs and win 1 if he checks. The gain by bluffing is 1. The probability Y has a hand in this range is 1x2. So, the expected increase in expected value by X bluffing with x1 is (1y1)*1 = y11.
 If Y has y1 to x1, then X will win 1 if he bluffs and win 1 if he checks. The gain by bluffing is 2. The probability Y has a hand in this range is y1x1. So, the expected increase in expected value by X bluffing with x1 is (y1x1)*2 = 2y12x1.
 If Y has x1 to 0, then X will win 1 if he bluffs and win 1 if he checks. In other words, X will win either way. So, the gain by bluffing is 0, and the expected value is also 0.
Here is a table the lays out the expected win by bluffing and checking if X has exactly X1, according to the value of Y1.
x1
x1 
Y 
Bluff 
Check 
Difference 
Expected Value 
1 to y1  2  1  1  y11 
y1 to x1  1  1  2  2y12x1 
x1 to 0  1  1  0  0 
The sum of the expected value column is 2x1+3y11. As stated above, we wish to set x1 to a value where the player is indifferent between bluffing and checking. So, we should set this expected value equation of the gain by bluff equal to zero. In other words, 2x1+3y11 = 0.
x2
Next, do the same thing, solving for x2. I will go right to the expected value table, which shows the expected value of raising, as opposed to checking, according to the value of Y.
x2 
Y 
Bluff 
Check 
Difference 
Expected Value 
1 to x2  2  1  1  x21 
x2 to y1  2  1  1  x2y1 
y1 to 0  1  1  0  0 
Setting the sum of the expected value column equal to 0 we get 2x2y11=0.
y1
Next, we do a table for y1, showing the expected gain by calling, as opposed to checking, for every possible value of X.
x2 
X 
Call 
Check 
Difference 
Expected Value 
1 to x2  2  1  1  x21 
x2 to y1  1  1  0  0 
y1 to x1  1  1  0  0 
x1 to 0  2  1  3  3x1 
Setting the sum of the expected value column equal to 0 we get 3x1+x21=0.
Now we have three equations and three unknowns. Again they are:
2x1+3y11=0
2x2y11=0
3x1+x21=0
With a little matrix algebra we can solve for x1, x2, and y1 as follows:
x1=0.1
x2=0.7
y1=0.4
In other words, X should raise with less than 0.1 or more than 0.7, otherwise check. Assuming X raises, Y should fold with 0.4 or less, otherwise call.
The next question is what is the expected value for X if both players follow this strategy. To answer that question, let's consider how much X wins according to every significant grouping of X and Y.
X Win Table 
X 
Y 
0 to 0.1  0.1 to 0.4  0.4 to 0.7  0.7 to 1.0 
0 to 0.1  1  1  2  2 
0.1 to 0.4  1  0  1  1 
0.4 to 0.7  1  1  0  1 
0.7 to 1  1  1  2  0 
Next, here is the probability of each combination of X and Y.
Proability Table 
X 
Y 
0 to 0.1  0.1 to 0.4  0.4 to 0.7  0.7 to 1.0 
0 to 0.1  0.01  0.03  0.03  0.03 
0.1 to 0.4  0.03  0.09  0.09  0.09 
0.4 to 0.7  0.03  0.09  0.09  0.09 
0.7 to 1  0.03  0.09  0.09  0.09 
Finally, here is the expected value of the how much X will win for each combination, which is equal to the product of the win and the probability.
X Return Table 
X 
Y 
0 to 0.1  0.1 to 0.4  0.4 to 0.7  0.7 to 1.0 
0 to 0.1  0.01  0.03  0.06  0.06 
0.1 to 0.4  0.03  0  0.09  0.09 
0.4 to 0.7  0.03  0.09  0  0.09 
0.7 to 1  0.03  0.09  0.18  0 
The sum of each cell in this table is 0.1. So, assuming the ante is $1 per player, X can expect to win 10 ¢ per hand.
Michael Shackleford, ASA — Sept. 6, 2012
  215.  Poker theory game #2
   Same problem as number 213, except if X checks then Y can bet. Here are the full rules.
 Player X and Y each ante $1.
 Both are given a random number uniformly distributed from 0 to 1. The higher number wins.
 Player X may bet $1 or check.
 If player X checks then Y may check or bet.
 If player X bets then Y may call or fold.
 If player X checks, then Y bets, then X may call or fold.
What is the optimal strategy for both players? What is the expected value for X under mutual optimal strategy?
Answer Problem 215 Answer
My method of solving this problem comes from The Mathematics of Poker by Bill Chen and Jerrod Ankenman.
First, we have to come up with a structure of the strategy for both players, without knowing yet the specifics. Let's define some variables to this strategy as follows.
 x1 = If x has below x1 then he will bluff.
 x2 = If x checks, and then y bets, then x will fold with under x2, otherwise call.
 x3 = If x has above x3 then he will bet.
 y1 = If x checks, then y will bluff with under y1.
 y2 = If x bets, then y will fold with under y2, otherwise call.
 y3 = If x checks, then y will bet with over y3.
Here are the values for these indifference points.
x1 = 0.111111111
x2 = 0.333333333
x3 = 0.666666667
y1 = 0.166666667
y2 = 0.333333333
y3 = 0.500000000
Assuming the ante is $1 per player, x can expect to lose 5.55556 ¢ per hand.
Michael Shackleford — Nov. 1, 2012
Solution Problem 215 Solution
Solution
My method of solving this problem comes from The Mathematics of Poker by Bill Chen and Jerrod Ankenman.
First, we have to come up with a structure of the strategy for both players, without knowing yet the specifics. Let's define some variables to this strategy as follows.
 x1 = If x has below x1 then he will bluff.
 x2 = If x checks, and then y bells, then x will fold with under x2, otherwise call.
 x3 = If x has above x3 then he will bet.
 y1 = If x checks, then y will bluff with under y1.
 y2 = If x bets, then y will fold with under y2, otherwise call.
 y3 = If x checks, then y will bet with over y3.
x1
Player x should be indifferent to checking and bluffing at x1. Let's take a look at how much x would win or lose both ways according to the value of y. The "difference" column shows the benefit to bluffing if x has exactly x1, and y has a value in the specified range. The "expected value" column shows the expected amount y can expect to win by bluffing, which is the product of the "difference" column and the probability that y has a value in the specified range.
x1 
Y 
Bluff 
Check 
Difference 
Expected Value 
1 to y2  2  1  1  y21 
y2 to y1  1  1  2  2y22y1 
y1 to x1  1  1  2  2y12x1 
x1 to 0  1  1  2  2x1 
The sum of the expected value column is 3y21. As stated above, we wish to set x1 to a value where the player is indifferent between bluffing and checking. So, we should set this expected value equation of the gain by bluff equal to zero. In other words, 3y21 = 0.
x2
Next, do the same thing, solving for x2. I will go right to the expected value table, which shows the expected value of calling, as opposed to foldling, according to the value of Y.
x2 
Y 
Call 
Fold 
Difference 
Expected Value 
1 to y3  2  1  1  y31 
y3 to y1  1  1  0  0 
y1 to 0  2  1  3  3y1 
Setting the sum of the expected value column equal to 0 we get 3y1+y31=0.
x3
Next, do the same thing, solving for x3. I will go right to the expected value table, which shows the expected value of raising, as opposed to checking, according to the value of Y.
x3 
Y 
Raise 
Check 
Difference 
Expected Value 
1 to x3  2  2  0  0 
x3 to y3  2  2  0  0 
y3 to y2  2  1  1  y3y2 
y2 to y1  1  1  0  0 
y1 to 0  1  2  1  y1 
Setting the sum of the expected value column equal to 0 we get y1y2+y3=0.
y1
Next, we do a table for y1, showing the expected gain by bluffing, as opposed to checking, for every possible value of X.
y1 
X 
Bluff 
Check 
Difference 
Expected Value 
1 to x3  1  1  0  0 
x3 to x2  2  1  1  x2x3 
x2 to y1  1  1  2  2x22y1 
y1 to x1  1  1  0  0 
x1 to 0  1  1  0  0 
Setting the sum of the expected value column equal to 0 we get 3x2x32y1=0.
y2
Next, we do a table for y2, showing the expected gain by calling, as opposed to folding, for every possible value of X.
y2 
X 
Call 
Fold 
Difference 
Expected Value 
1 to x3  2  1  1  x31 
x1 to 0  2  1  3  3x1 
Setting the sum of the expected value column equal to 0 we get 3x1+x31=0.
y3
Next, we do a table for y3, showing the expected gain by raising, as opposed to checking, for every possible value of X.
y3 
X 
Raise 
Check 
Difference 
Expected Value 
x3 to 1  2  2  0  0 
y3 to x3  2  1  1  y3x3 
x2 to y3  2  1  1  y3x2 
x1 to x2  1  1  0  0 
0 to x1  2  2  0  0 
Setting the sum of the expected value column equal to 0 we get x2x3+2y3=0.
Solving for x1, x2, x3, y1, y2, and y3.
Now we have six equations and six unknowns. Again they are:
3y21 = 0
3y1+y31 = 0
y1y2+y3 = 0
3x2x32y1 = 0
3x1+x31 = 0
x2x3+2y3 = 0
Using matrix algebra, we should solve for the following matrix.
0  0  0  0  3  0  1 
0  0  0  3  0  1  1 
0  0  0  1  1  1  0 
0  3  1  2  0  0  0 
3  0  1  0  0  0  1 
0  1  1  0  0  2  0 
Forgive me if I don't go through a matrix algebra lesson, and jump right to the solution, which is:
x1 = 0.111111111
x2 = 0.333333333
x3 = 0.666666667
y1 = 0.166666667
y2 = 0.333333333
y3 = 0.5
In other words, X should bluff with 0 to 0.111111, check with 0.111111 to 0.666667, and bet with 0.666667 to 1.0.
Assuming x bets or bluffs, then y should fold with 0 to 0.333333 and call with 0.333333 to 1.0.
If x checks, then y should bluff with 0.0 to 0.166667, check with 0.166667 to 0.500000, and bet with 0.500000 to 1.0.
Assuming x checks, and y bets, then x should fold with 0 to 0.333333 and call with 0.333333 to 1.0 (same strategy as y when forced to fold or call).
The next question is what is the expected value for X if both players follow this strategy. To answer that question, let's consider how much X wins according to every significant grouping of X and Y.
X Win Table 
X 
Y 
0 to x1  x1 to y1  y1 to x2  x2 to y3  y3 to x3  x3 to 1 
0 to x1  1  1  1  2  2  2 
x1 to y1  1  1  1  1  1  1 
y1 to x2  1  1  0  1  1  1 
x2 to y3  2  2  1  0  2  2 
y3 to x3  2  2  1  1  0  2 
x3 to 1  1  1  1  2  2  0 
Next, here is the probability of each combination of X and Y.
Proability Table 
X 
Y 
0 to x1  y1 to x1  x2 to y1  y3 to x2  x3 to y3  x3 to 1  Total 
0 to x1  0.012346  0.006173  0.018519  0.018519  0.018519  0.037037  0.111111 
y1 to x1  0.006173  0.003086  0.009259  0.009259  0.009259  0.018519  0.055556 
x2 to y1  0.018519  0.009259  0.027778  0.027778  0.027778  0.055556  0.166667 
y3 to x2  0.018519  0.009259  0.027778  0.027778  0.027778  0.055556  0.166667 
x3 to y3  0.018519  0.009259  0.027778  0.027778  0.027778  0.055556  0.166667 
x3 to 1  0.037037  0.018519  0.055556  0.055556  0.055556  0.111111  0.333333 
Total  0.111111  0.055556  0.166667  0.166667  0.166667  0.333333  1.000000 
Finally, here is the expected value of the how much X will win for each combination, which is equal to the product of the win and the probability.
X Return Table 
X 
Y 
0 to x1  y1 to x1  x2 to y1  y3 to x2  x3 to y3  1 to x3  Total 
0 to x1  0.012346  0.006173  0.018519  0.037037  0.037037  0.074074  0.111111 
y1 to x1  0.006173  0.003086  0.009259  0.009259  0.009259  0.018519  0.055556 
x2 to y1  0.018519  0.009259  0.000000  0.027778  0.027778  0.055556  0.138889 
y3 to x2  0.037037  0.018519  0.027778  0.000000  0.055556  0.111111  0.083333 
x3 to y3  0.037037  0.018519  0.027778  0.027778  0.000000  0.111111  0.000000 
1 to x3  0.037037  0.018519  0.055556  0.111111  0.111111  0.000000  0.333333 
Total  0.098765  0.049383  0.120370  0.064815  0.018519  0.370370  0.055556 
The sum of each cell in this table is 0.055556. So, assuming the ante is $1 per player, X can expect to lose 5.55556 ¢ per hand.
Michael Shackleford, ASA — Nov. 1, 2012
  225.  Poker theory game #3
   Consider a game with the following rules:
 A random number generator provides random numbers between 0 and 1 uniformly distributed.
 Two players each get a separate number. Each player can see his own number only.
 Player 1 may keep his initial number or swap for a new random number.
 Player 2, knowing player 1's action, has the same option to keep his original number or swap for a new one.
 Player with the higher number wins.
Answer the following questions about the game:  At what number is player 1 indifferent to standing and switching?
 Assuming player 1 switches, at what number should player 2 be indifferent to standing and switching?
 Assuming player 1 stands, at what number should player 2 be indifferent to standing and switching?
 Assuming optimal strategy by both players, what is the probability player 1 will win?
Answer Problem 225 Answer
 Player 1 is indifferent between standing and switching at (1/6)×(81+3×921^{1/2})^{1/3}+(813×21^{1/2})^{1/3} = 0.567364*.
 If player 1 hits, player 2 is indifferent between standing and switching at 0.5.
 If player 1 stands, player 2 is indifferent between standing and switching at 0.660951**.
 Assuming optimal strategy by bother players, the probability player 1 wins is 0.494333***.
* This is the solution to the equation 4x ^{3} + 4x  3 = 0.
** This is the solution to the equation 3/(8x).
*** This is the solution to the equation (1/12) + (1/96)×(2763×(921) ^{1/3}  19657) ^{1/3}  (2763×(921) ^{1/3} + 19657)) ^{1/3}
Michael Shackleford, Nov. 19, 2017
Solution Problem 225 Answer
There is some Nash equilibrium to be found between the two players. Let:
 Player 1 = player to act first.
 Player 2 = player to act second.
 x = Point at which player 1 is indifferent to standing and switching.
 y = Point at which player 2 is indifferent to standing and switching, assuming player 1 stood.
It should be obvious that y>x. If player 1 stood, player 2 knows that player 1 has a strong card, so he needs to be more aggressive to beat it. This is where positional advantage comes in. If this isn't obvious to you, then you probably should be attempting an easier problem.
First, let's calculate the probability of winning by standing.
 It is easy to see that the probability a given number x beats a random number is x.
 If player 2 has y or less, then the probability of player 1 winning is x.
 If player 2 has y or more, then the probability of player 1 winning is 0, since y>x.
 Taking the dot prodoct of the possible values for player 2 and the probability of player 1 winning by standing, the probability of winning by standing is xy.
Second, let's calculate the probability of winning by switching.
 It is easy to see that if player 1 switches, then player 2 will be indifferent to standing and switching at 0.5.
 If player 2 has 0.5 or less, then the probability of player 1 winning is 0.5, since both will have switched to a random card.
 If player 2 has 0.5 or more, then the probability of player 1 winning is 0.25. This can be found using geometry or integral calculus by looking at proption of the area player 1 wins over the area bounded by values of 0 to 1 for player 1 and 0.5 to 1 for player 2. Player 1 wins over 1/4 of that area.
 Taking the dot prodoct of the possible values for player 2 and the probability of player 1 winning by switching, the probability of winning by switching is 3/8.
Setting the expected value of standing and switching equal, we get xy = 3/8.
Next, solve for y using the same method. Forgive me if I don't through the details this time. Again, using simple geometry or integral calculus, you'll find the probability of player 2 winning by standing at y is (yx)/(1x). By the same method, the probability of player 2 winning by switching at y is [(1x)^{2}/2]/(1x).
Setting the two equations equal and doing some algebra we get x^{2}  2y + 1 = 0.
We now have two equations and two unknowns. With some simple algebra to solve for x we get to x^{3} + x 0.75 = 0. Using one of many cubic equaton solvers on the Internet, we find 0.567364. So, that is the value x is indifferent between standing and switching.
Putting that into xy=3/8, we solve for y as 0.660951.
Finally, find the probability of all four combinations of x and y standing and swithcing and the probability x wins for each one and take the dot product. The following table gives more details. The bottom right cell shows the probability player 1 wins is 0.494333.
Player 1 
Player 2 
Probability 
Player 1 Action 
Player 2 Action 
Probability Player 1 Wins 
Expected Value Player 1 Wins 
0 to x  0 to 0.5  0.283682  switch  switch  0.500000  0.141841 
0 to x  0.5 to 1  0.283682  switch  stay  0.250000  0.070921 
x to 1  0 to y  0.285951  stay  switch  0.783682  0.224095 
x to 1  y to 1  0.146685  stay  stay  0.391841  0.057477 
Total   1.000000     0.494333 
Final answers:
 Player 1 is indifferent between standing and switching at 0.567364.
 If player 1 hits, player 2 is indifferent between standing and switching at 0.5.
 If player 1 stands, player 2 is indifferent between standing and switching at 0.660951.
 Assuming optimal strategy by bother players, the probability player 1 wins is 0.494333.
Michael Shackleford, Nov. 19, 2017
  226.  Poker theory game #4
   Consider a game with the following rules:
 A random number generator provides random numbers between 0 and 1 uniformly distributed.
 Both players bet $1 to start.
 Two players each get a separate number. Each player can see his own number only.
 Player 1 may keep his initial number or swap for a new random number. If player 1 swaps, both players must bet an additional $1.
 Player 2, NOT knowing player 1's action, has the same option to keep his original number or swap for a new one.
 Player with the higher number wins. The winner keeps all money bet
Answer the following questions about the game:  At what number is player 1 indifferent to standing and switching?
 At what number should player 2 be indifferent to standing and switching?
 Assuming optimal strategy by both players, how much can player 1 expect to win or lose?
Answer Problem 226 Answer
 Player 1 is indifferent between standing and switching at 0.456260889.
 Player 2 is indifferent between standing and switching at 0.554423955.
 Assuming optimal strategy by bother players, the expected loss of player is 0.13162138 dollars.
My thanks to Joe Shipman for his help with this problem.
Michael Shackleford, Nov. 22, 2017
Solution Problem 226 Solution
There is some Nash equilibrium to be found between the two players. Let:
 Player 1 = player to act first.
 Player 2 = player to act second.
 x = Point at which player 1 is indifferent to standing and switching.
 y = Point at which player 2 is indifferent to standing and switching.
To begin, we must accept that y>x. Player 1 is going to be conservative about switching because he has to double his bet if he does. So, with a number around 0.5 he is better off to cut his losses and stick with it rather than double his bet and go against player 2 with a random card. Meanwhile, player 2 can be more aggressive because he can switch for free. With a number around 0.5, player 2 is apt to switch because he knows player 1 will have a number with an average over 0.5, because player 1 had the opportunity to switch out of the a bad number.
First, let's find x. We do so by finding the value of x where the expected value of standing and switching are equal.
At x, the expected value of standing is y*(2x1)(1x). The expected value of switching at x is 2*y*(1y).
Setting the two values equal to each other we get :
(1) 2*y^2  2xy  2y + 1 = 0.
Second, let's find y. We do so by finding the value of y where the expected value of standing and switching are equal.
At y, the expected value of standing is x*(2*(2y 1)) + y  x  (1y). The expected value of switching at x is 2x*(1  x).
Setting the two values equal to each other we get:
(2) x^2  2y  4xy +2x + 1 = 0.
We now have two equations and two unknowns. We can rewrite equation (1) as:
(3) x = y  1 + 1/(2y).
Substituting equation (3) into (2) gives us:
(4) (2y^2  2y + 1)^2  2y*(4y^2)  8y^2*(2y^2  2y + 1) + 4y*(2y^2  2y + 1) + 4y^2 = 0
After a lot of algebra, that reduces to:
(5) 12^4  8y^3  4y^2  1 = 0
You can use a fourorder calculator to solve for y as 0.554423955.
You can plug that into equation (3) to get x = 0.456260889.
Finally, find the probability of all four combinations of x and y standing and swithcing and the probability x wins for each one and take the dot product. The following table gives more details. The bottom right cell shows player 1 can expect to lose 0.131621 dollars.
Player 1 
Player 2 
Probability 
Player 1 Action 
Player 2 Action 
Probability Player 1 Wins 
Expected Return Player 1 
Expected Value Player 1 
Player 1  Player 2  Probability  Player 1 Action  Player 2 Action  Probability Player 1 Wins  ER player 1  EV player 1 
0 to x  0 to y  0.252962  switch  switch  0.500000  0.000000  0.000000 
0 to x  y to 1  0.203299  switch  stay  0.222788  1.108848  0.225428 
x to 1  0 to y  0.301462  stay  switch  0.728130  0.456261  0.137545 
x to 1  y to 1  0.242277  stay  stay  0.409733  0.180533  0.043739 
Total   1.000000      0.131621 
Final answers:
 Player 1 is indifferent between standing and switching at 0.456260889.
 Player 2 is indifferent between standing and switching at 0.554423955.
 Assuming optimal strategy by bother players, the expected loss of player is 0.13162138 dollars.
My thanks to Joe Shipman for his help with this problem.
Michael Shackleford, Nov. 22, 2017
  94.  Pothole problem
   Any given length of highway is equally likely to have as many potholes as any other length of equal size. The average number of potholes per mile of highway is 3. What is the probability that 2 miles of highway have 3 or fewer potholes?
Answer
Problem 94 Answer
Problem 94 Answer
The answer is 61*e ^{6} =~ 0.151204 .
Michael Shackleford, ASA, August 17, 1999
Solution
Problem 94 Solution
Problem 94 Solution
The number of potholes of miles of highway has a Poisson distristribution
with a mean of 3. The number of potholes in two miles of highway would have
a Poisson distristribution with a mean of 6. The probability of x potholes
in two miles of highway is e ^{6}*6 ^{x}/x!.
The probability of 0 potholes is e^{6}*6^{0}/0! = e^{6}.
The probability of 1 pothole is e^{6}*6^{1}/1! = 6*e^{6}.
The probability of 2 potholes is e^{6}*6^{2}/2! = 18*e^{6}.
The probability of 3 potholes is e^{6}*6^{3}/3! = 36*e^{6}.
Take the sum the answer is 61*e^{6} =~ 0.151204 .
This problem was taken from the May 1982 Actuary Exam 110, problem 20.
Michael Shackleford, ASA, August 17, 1999
  62.  Probabilities in poker problem
   Given an ordinary deck of cards, what is the probability of drawing each kind of hand in poker? No discards, just the first five cards. Remember, aces can be high or low.
Answer
Problem 62 Answer  Probabilities in Poker
Problem 62 Answer  Probabilities in Poker
Below are the probabilities of drawing
the following hands, without discarding:
Approximate
Hand Exact Probability Probability
  
Royal Flush: 4 in 2,598,960 1 in 649,740
Straight Flush: 36 in 2,598,960 1 in 72,193
Four of a Kind: 624 in 2,598,960 1 in 4,165
Full House: 3,744 in 2,598,960 1 in 694
Flush: 5,108 in 2,598,960 1 in 509
Straight: 10,200 in 2,598,960 1 in 255
Three of a Kind: 54,912 in 2,598,960 1 in 47
Two Pair: 123,552 in 2,598,960 1 in 21
One Pair: 1,098,240 in 2,598,960 1 in 2.4
Nothing: 1,302,540 in 2,598,960 1 in 2.0
Solution
Problem 62 Solution  Probabilities in Poker
Problem 62 Solution  Probabilities in Poker
Probabilities of drawing
Below are the probabilities of drawing
the following hands, without discarding:
Approximate
Hand Exact Probability Probability
  
Royal Flush: 4 in 2,598,960 1 in 649,740
Straight Flush: 36 in 2,598,960 1 in 72,193
Four of a Kind: 624 in 2,598,960 1 in 4,165
Full House: 3,744 in 2,598,960 1 in 694
Flush: 5,108 in 2,598,960 1 in 509
Straight: 10,200 in 2,598,960 1 in 255
Three of a Kind: 54,912 in 2,598,960 1 in 47
Two Pair: 123,552 in 2,598,960 1 in 21
One Pair: 1,098,240 in 2,598,960 1 in 2.4
Nothing: 1,302,540 in 2,598,960 1 in 2.0
Derivations
If you are not familiar with the factorial function, the factorial
of a number is the product of every integer from one to itself. For example
five factorial, denoted 5!, is 1*2*3*4*5 = 120.
If you had x cards and chose y of them, the number of
unique sets you could create would be x!/(y!*(xy)!). For the
purposes of this document I shall notate this is (x:y).
The number of ways to arrange 5 cards out of 52 is (52:5) = 2,598,960.
The odds of drawing any given hand are the number of ways it can
be arranged divided by the total number of ways to arrange five cards above.
Below are the number of ways to arrange each hand.
Royal Flush
The number of different royal flushes are four (one for each suit).
Straight Flush
The highest card in a straight flush can be 5,6,7,8,9,10,Jack,Queen, or King.
Thus there are 9 possible high cards, and 4 possible suits, creating 9 * 4 = 36 different
possible straight flushes.
Four of a Kind
There are 13 different possible ranks of the 4 of a kind. The fifth card could
be anything of the remaining 48. Thus there are 13 * 48 = 624 different four of
a kinds.
Full House
There are 13 different possible ranks for the three of a kind, and 12 left for the two
of a kind. There are 4 ways to arrange three cards of one rank (4 different cards to
leave out), and (4:2) = 6 ways to arrange two cards of one rank. Thus there are
13 * 12 * 4 * 6 = 3,744 ways to create a full house.
Flush
There are 4 suits to choose from and (13:5) = 1,287 ways to arrange five cards in the same
suit. Then subtract the royal and straight flushes to avoid double counting. The
total number of flushes is 4 * 1,287  4  36 = 5,108.
Straight
The highest card in a straight flush can be 5,6,7,8,9,10,Jack,Queen,King, or Ace.
Thus there are 10 possible high cards. Each card may be of four different suits.
Then subtract the royal and straight flushes to avoid double counting. Thus the
number of ways to arrange a straight is 10 * 4^{5}  4  36 = 10,200.
Three of a Kind
There are 13 ranks to choose from for the three of a kind and 4 ways to arrange
3 cards among the four to choose from. There are (12:2) = 66 ways to arrange the other
two ranks to choose from for the other two cards. In each of the two ranks there are four
cards to choose from. Thus the number of ways to arrange a three of a kind
is 13 * 4 * 66 * 4^{2} = 54,912.
Two Pair
There are combin(13,2)=78 ways to choose 2 ranks out of 13 for the two pair. Then there are 11 ranks left for the singleton. For each pair there are combin(4,2)=6 ways to choose 2 suits out of 4. For the singleton there are 4 possible suits to chose from. So the total two pair combinations are 78*11*6*6*4 = 123,552.
One Pair
There are 13 ranks to choose from for the pair and (4:2) = 6 ways to arrange the two
cards in the pair. There are (12:3) = 220 ways to arrange the other three ranks
of the singletons, and four cards to choose from in each rank. Thus there are
13 * 6 * 220 * 4^{3} = 1,098,240 ways to arrange a pair.
Nothing
There must be five different ranks represented, of which there are (13:5) = 1,287
possible combinations. Each rank has four cards to choose from. Finally subtract
the number of straights, flushes, straight flushes, and royal flushes, to
avoid double counting. Thus the
number of ways to arrange nothing is 1,287 * 4^{5}  4  36  5,108  10,200
= 1,302,540.
  77.  Probability of gender problem
   Assume that a pregnant woman's probability of giving birth to a girl is p, where p is determined at the mother's birth according to a uniform distribution from 0.4 to 0.6 . If the woman's first child is a girl what is the probability her next child will also be a girl?
Answer
Problem 77 Answer
Problem 77 Answer
The answer is 38/75 =~ 50.7%
Michael Shackleford, A.S.A.
Solution
Problem 77 solution
Problem 77 Solution
The probability of event A happening given that event B already
happened is the probability of A and B happening divided by the
probability that B happened. This can be pressed as Pr(AB)=Pr(A and B)/Pr(B).
In this case Pr(A and B) = 5*(The integral from 0.4 to 0.6 of p^{2}) =
5*(p^{3}/3 from 0.4 to 0.6) =
(5/3)*(.216  .064) =
19/75 =~ 0.2533333
Pr(B) = 5 * The integral from 0.4 to 0.6 of p = 0.5 .
Thus the answer is 38/75 =~ .5066667
Michael Shackleford, A.S.A.
  1.  Probability of n points in a semicircle problem
   Given n points drawn randomly on the circumference of a circle, what is the probability they will all be within any common semicircle?
Answer
Problem 1 Answer
Problem 1 Answer
For n points the answer is n/(2 ^{n1}).
Michael Shackleford, A.S.A.
Solution
Problem 1 Solution
Problem 1 Solution
I have had a lot of mail about this problem over the years. Some people question the answer, and some admit the answer is right, but that the solution is wrong. Here is my original solution:
For n points choose any given point and evaluate the
probability that the other n1 lie within a semicircle
going clockwise. This probability is (1/2)^{n1}.
Given that there are n points to start with the overall
probability is n/2^{n1}. This may seem like an
abuse of taking the sum of probabilities, but in this case
only zero or one of the events may be true, which eliminates
the problem of joint probabilities.
To prove at least the answer is correct, I did ten million trials for various numbers
of points on the circle. I assumed the circle to have 32,767
units or degrees. The first point was defined to be at a halfway
point, or 16,383. Then a maximum and minimum were taken for the rest of
the points. If the difference between the maximum and minimum were
less than 16,383 then the trial was considered to
be a success. Note that 16,383 is half of 32,767, rounded down.
Because of the rounding we should expect a slightly lower number
of successes. Here are the results:
Problem 1 Solution

Number of Points 
Actual Successes 
Expected Successes 
3  7,499,013  7,500,000 
4  4,999,960  5,000,000 
5  3,123,456  3,125,000 
6  1,873,092  1,875,000 
10  194,739  195,312.5 
20  373  381.4697 
Several people have written, saying my answer is right, but the solution is wrong. Here is what David Beim of Columbia University wrote to me about it:
There is no doubt that your answer is right, but your reasoning is unsettling because it confounds ex ante with ex post information. We cannot construct the semicircles until the n points have been selected, and then it is too late to determine ex ante probabilities. This kind of mixing can lead to all kinds of paradoxes in probability, as I am sure you know better than I. For one simple example, select a point on a line. Ex ante, the probability of selecting that point is zero; ex post, the probability must have been greater than zero because it happened.
I attach my own solution (PDF 33K) to the problem, in case it is of interest. Incidentally, I would give this problem four stars rather than three took me days to get it clear!
Per his comment, I did add a star to the difficulty level, it was 3 starts until November, 2008.
This problem has been published in book form in "Contests in Higher Mathematics" by Gabor Szekely, which is a collection of math problems given to the best mathematicians of Poland over the last several decades. The problem (1963 problem 10) is stated as, "Select n points on a circle independently with uniform distribution. Let P_{n} be the probability that the center of the circle is in the interior of the convex hull of these n points. Calculate the probabilities P_{3} and P_{4}. I'm told the answer agrees with mine.
Michael Shackleford, A.S.A.
  100.  Pythagorean problem
   This problem has been removed.
Answer
Problem 100 Answer
Problem 100 Answer
Given any right triangle you can create a figure like the one above using four
of them. This is because the sum of the three angles
always is 180 degrees. The area of the outside square is c^{2}.
We can also find the area by summing the area of the individual
parts: 4*(a*b/2)+(ab)^{2} = 2ab + a^{2} 2ab + b^{2}
= a^{2} + b^{2}. Thus a^{2} + b^{2} =
c^{2}.
I didn't assign a diffulty level because I saw Carl Sagan explain this
proof on television so I never had a chance to solve it myself.
Michael Shackleford, A.S.A.
  192.  Rock/Paper problem
   In a game, players A and B must each secretly pick Rock or Paper. The matrix below shows how much player B must pay player A, according to each combination of choices. Player A wishes to maximize his expected win, and player B wishes to minimize his expected loss. What is the optimal strategy for each player? Problem 192  Player A  Player B  Rock  Paper  Rock  7  3  Paper  2  11 
Answer
Problem 192 Answer
Problem 192 Answer
Problem 192 
Player A 
Player B 
Rock 
Paper 
Rock 
7 
3 
Paper 
2 
11 
Player A should pick Rock with probability 9/13, and paper with probability 4/13.
Player B should pick Rock with probability 8/13, and paper with probability 5/13.
For help with this problem I recommend 'The Complete Strategyst' by J.D. Williams.
Michael Shackleford, A.S.A.
Solution
Problem 192 Solution
Problem 192 Solution
Assume the matrix looks like this.
Problem 192 
Player A 
Player B 
Rock 
Paper 
Rock 
a 
b 
Paper 
c 
d 
First, verify neither player is always better off with one strategy. If a>c and b>d then player A will always choose rock. Likewise, if a>b and c>d, the player B will always choose paper. Assuming there is no obvious pick for either player then it becomes a random strategy.
Player A should pick rock with a probability propotional to abs(cd), and paper with probability proportional to abs(ab).
Player B should pick rock with a probability propotional to abs(bd), and paper with probability proportional to abs(ac).
Here, again is the matrix for this problem.
Problem 192 
Player A 
Player B 
Rock 
Paper 
Rock 
7 
3 
Paper 
2 
11 
Player A should pick rock with probability abs(211)/(abs(211)+abs(73)) = 9/(9+4) = 9/13.
Player A should pick paper with probability abs(73)/(abs(211)+abs(73)) = 9/(9+4) = 4/13.
Player B should pick rock with probability abs(311)/(abs(311)+abs(72)) = 8/(8+5) = 8/13.
Player B should pick paper with probability abs(72)/(abs(311)+abs(72)) = 5/(8+5) = 5/13.
Player A's average win will be (9/13)×(8/13)×7 + (9/13)×(5/13)×3 +
(4/13)×(8/13)×2 + (4/13)×(5/13)×11 =
2.982248521 + 0.798816568 + 0.378698225 + 1.301775148 = 5.461538462
For help with this problem I recommend 'The Complete Strategyst' by J.D. Williams.
Michael Shackleford, A.S.A.
  193.  Rock/Paper/Scissors problem
   In a game, players A and B must each secretly pick Rock, Paper, or Scissors. The matrix below shows how much player B must pay player A, according to each combination of choices. Player A wishes to maximize his expected win, and player B wishes to minimize his expected loss. What is the optimal strategy for each player? Problem 193  Player A  Player B  Rock  Paper  Scissors  Rock  6  0  6  Paper  8  2  0  Scissors  4  6  5 
Answer
Problem 193 Answer
Problem 193 Answer
Here, again is the matrix for this problem.
Problem 193 
Player A 
Player B 
Rock 
Paper 
Scissors 
Rock 
6 
0 
6 
Paper 
8 
2 
0 
Scissors 
4 
6 
5 
Player A should pick rock with probability proportional to 6.
Player A should pick paper with probability proportional to 6.
Player A should pick scissors with probability proportional to 48.
Player B should pick rock with probability proportional to 38.
Player B should pick paper with probability proportional to 14.
Player B should pick scissors with probability proportional to 8.
For help with this problem I recommend 'The Complete Strategyst' by J.D. Williams.
Michael Shackleford, A.S.A.
Solution
Problem 193 Solution
Problem 193 Solution
Assume the matrix looks like this.
Problem 193 
Player A 
Player B 
Rock 
Paper 
Scissors 
Rock 
a 
b 
c 
Paper 
d 
e 
f 
Scissors 
g 
h 
i 
First, verify neither player is always better off with one or two strategies. If there is an obviously inferior pick for either player then the problem can be reduced to one such as problem 192. Assuming there is no obvious pick for either player then it becomes a random strategy.
Player A should pick rock with a probability propotional to abs[(de)(hi)  (ef)(gh)].
Player A should pick paper with a probability propotional to abs[(ab)(hi)  (bc)(gh)].
Player A should pick scissors with a probability propotional to abs[(ab)(ef)  (bc)(de)].
Player B should pick rock with a probability propotional to abs[(be)(fi)  (eh)(cf)].
Player B should pick paper with a probability propotional to abs[(ad)(fi)  (dg)(cf)].
Player B should pick scissors with a probability propotional to abs[(ad)(eh)  (dg)(be)].
Here, again is the matrix for this problem.
Problem 193 
Player A 
Player B 
Rock 
Paper 
Scissors 
Rock 
6 
0 
6 
Paper 
8 
2 
0 
Scissors 
4 
6 
5 
Player A should pick rock with probability proportional to abs[(10*1)  (2)*(2)] = abs[104] = 6.
Player A should pick paper with probability proportional to abs[(6*1)  (6)*(2)] = abs[612] = 6.
Player A should pick scissors with probability proportional to abs[(6*2)  (6)*(10)] = abs[12+60] = 48.
Player B should pick rock with probability proportional to abs[(2*5)  (8)*(6)] = abs[10+48] = 38.
Player B should pick paper with probability proportional to abs[(2*5)  (4)*(6)] = abs[1024] = 14.
Player B should pick scissors with probability proportional to abs[(2*8)  (4)*(2)] = abs[168] = 8.
Player A's average win will be (6/60)×(38/60)×6 + (6/60)×(14/60)×0 + (6/60)×(8/60)×6 +
(6/60)×(38/60)×8 + (6/60)×(14/60)×2 + (6/60)×(8/60)×0 +
(48/60)×(38/60)×4 + (48/60)×(14/60)×6 + (48/60)×(8/60)×5 =
0.38 + 0+ 0.08+0.506666667 0.046666667+ 0+2.026666667 +1.12 +0.533333333 = 4.6
The solution for the general case for an n × n matrix:
 Call the original matrix m
 For the row weights, make submatrix (r) of row differences, where r_{i,j} = m_{i+1,j}  m_{i,j}.
 The weighting for row i is abs(mdeterm(r')), where r' is the submatrix s, omitting row i. Mdeterm is the Excel function for the determinant.
 For the column weights, make submatrix (c) of row differences, where c_{i,j} = m_{i,j+1}  m_{i,j}.
 The weighting for column i is abs(mdeterm(c')), where c' is the submatrix c, omitting column i.
Consider the matrix:
1 7 0 3
0 0 3 5
1 2 4 1
6 0 2 0
The weighting for row 1 is the absolute value of the determinant of:
0 3 2
1 2 3
6 2 2
=88
The weighting for row 2 is the absolute value of the determinant of:
6 7 3
1 2 3
6 2 2
=86
Here is the full solution:
1 7 0 3 88
0 0 3 5 86
1 2 4 1 78
6 0 2 0 101
89 62 119 83
For help with this problem I recommend 'The Complete Strategyst' by J.D. Williams.
Michael Shackleford, A.S.A.
  233.  Rook in the corner
   A rook is placed at the corner of a chess board. Draw a path that goes through every square exactly once and end at the opposite corner, or prove it is impossible.
Answer
Problem 233 Answer
It is impossible. See the solution for why.
Michael Shackleford
Solution
Problem 233 Solution
Since the rook can go vertical and horizontal only, it must alternate going through black and white squares. There are 8*8=64 total squares on the chess board. After an odd number of moves, it must be on an opposite color square as that it started on. After an even number, the same color. There are 64 squares on the chess board, meaning 63 moves movements from one square to the other. Since 63 is odd, it must end up on a square of the opposite color. Opposite corners will be the same color, thus it is impossible.
Michael Shackleford
  116.  Roulette problem
   A player sits down at a roulette table with $20. He bets $1 at a time on either red or black. Either bet pays even money and has a probability of 9/19 of winning. What is the probability that the player wins $10 before losing all of his $20?
Answer
Problem 116 Answer
Problem 116 Answer
The answer is ((10/9) ^{20}  1) / ((10/9) ^{30}  1) =~ .3198
Michael Shackleford, A.S.A.
Solution
Problem 116 Solution
Problem 116 Solution
Let a _{i} denote the probability if the player has $i he will reach $30 before
losing everything. Let p the probability of winning any given bet = 9/19.
a_{0} = 0
a_{1} = 9/19*a_{2}
a_{2} = 9/19*a_{3} + 10/19*a_{1}
a_{3} = 9/19*a_{4} + 10/19*a_{2}
.
.
.
a_{27} = 9/19*a_{28} + 10/19*a_{26}
a_{28} = 9/19*a_{29} + 10/19*a_{27}
a_{29} = 9/19*a_{30} + 10/19*a_{28}
a_{30} = 1
Divide the left side into two parts:
9/19*a_{1} + 10/19*a_{1} = 9/19*a_{2}
9/19*a_{2} + 10/19*a_{2} = 9/19*a_{3} + 10/19*a_{1}
9/19*a_{3} + 10/19*a_{3} = 9/19*a_{4} + 10/19*a_{2}
.
.
.
9/19*a_{27} + 10/19*a_{27} = 9/19*a_{28} + 10/19*a_{26}
9/19*a_{28} + 10/19*a_{28} = 9/19*a_{29} + 10/19*a_{27}
9/19*a_{29} + 10/19*a_{29} = 9/19*a_{30} + 10/19*a_{28}
Rearange with 10/19 terms on the left side and 9/19 terms on the right:
10/19*(a_{1}) = 9/19*(a_{2}  a_{1})
10/19*(a_{2}  a_{1}) = 9/19*(a_{3}  a_{2})
10/19*(a_{3}  a_{2}) = 9/19*(a_{4}  a_{3})
.
.
.
10/19*(a_{27}  a_{26}) = 9/19*(a_{28}  a_{27})
10/19*(a_{28}  a_{27}) = 9/19*(a_{29}  a_{28})
Next multiply both sides by 19/9:
10/9*(a_{1}) = (a_{2}  a_{1})
10/9*(a_{2}  a_{1}) = (a_{3}  a_{2})
10/9*(a_{3}  a_{2}) = (a_{4}  a_{3})
.
.
.
10/9*(a_{27}  a_{26}) = (a_{28}  a_{27})
10/9*(a_{28}  a_{27}) = (a_{29}  a_{28})
Next telescope sums:
(a_{2}  a_{1}) = 10/9*(a_{1})
(a_{3}  a_{2}) = (10/9)^{2}*(a_{1})
(a_{4}  a_{3}) = (10/9)^{3}*(a_{1})
.
.
.
(a_{29}  a_{28}) = (10/9)^{28}*(a_{1})
(a_{30}  a_{29}) = (10/9)^{29}*(a_{1})
Next add the above equations:
(a_{30}  a_{1}) = a_{1} * ((10/9) + (10/9)^{2} + (10/9)^{3}
+ ... + (10/9)^{29})
1 = a_{1} * (1 + (10/9) + (10/9)^{2} + (10/9)^{3}
+ ... + (10/9)^{29})
a_{1} = 1 / (1 + (10/9) + (10/9)^{2} + (10/9)^{3}
+ ... + (10/9)^{29})
a_{1} = (10/9  1) / ((10/9)^{30}  1)
Now that we know a_{1} we can find a_{20}:
(a_{2}  a_{1}) = 10/9*(a_{1})
(a_{3}  a_{2}) = (10/9)^{2}*(a_{1})
(a_{4}  a_{3}) = (10/9)^{3}*(a_{1})
.
.
.
(a_{19}  a_{18}) = (10/9)^{18}*(a_{1})
(a_{20}  a_{19}) = (10/9)^{19}*(a_{1})
Add the above equations together:
(a_{20}  a_{1}) = a_{1} * ((10/9) + (10/9)^{2} + (10/9)^{3}
+ ... + (10/9)^{19})
a_{20} = a_{1} * ((10/9)^{20}  1)) / (10/9  1))
a_{20} = [ (10/9  1) / ((10/9)^{30}  1) ] *
[ ((10/9)^{20}  1) /
(10/9  1) ]
a_{20} = ((10/9)^{20}  1) / ((10/9)^{30}  1) =~ .3198
Michael Shackleford, A.S.A.
  191.  Roulette problem
   In doublezero roulette, what is the probability that any number will not have hit by the 200th spin?
Answer
Problem 191 Answer
Problem 191 Answer
38×(37/38) ^{200}  combin(38,2)×(36/38) ^{200} + combin(38,3)×(35/38)^{200} = 16.9862%.
Thanks to J.F.W. from Marshall for this problem.
Michael Shackleford, A.S.A.
Solution
Problem 191 Solution
Problem 191 Solution
The probability that any given number will not have hit is (37/38)^{200} = 0.48%.
With 38 numbers, we could incorrectly say that the probability that any one of them would not be hit is 38 × (37/38)^{200} = 18.34%.
The reason this is incorrect is it double counts two numbers not being hit. So we need to subtract those probabilities out. There are combin(38,2) = 703 sets of 2 numbers out of 38. The probability of not hitting any two given numbers is (36/38)^{200} = 0.000020127. We need to subtract the probability of avoiding both numbers. So we are at:
38×(37/38) ^{200}  combin(38,2)×(36/38) ^{200} = 16.9255%.
However, now we have canceled out the probability of three numbers not hitting. For any given group of three numbers we triple counted the probability of any single number not being hit. We then triple subtracted for each way to choose two numbers out of the three, leaving with zero for the probability that all three numbers were not hit. There are combin(38,3)=8,436 such groups. Adding them back in we are now at:
38×(37/38) ^{200}  combin(38,2)×(36/38) ^{200} + combin(38,3)×(35/38)^{200} = 16.9862%.
Yet, now we have overcounted the probability of four numbers not hitting. For each of the combin(38,4)=73,815 groups of four numbers, each was originally quadruple counted. Then we subtracted each of the combin(4,2)=6 groups of 2 out of the 4. Then we added back in the 4 groups of 3 out of the 4. So, for each union of four numbers, it was counted 4 – 6 + 4 = 2 times. To adjust for the double counting we must subtract for each group. Subtracting them out we are now at:
38×(37/38) ^{200} 
combin(38,2)×(36/38) ^{200} +
combin(38,3)×(35/38)^{200} 
combin(38,4)×(34/38)^{200}
= 16.9845%.
Continuing in the process we would keep alternating adding and subtracting, all the way until missing 37 numbers. Thus the probability of at least one number never being hit is:
Sum i=1 to 37 [(1)^{(i+1)} × combin(38,i) × ((38i)/38)^{38}] = 16.9845715651245%
Here are the results of a random simulation of 126,900,000 such 200spin experiments.
Numbers Hit in 200 Roulette Spins 
Numbers Hit 
Observations 
Ratio 
31 or Less  0  0 
32  1  0.00000001 
33  33  0.00000026 
34  1812  0.00001428 
35  68845  0.00054251 
36  1577029  0.01242734 
37  19904109  0.15684877 
38  105348171  0.83016683 
Total  126900000  1 
The ratio of times at least one number was not hit was 0.169833.
Thanks to J.F.W. from Marshall for this problem.
Michael Shackleford, A.S.A.
  139.  Roulette promotion problem
   A casino offers to refund half your net losses if you bet only on red or black in roulette. The roulette wheel is the standard American variety that has both a zero and double zero and pays even money. You must bet the same amount every time. How many times should you bet to maximize your expected return?
Answer
Problem 139 Answer
Problem 139 Answer
The maximum expected gain occurs with 7 bets, with an expected gain of 0.2751 bets.
Thanks to
Extra Stuff: Gambling Rambling by Peter Griffin for this problem. See chapter 7.
Michael Shackleford, ASA, August 19 1999
Solution
Problem 139 Solution
Problem 139 Solution
Denote the probability of winning any given get to be p=18/38.
Denote the number of bets made to be n.
Denote the number of winning bets to be w.
The probability of a w wins out of n bets is (n!/(w!*(nw)!)) * p^{w} * (1p)^{nw}
The expected return of n bets is $\sigma $ (for w=0 to n) (n!/(w!*(nw)!)) * p^{w} * (1p)^{nw} * if(w>n/2,2wn,wn/2)
The table below shows the expected gain given the number of wagers made.
Problem 139 Expected Gain 
Number of Bets 
Expected Gain 
1  0.210526 
2  0.171745 
3  0.257618 
4  0.219182 
5  0.273977 
6  0.235800 
7  0.275100 
8  0.237137 
9  0.266779 
10  0.229003 
11  0.251903 
12  0.214296 
13  0.232154 
14  0.194700 
15  0.208607 
16  0.171295 
17  0.182001 
18  0.144822 
19  0.152868 
20  0.115815 
21  0.121605 
22  0.084671 
23  0.088519 
24  0.051698 
25  0.053852 
26  0.017138 
27  0.017797 
28  0.018813 
29  0.019482 
30  0.055994 
From the table it can be seen that the maximum profit occurs when the
number of bets is 7.
Thanks to
Extra Stuff: Gambling Rambling by Peter Griffin for this problem. See chapter 7.
Michael Shackleford, ASA, August 19 1999
  176.  Roulette promotion problem
   You are given $250 in promotional roulette chips. You may play roulette as long as you like but may only keep your winnings above the initial $250. You also may only play the even money bets like red and black. You also may only bet $10 at a time. With two zeros on the wheel the probability of winning any given bet is 18/38. What should your strategy be to maximize your expected winnings and what is the most you can expect to win?
Answer
Problem 177 Answer
Problem 177 Answer
Your strategy should be to play until you reach $340 or bust.
Michael Shackleford, A.S.A.
Solution
Problem 176 Solution
Problem 176 Solution
From problem 116 we know that with the probability of success of reaching $x before going broke is ((10/9) ^{25}1)/((10/9) ^{x}1). The following table shows this probability for various values of x, as well as the expected return.
Problem 176 
Goal 
Probability of Success 
Expected Return 
$260  0.893093  $8.93 
$270  0.798269  $15.97 
$280  0.714034  $21.42 
$290  0.639101  $25.56 
$300  0.572362  $28.62 
$310  0.512855  $30.77 
$320  0.459746  $32.18 
$330  0.412306  $32.98 
$340  0.369895  $33.29 
$350  0.331956  $33.20 
$360  0.297996  $32.78 
$370  0.267579  $32.11 
$380  0.240324  $31.24 
$390  0.215890  $30.22 
$400  0.193977  $29.10 
$410  0.174318  $27.89 
$420  0.156675  $26.63 
$430  0.140837  $25.35 
$440  0.126615  $24.06 
$450  0.113842  $22.77 
The expected value is maximized at a goal of $340.
Michael Shackleford, A.S.A.
  119.  Round table in a square room problem
   A round table sits flush in a corner of a square room. One point on the edge of the table is 5" from one wall and 10" from the other. What is the radius of the table?
Answer
Problem 119 Answer
The answer can be either 5" or 25".
Michael Shackleford, A.S.A.
Solution
Problem 119 Solution
Let the radius of the table be r.
Denote the corner of the room (0,0), the point of
contacts (0,r) and (r,0) and the point on the edge
in question (10,5).
Draw a triangle with these points (10,5), (r,5), and (r,r).
This is a right triangle with sides of length r10, r5, and
r. The pythagorean formula tells us that (r10)^{2} +
(r5)^{2} = r^{2}. Solving for r yields
r=5 and 25.
Thanks to Nick Hobson for this one.
Michael Shackleford, A.S.A.
  207.  Rubik's Cube problem
   Without taking it apart, how many permutations does a Rubik's Cube have?
Answer Problem 207 Answer
Question
Without taking it apart, how many permutations does a Rubik's Cube have?
8! × 12! × 3^{8} × 2^{12} = 519,024,039,293,878,000,000
Michael Shackleford, ASA — Apr. 9, 2011
Solution Problem 207 Answer
Question
Without taking it apart, how many permutations does a Rubik's Cube have?
The six center faces of the cube are fixed. By turning the faces all you can do is rearrange the corners and edges. If you took the cube apart, then there would be 8!=40,320 ways to arrange the eight corners, without respect to the orientation of each piece. Likewise, there are 12!=479,001,600 ways to arrange the 12 edges without regard to orientation.
There are 3 ways each corner can be oriented, for a total of 3^{8}=6,561 corner orientations. Likewise there are two ways each edge piece can be oriented, for a total of 2^{12}=4,096 edge orientations.
So if we could take the cube apart, and rearrange the edge and corner groups, then there would be 8! × 12! × 3^{8} × 2^{12} = 519,024,039,293,878,000,000 possible permutations. However, not all of these permutations can be arrived at from the starting position by rotating the faces.
First, it is impossible to rotate just one corner and leave everything else the same. No combination of turns will achieve that. Basically, every action has to have a reaction. If you wish to rotate one corner, it would disturb the other pieces somehow. Likewise, it is impossible to flop just one edge piece. For these reasons, we have to divide the number of permutations by 3 × 2 = 6.
Second, it is impossible to switch two edge pieces without disturbing the rest of the cube. This is the hardest part of this answer to explain. All you can do with a Rubik's Cube is rotate one face at a time. Each movement rotates four edge pieces and four corner pieces for a total of eight pieces moved. A sequence of rotations can be represented by a number of piece movements divisible by 8. Often a sequence of moves will result in two movements canceling each other out. However, there will always be an even number of pieces moved with any sequence of rotations. To swap two edge pieces would be one movement, an odd number, which can not be achieved with the sum of any set of even numbers. Mathematicians would call this a parity problem. So we have to divide by another 2 because two edge pieces cannot be swapped without other pieces being disturbed.
So there are 3 × 2 × 2 = 12 possible groups of Rubik's Cube permutations. If you disassembled a Rubik's Cube and put it back together randomly, there is a 1 in 12 chance that it would be solvable. So the total number of permutations in a Rubik's Cube is 8! × 12! × 3^{8} × 2^{12} / 12 = 43,252,003,274,489,900,000. If you had seven billion monkeys, about the human world population, playing randomly with the Rubik's cube, at a rate of one rotation per second, a cube will pass through the solved position on average once every 196 years.
Michael Shackleford, ASA — Apr. 9, 2011
  33.  Russian roulette problem
   Assuming both players take turns what is the probability the player who goes first will lose at Russian roulette using a gun with six chambers?
Answer
Problem 33 Answer
Problem 33 Answer
The answer is 6/11 =~ 54.5%
Michael Shackleford, A.S.A.
Solution
Problem 33 Solution
Problem 33 Solution
The probability the first player will lose is
1/6 * Sum for i=0 to infinity of (5/6) ^{2i} = 1/6 * 1/(125/36) = 6/11.
Mark Perkins suggested an alternative solution:
p = 1/6 + (5/6)*(1p)
6p = 1 + 5  5p
11p = 6
p = 6/11
Michael Shackleford, A.S.A., Jan. 6, 1999
  138.  SEND+MORE=MONEY problem
   A college student send the following letter to his parents: SEND +MORE  MONEY
Each letter in this letter represents a specific and unique integer. What does each letter stand for?
Answer
Problem 138 Answer
Problem 138 Answer
Posted June 25, 1999
9567
+1085

10652
Michael Shackleford, ASA
Solution
Problem 138 Solution
Problem 138 Solution
Posted June 25, 1999
The solution to this one is rather hard to explain so be patient with me.
At first we have SEND+MORE=MONEY.
The most MONEY can be is 9753+8642=18395, thus the M must be a 1.
SEND+1ORE=1ONEY.
What is the most SEND+1ORE can be? 9864+1753=11,617. The second digit in the sum (O) can't be a 1 because that is
already taken by M. Since it can't be greater than 1 it must be 0.
SEND+10RE=10NEY.
S is obviously a large number. We have less than 1100 to add to SEND to get a five digit number. Lets
consider that possibility that S=8, the most SEND+10RE could be is 8975+1064 = 10039. However N cannot
be 0 since that was already taken by O, thus S can not be 8, leaving the only plausible number to be 9.
9END+10RE=10NEY.
Subtract 10000 from both sides (9000 from SEND, 1000 from 10RE, and 10000 from 10NEY).
Now we have END+RE=NEY.
It is obvious that N=E+1. ND+RE can not add not equal or exceed 200, thus N can not exceed E by more than 1, and also
can not be equal to 1.
EED+RE+10=EEY+100.
EED+RE=EEY+90.
Next subtract 100E from both sides.
ED+RE=EY+90.
Lets rearrange that a little.
ED+REEY=90.
RE+DY=90.
R must be 8 for the sum to have hope to reach 90.
8E+DY=90.
E+DY=10.
Recall that N=E+1 so E can not be 7, since 8 is already
taken. 0, 1, 8, and 9 have already been taken, leaving
us with the middle numbers. For this to work E and D
will need to be on the large end and Y on the small end.
Without considering the N=E+1 constraint here are the
possibilities given the remaining numbers:
 E=7, D=6, Y=3
 E=6, D=7, Y=3
 E=5, D=7, Y=2
The first possiblility is not an option because E=7, which would make N=8, but N is already
taken. The second possibility is not an option because N=E+1, or N=7, which is also taken
by D. The third option is the one that works.
To put it all together:
9567
+1085

10652
Michael Shackleford, ASA
  60.  Screw your neighbor problem
   Two players are each dealt a card face down. Each player may look at his own card. The highest card wins. Cards are valued as in poker with aces being low. The first player may either keep his card or switch with the second player. The second player may keep his card, whether it be his original card or one that the first player gave him after switching, or trade it with the next card on the deck, which is also face down. The loser pays the winner $1 and if both cards are equal then no money exchanges hands. For the sake of simplicity assume an infinite number of decks. Both players are infinitely logical. At what point should the first player switch? At what point should the second player switch if the first player doesn't switch? What is the expected gain of the first player?
Answer
Problem 60 Answer
Problem 60 Answer
The first player should switch on 7 or less.
The second player should switch on 8 or less.
The expected return of the first player is ~ 0.081930 .
Michael Shackleford, ASA
Solution
Problem 60 Solution
Problem 60 Solution
Regardless of the reasonable point at which player 2 will switch with the deck player 1 will maximize his expected outcome by switching on 7 or less. Assuming player 1 switches on 7 or less
player 2 will maximize his expected outcome by switching on 8 or less. Given that the
first player swithces on 7 or less and the second player switches on 8 or less the
expected gain of the first player is ~ 0.081930. This was all calculated using a spreadsheet.
Following is an email sent by Steve Schaefer, arguing that a random strategy on the part of player 1 is better than a fixed on.
Mike,
I agree with your answer to #60, but I'm unconvinced by your solution. I'm
pretty sure that you didn't sufficiently verify the answer.
I agree that you found the best answer with integer solutions using a
spreadsheet, but I don't think you ruled out noninteger solutions.
I found the integer solution P1 trades when holding 7 or less and P2 draws
when holding 8 or less, but I wasn't convinced that P1's optimum might not
involve trading _sometimes_ when holding an 8.
As it turns out, the integer solution is the best for that choice of rules.
If P1 trades the 8 with probablility p, while P2 continues to draw for 8 or
less, then P1's expected outcome is (1802p)/2197. If P1 holds the 7 with
probability q, while P2 continues to draw for 8 or less, then P1's expected
outcome is (18029q)/2197. In this case, minor variations from the optimum
integer solution result in a lower expected outcome for P1.
If, however, we change the rules for ties, we can get a different situation.
Consider rules where P2 wins all ties, but P2 is required to draw to break a
tie if P1 and P2 just traded equal cards. (Incidentally, this seems to be
the best game on several different levels.)
The best integer solution for P1 is to trade 7 and below. If P1 doesn't
trade, then P2 will draw to 8 and below. The expected outcome is in P2's
favor, but it is only 3/2197. If P1 trades eights as well, then P2 will
change strategies and draw to a nine; then P2's expected outcome improves to
5/2197. But, it is wrong to conclude that P1 should never trade eights!
If P1 occasionally trades eights, then P2 can't afford to shift strategy.
In fact if P1 trades eights five out of seven times (at random), then P2's
expected outcome is reduced to 1.571/2197. (That's 11/15379).
If P1 executes this strategy, it doesn't matter whether P2 draws to a nine
or not. Except that if P2 always draws to a nine, then P1 might want to try
cheating a little to trade eights less than five out of seven times. If P1
could cheat completely and stop trading eights altogether, then P1 could
have an advantage of 7/2197! Similarly, if P2 never draws to a nine, then
P1 might want to try cheating the other way. If P1 could cheat completely
and always trade eights, then P1's advantage would only be 1/2197.
So, P2 needs a randomized strategy, too! In fact, the equation for P2's
expected outcome is:
2197*y = x2*(7+12*x1)+(1x2)*(32*x1),
where x1 is the probability that P1 will trade an eight and x2 is the
probability that P2 will draw to a nine. If each player tries to maximize
their own outcome against an opponent doing the same, then the two partial
derivatives will be zero:
0 = 2197* dy/dx1 = 14*x2  2
0 = 2197* dy/dx2 = 14*x1  10
So, we get x1 = 5/7 and x2 = 1/7. That is, player 1 will always trade a
seven or less and will never trade a nine or more, but will trade an eight
five times out of seven at random. When player 1 elects not to trade, then
player 2 will always draw to an eight or less and will always keep a ten or
more, but will draw to a nine one time out of seven at random.
As it turns out, this variation on the rules gives the smallest margin for
either player that I could find.
In short, your solution sounded like you only considered integer strategies,
whereas noninteger strategies are definitely possible. If we pick more
interesting rules, then we get a more interesting solution.
Steve
Michael Shackleford, ASA
  243.  Secret Santa
   Your office of 100 workers does a Secret Santa gift exchange. This where you write down everybody's name on individual pieces of paper, put them in a hat, and everybody draws a name at random to give a gift to.
The question is, how many closed loops will there be, on average? For example of a closed loop, Gordon gives to Don, who gives to Jon, who gives to Nathan, who gives to Gordon. Or drawing your own name.
Answer
Problem 243 Answer
Problem 243 Answer
The answer is 1/1 + 1/2 + 1/3 + ... + 1/100 =~ 5.187377518.
Michael Shackleford
Solution
Problem 243 Solution
Problem 243 Solution
Consider everybody choosing one at a time.
Let f(n) equal the average number of closed loops with n people.
When the first person picks, there is a 1/100 chance he picks himself and 99/100 he doesn't. In other words, there is a 1/100 chance he closes a loop and 99/100 he doesn't. Either way, it reduces the problem to one of 99 people. In other words, f(100) = (1/100) + f(99).
When the second person goes, he will have a 1/99 chance of closing a loop and 98/99 he doesn't. Whatever happens with the first person, there can be only one name left that started an unclosed loop. Thus, f(99) = 1/99 + f(98). Or, f(100) = (1/100) + (1/99) + f(98).
This process keeps repeating until there is just one person left, who will close the last loop, since the person who started it must be the only name left.
Thus the answer is 1/100 + 1/99 + 1/98 + ... + 1/1 =~ 5.187377518.
An estimate for any sufficiently large n is ln(n).
Michael Shackleford
  231.  Semicircle inscribed in a quartercircle problem
   A semicircle is inscribed in a quartercircle, as shown in this diagram.
What is the ratio of the area of the semicircle to the quartercircle?
Answer
Problem 231 Answer
The answer is 2/3.
Michael Shackleford
Solution   247.  Semicircle in a rectangle
   What is the area in green?.
Answer
Problem 247 Answer
Problem 247 Answer
The answer is 10  12.5×cos^{1}(4/5) =~ 1.95624.
Michael Shackleford
Solution
Problem 247 Solution
Problem 247 Solution
Please see my problem 247 solution (PDF).
Michael Shackleford — May 6, 2019
  178.  Serial number problem
   You pick three different numbers from 0 to 9. What is the probability that a dollar bill chosen at random will contain all 3 of your numbers in the serial number? Note: the serial number has 8 digits, you may assume that an equal number of each combination is in circulation.
Answer
Problem 178 Answer
Problem 178 Answer
The answer is 1  (3*9 ^{8}3*8 ^{8}+7 ^{8})/10 ^{8} = 0.15426684.
Michael Shackleford, A.S.A.
Solution
Problem 178 Solution
Problem 178 Solution
Let's answer the question what is the probability that the bill does NOT contain all three numbers. Let's assume that our three chosen numbers are 7, 8, and 9.
The number of ways the serial number can contain none of the three numbers, or be composed of only digits 0 to 6 is 7^{8}.
The number of ways the serial number can contain at least one 7 but no 8 or 9 is 8^{8}7^{8}. We must subtract 7^{8} because there are that many numbers that contain only the digits 0 to 6 but no 7. Likewise the number of ways to add at least one 8 or one 9 is also 8^{8}7^{8}. So the number of ways the serial numbers will contain just one digit out of {7,8,9} is 3*(8^{8}7^{8}).
Now let's consider the number of ways the serial number can have at least one 7 and 8 but no 9. The number of all combinations leaving out the 9 is 9^{8}. However some of those have no 7 or 8 digits. The number of combinations with 7 but no 8 is 8^{8}7^{8}. This is also the number of combinations with 8 but no 7. We must also subtract the number of combinations with only 06, which is 7^{8}. So the number of combinations with at least one 7 and one 8, but no 9, is 9^{8}2*(8^{8}7^{8})7^{8}. Mutliply this by three for the number of ways the serial number can omit the 7 and 8 but leave the other two chosen numbers. So the probability the serial number will have 2 out of our 3 numbers is 3*(9^{8}2*(8^{8}7^{8})7^{8}).
So the probability that the serial number will not contain all three chosen numbers is:
(7^{8} +
3*(8^{8}7^{8}) +
3*(9^{8}2*(8^{8}7^{8})7^{8})))/10^{8} =
(3*9^{8}3*8^{8}+7^{8})/10^{8} = 0.84573316.
To get the probability will contain all three chosen numbers just subtract the number above from one, or 1  (3*9^{8}3*8^{8}+7^{8})/10^{8} = 0.15426684.
Michael Shackleford, A.S.A.
  129.  Sicherman dice problem
   Create two sixsided dice, such that the probability of each sum from 2 to 12 is the same as two standard dice. Each side must have at least one dot. Negative numbers are not allowed. There is another answer besides two standard {1,2,3,4,5,6} dice.
Answer
Problem 129 Answer
Problem 129 Answer
The answer is {1,2,2,3,3,4} and {1,3,4,5,6,8}.
Michael Shackleford, A.S.A., March 4, 2003
Solution
Problem 129 Solution
Problem 129 Solution
The following table shows the frequency of each total.
Two dice totals 
Die 1 
Die 2 
1  2  3  4  5  6 
1  2  3  4  5  6  7 
2  3  4  5  6  7  8 
3  4  5  6  7  8  9 
4  5  6  7  8  9  10 
5  6  7  8  9  10  11 
6  7  8  9  10  11  12 
Note that the manner of systematically taking the sums of each set of faces is the same as adding exponents when taking the product of (x^{6}+x^{5}+x^{4}+x^{3}+x^{2}+x^{1})^{2} =
x^{12}+
2x^{11}+
3x^{10}+
4x^{9}+
5x^{8}+
6x^{7}+
5x^{6}+
4x^{5}+
3x^{4}+
2x^{3}+
x^{2}. The coefficient for x^{n} corresponds to the number of ways a total of n can be thrown.
The challenge is to find two other polynomials whose product is also x^{12}+
2x^{11}+
3x^{10}+
4x^{9}+
5x^{8}+
6x^{7}+
5x^{6}+
4x^{5}+
3x^{4}+
2x^{3}+
x^{2}. Let's start by factoring each standard die.
x^{6}+x^{5}+x^{4}+x^{3}+x^{2}+x^{1} =
x*(x^{5}+x^{4}+x^{3}+x^{2}+x^{1}+1) =
x*(x^{6}1)/(x1) =
x*(x^{3}1)*(x^{3}+1)/(x1) =
x*(x1)*(x^{2}+x+1)*(x+1)*(x^{2}x+1)/(x1) =
x*(x^{2}+x+1)*(x+1)*(x^{2}x+1)
So the product of two dice is
(x^{6}+x^{5}+x^{4}+x^{3}+x^{2}+x^{1})^{2} =
(x*(x^{2}+x+1)*(x+1)*(x^{2}x+1))^{2} =
x^{2}*(x^{2}+x+1)^{2}*(x+1)^{2}*(x^{2}x+1)^{2}
So we want to divide these terms into two sets, one for each die. There are lots of ways to do this so lets try to narrow down our options. We know each face has to have at least one dot. In other other words the polynomial for each die must not have a 1=x^{0} term, which would correspond to a face with zero dots. So we have to put on x on each side, otherwise one side would be left with a 1.
We also know the number of faces on each die is six. So each die must have six factors. If we let x=1 then the polynomial for each die should equal 6. Let's look at the value of each factor when x=1.
x: 1
x+1: 2
x^{2}x+1: 1
x^{2}+x+1: 3
So we want to divide these up into two groups such that the product of the terms equals 6. We already know we have to put one x on each side. There is only one way to get to a product of 6 with these constaints: 2*3 = (x+1)*(x^{2}+x+1).
So we know each side must contain x*(x+1)*(x^{2}+x+1). With the (x^{2}x+1) we are free to do as we wish. Putting one set on each die results in two standard dice. So putting two sets on one side and zero on the other results in:
x*(x+1)*(x^{2}+x+1) and x*(x+1)*(x^{2}+x+1)*(x^{2}x+1) ^{2} =
x + 2x^{2} + 2x^{3} + x^{4} and x + x^{3} + x^{4} + x^{5} + x^{6} + x^{8}.
So the two dice are {1,2,2,3,3,4} and {1,3,4,5,6,8}.
I'd like to thank Nick's Math Puzzles for helping with the solution to this problem. Nick's solution can be found at www.qbyte.org/puzzles/puzzle05.html. There are also two alternative answers for 8sided dice and one for 35sided dice. For more on that visit Mathnerds of do a search on "Sicherman Dice".
Michael Shackleford, A.S.A., March 4, 2003
  18.  Sideways tank of gas problem
   The gas tank of a truck is cylindrical in shape with a radius of r inches and a lengh of l inches, and tank lays on its side. With a measuring stick you note how many inches of gas are in the tank. Given this depth of gas, how many cubic inches of gas are in the tank? You may assume the tank is less than half full.
Answer
Problem 18 Answer
Problem 18 Answer
If g is the depth of the gas then g=rx. The solution is
l *
((r ^{2} * (cos ^{1}((rg)/r))  (rg)*(2rgg ^{2}) ^{1/2}).
Note: To convert into gallons divide cubic inches by 231 .
Michael Shackleford, A.S.A.
Solution
Problem 18 Solution
Problem 18 Solution
First, ignore the length of the tank until the end of the problem.
Next, define x as the shortest distance from the level of gas line segment to the center of the
a circular cross section of the tank. By the pythagorean formula the length of the level of
gas line is 2×(r^{2}x^{2})^{1/2}.
Next measure the area of the triangle with the level of gas segment as a base and the center of
the circle as the opposite angle, which is the the sum of two right triangles, or
x×(r^{2}x^{2})^{1/2}.
Next, determine the degree of the angle opposite of the base of gas line segment in the triange
above. With a little trigonometry it is found that this angle is 2×cos^{1}(x/r).
Next, find the area of the slice of the circle made between the two equal sides of the triangle
mentioned above, which is the product of the area of the circule and the ratio of the angle of
the wedge to 2×pi: pi×r^{2} × (2×cos^{1}(x/r)/2×pi) =
r^{2} × (cos^{1}(x/r)) .
From this wedge subtract the area of the triangle, leaving
r^{2} × (cos^{1}(x/r))  x×(r^{2}x^{2})^{1/2}.
Next, multiply by the length of the tank: l ×
((r^{2} × (cos^{1}(x/r))  x×(r^{2}x^{2})^{1/2}).
If g is the depth of the gas then g=rx. The solution in terms of g is
l ×
((r^{2} × (cos^{1}((rg)/r))  (rg)×(2rgg^{2})^{1/2}).
Thanks for Craig Katz for this solution.
Here is my old calculus solution:
First, ignore the length of the tank until the end of the problem.
Next, define x as the shortest distance from the level of gas line segment to the center of the a circular cross
section of the tank. By the pythagorean formula the length of the level of gas line is
2×(r^{2}x^{2})^{1/2}.
Next, take the integral from 0 to the depth of the gas, measured in distance from the center of the side.
For example if the level of gas is y (where y is less than x) then the answer is the integral from ry to r of
2×(r^{2}x^{2})^{1/2}.
To solve this integral it is helpful to know that the integral of (r^{2}x^{2})^{1/2}
is 1/2×[x×(r^{2}x^{2})^{1/2} + r^{2}×sin^{1}(x/r).
I'm sure if you can get this far you can do the rest yourself.
Michael Shackleford, A.S.A.
  124.  Small town election problem
   There exists a small town of 101 people in which John and Richard are running for mayor. Each person in the town (including the two running) cast their vote based on the toss of a fair coin. As the ballots are counted the results are reported vote by vote. The final result is 51 votes for Richard and 50 for John. What is the probability that as the votes were being counted Richard was always ahead?
Answer
Problem 124 Answer
The answer is 1/101.
Michael Shackleford, A.S.A., 9/4/1998
Solution
Problem 124 Solution
Consider a path drawn as the votes are counted. A vote for Richard
constitutes a movement 1 unit up and 1 unit to the right, a vote for John
constitutes a movement 1 unit down and 1 unit to the right. Let the point of
origin be (0,0). The question is asking what is the probability that the
path ends at (101,1) and never crosses the x axis (although it may touch it).
First consider the total number of paths, including those crossing
the x axis, which is 50 choose 101 or 101!/(50!*51!). I will this (101:50).
Next consider the number of paths which never touch the x axis, except at the
beginning. Another way to express this is the number of paths from (1,1) to (101,1) minus
the number of those ways that every touch the x axis. Remember the first vote
must be for Richard and the path must go to (1,1).
The number of paths from (1,1) to (101,1) that ever touch the x axis is
the same number of paths that go from (1,1) to (101,1). This is the
hardest step of this problem to explain. The reason is that any path from
(1,1) to (101,1) has a 1:1 counterpart of paths from (1,1) to (101,1) that
touch the x axis at least once. To find the counterpart for any given path just reflect
the path from (1,1) to the point of first contact with the x axis over the x axis
and keep the rest of the path the same.
So the number of paths from (1,1) to (101,1) that never touch the x axis
is (100:50)(100:49), where (100:50) is the number of paths from (1,1) to (101,1)
and (100:49) is the number of paths from (1,1) to (101,1). Remember that (100:50)
is the number of ways to have a 50/50 tie and (100:49) is the number of ways to
have a specific person win by two by a vote of 51 to 49.
(100:50)(100:49) =
100!/(50!*50!)  (100!/(49!*51!) =
(100!*51)/(51!*50!)  (100!*50)/(50!*51!) =
100!/(50!*51!) =
101!/(50!*51!*101) =
(101:50)/101 .
Remember that the total number of possible paths from (0,0) to (101,1)
was (101:50) from the second paragraph. The final answer is the number
of valid paths divided by the number of total paths =
(101:50)/101 / (101:50) = 1/101.
In general the probability that a path goes from (0,0) to (x,y) without
ever touching the x axis after (0,0) is y/x.
Michael Shackleford, A.S.A., 9/4/1998
  2.  Snow plow problem
   One morning is starts to snow at a constant rate. Later, at 6:00am, a snow plow sets out to clear a straight street. The plow can remove a fixed volume of snow per unit time, in other words its speed it inversely proportional to the depth of the snow. If the plow covered twice as much distance in the first hour as the second hour, what time did it start snowing?
Answer
Problem 2 Answer
Problem 2 Answer
It started snowing at (5 ^{1/2}1)/2 hours before 6:00am,
or 5:22:55am.
Michael Shackleford, A.S.A.
Solution
Problem 2 solution
Problem 2 Solution
Let the depth of snow at time t to be t units. The speed of
the plow at time t will be 1/t. Define t=0 as the time it
started snowing and t=x the time the plow started.
The distance covered in the first hour is the integral from
x to x+1 of 1/t dt. The antiderivative of 1/t is ln(t) so
the total distance covered in the first hour is ln((x+1)/x).
By the same reasoning the distance covered in the second
hour in ln((x+2)/(x+1)).
Using the fact that it the plow traveled twice as
far in the first hour as the second: ln((x+1)/x) = ln((x+2)/(x+1))^{2}
Exp both sides and you have (x+1)/x = ((x+2)/(x+1))^{2}.
Solving for x you get x=(5^{1/2}1)/2, which is the number
of hours that elapsed between the time it started snowing and the snow plow
left.
This problem was taken from the Actuarial Review, although
I heard it somewhere else before.
Michael Shackleford, A.S.A.
  234.  Solve (x^215x+55)^(x^29x+20)=1
   Find all solutions of (x^{2}  15x + 55)^{(x2 9x + 20)}.
Answer
Problem 234 Answer
There are six solutions. Scroll down 100 lines for them.
The solutions are 4, 5, 6, 7, 8, and 9.
Michael Shackleford
Solution
Problem 234 Solution
The question is to solve for all values of x where (x^{2}  15x + 55)^{(x2 9x + 20)} = 1.
Let's take the log of both sides...
(x^{2} 9x + 20) ln(x^{2}  15x + 55) = 0
One way this can be true is if x^{2} 9x + 20 = 0
Let's factor the left side.
(x4) × (x5) = 0
So, the equation seems to be true for x = 4 and x = 5. Let's plug them into the original equation, to make sure we don't run into a 0^0 problem.
Let's do x=4 first:
(4^{2}  15×4 + 55)^{(42 9×4 + 20)} = 11^0 = 1, so that's good.
Let's do x=5 next:
(5^{2}  15×5 + 55)^{(52 9×5 + 20)} = 5^0 = 1, so that's good too.
But are there other values of x that work?
We also know (or at least we should know) that ln(1) = 0.
This will be true if x^{2}  15x + 55 = 1.
Subtract 1 from both sides:
x^{2}  15x + 54 = 0.
Factor that:
(x9) × (x6) = 0
So, the equation will also be true for x = 6 and x = 9. But any other values of x?
Recall that (1)^{x} = 1, if x is any positive integer.
The original equation will be true if x^{2}  15x + 55 = 1 and x^{2} 9x + 20 is a positive integer.
First, let's try to find values where x^{2}  15x + 55 = 1. Adding one to each side:
x^{2}  15x + 56 = 0
Factoring:
(x7) × (x8) = 0
We get x=7 and x=8. Let's plug those into x^{2} 9x + 20, to make sure it is positive.
7^{2} 9*7 + 20 = 6, so x=7 is good.
8^{2} 9*8 + 20 = 12, so x=8 is good.
So, there are six values for x that work: 4, 5, 6, 7, 8, and 9.
This problem was inspired by a similar one by Presh Talwalker, which he goes over in the video Most Computers Can't Solve This, But You Can.
Michael Shackleford
  217.  Solve e^{pi*i}
   Using Taylor's Formula, solve e^{pi*i}. hint.
Answer Problem 216 Answer
The answer is 1.
Michael Shackleford, ASA — Oct. 11, 2013
Solution   248.  Sphere in a tetrahedron    Consider a sphere inscribed in a tetrahedron, where each side of the tetrahedron is an equilateral triangle with sides of length 1. What is the ratio of the area of the sphere to the tetrahedron?
Answer
Problem 248 Answer
Problem 248 Answer
The answer is approximately 0.302299894.
Michael Shackleford
Solution
Problem 248 Solution
Problem 248 Solution
 Let's start by finding the height of any of the sides. Consider the right triangle formed by that height, the side of a triangle, and half of another side. Let the height be x. So we have:
1^2 = 0.5^2 + x^2
x^2 = 0.75
x = sqrt(0.75) =~ 0.866025404
 Next, let's find the distance from the center of that triangular side to the middle of any edge. Call that distance x. Consider the right triangle with legs x and 0.5. The hypotenuse equals the height less x. Let's call the height h. So, we have a right triangle with sides of 0.5, x, and hypotenuse hx. Using the Pythagorean Theorem, we can solve for x as (h^20.25)/(2*h). We know h from the last step, so the distance from the center of that triangular side to the middle of any edge =~ 0.288675135.
 Next, let's find the distance from the center of the triangle to a corner. Consider the right triangle formed by the corner of the side of the equilateral triangular side of the tetrahedron, center of that triangle, and the middle of any side of that equilateral triangular. We know the two legs of that right triangle are 0.5 and 0.288675135, so using the Pythagorean Theorem we can find the distance from the corner of the triangle to the center is approximately 0.577350269.
 Next, let's find the height of the tetrahedron. Imagine the tetrahedron laying flat. Consider the right triangle formed by a corner of the tetrahedron on the base, center of the base, and top of the tetrahedron. We know the hyppotenuse of that triangle is 1 and one leg is the distance from the corner of the triangle to the center, which we solved for in the last step. Using the Pythagorean Theorem yet again, we can find the height of the tetrahedron is 0.816496581.
 Next, let's find the distance from the center of any side of the tetrahedron to the center of the tetrahedron. Here is a trick to easily finding that. Continue to think of the tetrahedron with a side flat on the ground. The height from the ground of three of the corners is 0. The height of the fourth corner is the height of the tetrahedron, which we just solved for.
 The height of the center will be the average of the height of each corner. In other words the height is (0+0+0+0.816496581)/4 = 0.816496581/4 = 0.204124145. This is the radius of the sphere inside the tetrahedron.
 Next, let's find the area of one of the sides of the tetrahedron. The area is (1/2)*side*height = (1/2)*1*0.816496581 = 0.433012702.
 Next, let's find the area of the tetrahedron. The formula is (1/3)*base*height = (1/3)*0.433012702*0.816496581 = 0.117851130.
 Next, let's find the volume of the sphere. The formula for that is (4/3)*pi*r^3 = (4/3)*pi*0.204124145^3 = 0.035626384.
 Finally, our answre is the area of the sphere divided by the area of the tetrahedron = 0.035626384/0.117851130 = 0.302299894.
Michael Shackleford — May 6, 2019
  65.  Spirograph problem
   In a spirograph toy you first select two circular gears. Keeping one gear stationary, you let the second gear travel around the first one until it arrives back at its initial position and orientation. By tracing the path with a pen through a hole in the second gear you can make designs like the one above. Assuming the notches in the gears are equally spaced apart, and that there are x notches in the stationary gear, and y notches in the moving gear, how many trips around the first gear will the second gear need to make before it arrives back at its initial position and orientation?
Answer
Problem 65 Answer
Problem 65 Answer
Let z equal the greatest common denominator of x and y.
If the y/x is an integer then the answer is y/x, otherwise
it is y/z.
Michael Shackleford, A.S.A.
  209.  Square wheel problem
   A square wheels rolls over a curved track such that the center of the square is always at the same height relative to the xaxis. If the square of side 2 makes a full revolution, how much horizontal distance will it cover? Here are some equations you may find useful: (cosh^1(x))' = 1/sqrt(x^2  1) cosh^1(x) = ln(x +/ sqrt(x^21))
Answer
Problem 209 Answer
The answer is 8 × ln(2^0.5 + 1) =~ 7.051.
Michael Shackleford
Solution   185.  Stick and buz saw problem
   A 10 inch stick is thrown into a buzz saw and cut in two pieces at a random point. Each resulting piece is thrown into the buzz saw again, and each is again cut at a random point. What is the probability that all four remaining pieces are one inch long or greater?
Answer
Problem 185 Answer
Problem 185 Answer
The answer is (1516*ln(2))/25 ~ 0.1564
Michael Shackleford, A.S.A, January 18, 2002
Solution
Problem 185 Solution
Problem 185 Solution
Instead of 10 inches lets give the stick a length of 1. Let x be the point of the original cut. The probability given x that all four pieces are 0.1 long or greater is (x ^{2}x+0.16)/(x ^{2}x).
Next take the integral over this probability for the larger initial cut ranging from 0.5 to 0.8. Note that if it were greater than 0.8 the smaller piece could not have two cuts more than 0.1. So we have 2 times the integral from 0.5 to 0.8 of
(1) x^{2}/(x^{2}x) + ...
(2) x/(x^{2}x) + ...
(3) 0.16/(x^{2}x)
Integral (1) works out to 0.3+ln(.4)
Integral (2) works out to ln(.4)
Integral (3) can also be expressed as 0.16*(1/(x1)  1/x). The integral of 1/x from 0.5 to 0.8 is ln(0.8)ln(0.5). The integral from 1/(x1) from 0.5 to 0.8 = ln(x1) from 0.5 to 0.8 =
ln(0.2)ln(0.5) = ln(1)+ln(.2)(ln(1)+ln(.5)) = ln(.2)ln(.5). So the initial expression equals 0.16*(ln(0.2)ln(0.5)(ln(0.8)ln(0.5))) = 0.16*(ln(0.2)ln(0.8)) = 0.16*ln(.2/.8) = 0.16*ln(1/4) = 0.16*ln(4)=0.32*ln(2).
So the final answer is 2*(0.3+ln(.4)ln(.4)0.32*ln(2)) = 0.60.64*ln(2) = (1516*ln(2))/16 =~ 0.1564
To check my answer I ran a simulation with the following results:
Outcome Number Probability
Success: 49883201 0.156374
Failure: 269116799 0.843626
Total: 319000000 1.000000
Here is my old and much harder solution to solving the third integral...
Integral (3) is the tough one. My copy of C.R.C. Standard Mathematical Tables, which I got on Ebay, says that the integral of 1/X, where X=a+bx+cx^{2} and q=4acb^{2}, is 2/sqr(q)*tanh^{1}((2cx+b)/sqr(q)).
In the case of a=0, b=1, c=1 the integral works out to 0.16*tanh^{1}x. Given the bounds of 0.5 to 0.8 the area under the curve is 0.16*2*tanh^{1}0.6
My copy of C.R.C. Standard Mathematical Tables also reminds me that tanh^{1}(x) = ln(1+x)/2  ln(1x)/2. So (3) equals 0.16*2*(ln(1.6)/2  ln(0.4)/2) = 0.16*(ln(1.6)ln(0.4)) = 0.16*ln(4)
So the final answer is 2*(0.3+ln(0.4)ln(0.4)0.16*ln(4)) = 2*(0.30.16*ln(4)) = (158*ln(4))/25 = (1516*ln(2))/25 =~ 0.156386
Michael Shackleford, A.S.A, January 16, 2002
  45.  Surface area of Colorado problem
   What is the ratio of the surface area of the state of Colorado to the surface area of the earth? You may assume that the lower and upper lines of latitude are 37 degrees and 41 degrees, and that the east and west lines of longitude are 102 and 109 degrees (the latitude lines I believe are exactly right and the longitude lines are very close and both appear to be off by about the same amount in the same direction). You may also assume the earth is perfectly spherical (which it isn't).
Answer
Problem 45 Answer
Problem 45 Answer
The answer is (7/360) * ((sin(41)sin(37))/2 =~ .0005274 .
Michael Shackleford, A.S.A.
Solution
Problem 45 Solution
Problem 45 Solution
Take the sum of the rings going around the earth from the 37th to the 41st lines of latitude.
This is the integral from 37 to 41 of the cosine of x dx. Integrate this and you get
sin(41)sin(37). Divide this by 2 since this is only relative to the northern hemisphere. Then
multiply by (109102)/360 for its share of the earth's longitude.
Michael Shackleford, A.S.A.
  93.  Swimming pool dilution problem
   You are given:
 A swimming pool has a capacity of 7 million cubic inches.
 Clean water flows into the pool at a rate of 30 cubic inches per second.
 Pool water exits the pool at the same rate.
 A kid pees in the pool. The volume of pee is 8 cubic inches, which displaces the same volume of clean water.
 As fresh water pours in the pool, it is immediately diluted. In other words, the concentration of pee is the same throughout the pool.
 You deem the pool safe to swim in if the total volume of pee is less than or equal to one cubic inch.
From the time the kid pees, how long do you have to wait to swim?
Answer Problem 93 Answer
The answer is (7,000,000/30) × ln(8/1) seconds = 485,203 seconds = 5.6158 days.
Michael Shackleford, ASA — Oct. 11, 2013
Solution Problem 93 Solution
Solution
Let p = volume of pee in the pool, in cubic inches.
Let t = time since the kid peed in the pool, in seconds.
dp/dt = (30/7000000)×p
dp = (30/7000000)×p dt
dp (7000000/30p)= dt
Taking the integral of each side...
(1) (7000000/30)×ln(p)= t + c, where c is the constant of integration.
We're given that when t=0, v=8. Plug that into the equation to find c.
(7000000/30)×ln(8)= 0 + c
c = (7000000/30)×ln(8)
Putting that value of c in equation (1) ...
(7000000/30)×ln(p)= t (7000000/30)×ln(8)
t = (7000000/30)×ln(p) + (7000000/30)×ln(8)
t = (7000000/30)×(ln(8)ln(p))
t = (7000000/30)×ln(8/p)
The question at hand is what is t when p = 1
t = (7000000/30)×ln(8/1)
t = 485,203 seconds = 8,087 minutes = 134.8 hours = 5.6158 days.
The general formula is t=(w/r)×ln(p _{1}/p _{2}), where
t = time
w = total volume in pool.
r = rate water flows in and out of pool.
p _{1}=Initial volume of pee.
p _{2}=Volume of pee at time t.
Michael Shackleford, ASA — Nov. 4, 2013
  126.  The Price is Right bidding problem
   On 'The Price is Right' there is a game in which each contestant makes a guess as to the value of a particular item. The contestant who comes closest, without going over, wins. If all contestants overbid then they start over until one of them does win. Lets assume there are only three contestants (call them x, y, and z) and that x goes first, then y, and then z. Each player can hear the bids of all previous players. Also assume that it is common knowledge that the value of the item is a random variable with a uniform distribution over the range of $1000 to $2000 (unless everyone overbids in which case the range would be narrowed). Also assume that if everyone overbids then all players will bid the same amount as before but relative to the new interval. Also assume that bidding is done to the nearest penny. What should the strategy of each player be given that x assumes y and z are mathematicians and y assumes z is a mathematician.
Answer
Problem 126 Answer
Problem 126 Answer
Strategy of z
After x and y bid the $1000 range will be divided into three intervals. Call
A the interval from $1000 to the lesser of the two prior bids, B the interval
between the two prior bids, and C the interval between $2000 and the greater of the
two prior bids. If max(B,C)/(B+C) > (A/$1000) then bid $.01 more than x or y, depending on
which will give you the greater interval. If max(B,C)/(B+C) < (A/$1000) then bid
$1000.
Strategy of y
Call x's bid X.
If x bids more than $1750 then y should bid X  X*(sqr((2000X)/1000)(2000x)/2000)+ $.01 .
If x bids less than $1750 then y should bid 2X$1999.99 .
Strategy of x
x should bid $1750.01 .
Michael Shackleford, A.S.A., 10/20/1998
  69.  The Price is Right wheel problem #1
   The object of a game is to come as close to 1 point as possible without going over. On a player's first turn he is given a random number of points, taken from a uniform distribution from 0 to 1. He may then keep that number, or take another from the same distribution, adding the second number of the first. The second player, knowing the first player's score, takes one number and has a chance for a second. Assuming the first player follows an optimal strategy in determining whether or not to take a second number, what is the probability that the first player will win? You may express your answer in terms of x, where x3 + 3x2  1 = 0. (Note that this game is the same as the Showcase Showdown in The Price is Right, only using a wheel with an infinite number of intervals).
Answer
Problem 69 answer
Problem 69 answer
The answer is
(x ^{4}/4 x ^{3} + x + 1)/3 =~ .45380187 .
Michael Shackleford, A.S.A.
Solution
Problem 69 solution
Problem 69 Solution
If the first player's score is y his chance of winning is y ^{2}
(if you can't figure this out on your own this problem is too advanced
for you).
Let x stand for the point at which the first player is indiffent
between standing on one turn or taking another. Let z be the
score on the first spin. The probability of winning by taking
another turn is the integral from z to 1 of t^{2} which
equals (1z^{3})/3. By equating this with the probability
of winning by not taking another turn, z^{2}, you find
the indifference point: (1z^{3})/3= z^{2}.
So the indifference point is x where x^{3}+x^{2}1=0,
x =~ 0.53208889 .
The probability of winning is the integral from 0 to x of (1z^{3})/3
plus the integral from x to 1 of z^{2}. After
integration this answer is (x  x^{4}/4)/3 + (1 x ^{3})/3 =
(x^{4}/4 x^{3} + x + 1)/3 =~ .45380187 .
Michael Shackleford, A.S.A.
  104.  The Price is Right wheel problem #2
   In the Price is Right three players compete for one place in the showcase showdown. Each player takes his turn spinning a wheel which has an equal probability of stopping on every amount evenly divisible by .05 from .05 to 1.00 . If the player does not like their first spin they may spin again, adding the second spin to their first, however if they go over 1.00 they are immediately disqualified. In the event of a tie the winner is determined randomly with each player left having an equal probability of winning. What is perfect strategy for all players in this game, assuming that all other players also play by perfect strategy?
Answer
Problem 104 Answer
Problem 104 Answer
The first player should spin again if his first spin is 65 cents or less.
If any of the following conditions are true the second player should spin again.
 His score is less than the first player's score.
 His score is 50 cents or less.
 His score is 65 cents or less and he has tied the first player.
If there is only one person to beat, then the third player should spin again if his score is less than the current highest score. If his first spin ties the highest score, then he should spin again if the tie is at 45 cents or less. If the tie is at 50 cents he is indifferent to spinning to staying pat.
If there are two tied players to beat, then the third player should spin again if his score is less than the current high scores. If his first spin ties the high scores, then he should spin again if the tie is at 65 cents or less.
The following table shows the probability of each player winning, according to player 1's initial spin. The bottom row shows the overal probabilities for each player winning before the spinoff begins.
Probabilities in the Price is Right SpinOff 
Spin 1  Strategy  Player 1  Player 2  Player 3 
0.05  spin  20.59%  37.55%  41.85% 
0.1  spin  20.59%  37.55%  41.86% 
0.15  spin  20.57%  37.55%  41.87% 
0.2  spin  20.55%  37.55%  41.9% 
0.25  spin  20.5%  37.56%  41.94% 
0.3  spin  20.43%  37.56%  42.01% 
0.35  spin  20.33%  37.58%  42.1% 
0.4  spin  20.18%  37.6%  42.22% 
0.45  spin  19.97%  37.64%  42.39% 
0.5  spin  19.68%  37.71%  42.61% 
0.55  spin  19.26%  37.81%  42.93% 
0.6  spin  18.67%  37.96%  43.36% 
0.65  spin  17.86%  38.21%  43.93% 
0.7  stay  21.56%  38.28%  40.16% 
0.75  stay  28.42%  35.21%  36.38% 
0.8  stay  36.82%  31.26%  31.92% 
0.85  stay  46.99%  26.35%  26.66% 
0.9  stay  59.17%  20.36%  20.47% 
0.95  stay  73.61%  13.19%  13.21% 
1  stay  90.57%  4.72%  4.72% 
Average   30.82%  32.96%  36.22% 
Here are the winning number of combinations out of 6*20^{6} possible.
Player 1: 118,331,250
Player 2: 126,566,457
Player 3: 139,102,293
Michael Shackleford, A.S.A., 9/4/98
  88.  Three Surgeons, Two Gloves, and a Patient
   How can three surgeons, with open wounds on their hands, operate on a patient with only two surgical gloves so that nobody is directly exposed to anybody else's blood?
Answer
Problem 88 Answer
Problem 88 Answer
Call the gloves A and B, and the surgeons 1, 2, and 3.
First, surgeon 1 operates with glove A inside glove B.
Second, take glove A out of glove B.
Third, surgeon 2 operates using glove B.
Fourth, turn glove A inside out.
Fifth, surgeon 3 operates with glove A inside glove B.
Note: Another way to ask this is how do you have safe
sex with three men, one woman, and two condoms?
Michael Shackleford, A.S.A.
  237.  Three ants problem
   Three ants go around the circumference of a circle, each at his own contant rate and in the same direction. It takes ant A three minutes to make a revolution, ant B five minutes, and ant C seven minutes. They all start out at random points. A point is picked at random on the circumference. What is the probability each ant reaches that point first?
Answer
Problem 237 Answer
Problem 237 Answer
Here are the probabilities each ant arrives first:
 Ant A: 20/35 = 57.14%
 Ant B: 9/35 = 25.71%
 Ant C: 6/35 = 17.14%
Michael Shackleford
Solution   173.  Three card problem
   Three playing cards, removed from an ordinary deck, lie face down in a horizontal row. Immediately to the right of the King there's a Queen or two. Immediately to the left of a Queen there's a Queen or two. Immediately to the left of a Heart there's a Spade or two. Immediately to the right of a Spade there's a Spade or two. Name the three cards in order.
Answer
Problem 173 Answer
Problem 173 Answer
The three cards are:
Or, if you can't see the pictures: the king of spades, queen of spades, and queen of hearts.
Credit for this problem goes to Martin Gardner who published it in the November 1965 Scientific American.
Michael Shackleford, A.S.A.
Solution
Problem 173 Solution
Problem 173 Solution
Let's label the clues as follows:
 To the right of the King there's a Queen or two.
 To the left of a Queen there's a Queen or two.
 To the left of a Heart there's a Spade or two.
 To the right of a Spade there's a Spade or two.
From clues 1 and 2 we can infer that a king is on the left and a queen on the right and the middle card is either a queen or 2.
First consider the possibility that a 2 is in the middle. If there are two consecutive spades (from clue 4) then then three cards must be the king of spades, 2 of spades, queen of hearts. If to the right of a spade there is a 2 (again from rule 4) then the three cards are the kings of spades, a two of unknown suit, and the queen of hearts. So if we assume a 2 in the middle it is not certain what the suit of the 2 is.
Next consider the possibility that a queen is in the middle. Clue 4 tells us there must be two consecutive spades, since there is no 2. Both queens can not be spades since they are from the same deck, thus the left and center cards must be spades. Clue 3 tells us the right queen is a heart. So the three cards would be the king of spades, queen of spades, and queen of hearts.
So a 2 in the middle does not lead to a definitive answer, but a queen in the middle does. The question implies there is just one possible answer. Therefore there is a queen in the middle, leading to the final answer of king of spades, queen of spades, and queen of hearts.
Credit for this problem goes to Martin Gardner who published it in the November 1965 Scientific American.
Michael Shackleford, A.S.A.
  106.  Three circles and a line problem
   The three circles in the diagram above all are tangent to the line in the picture. The radius of circle A is a, of circle B is b, and of circle C is c. All three circles are tangent to each other. What is c as a function of a and b?
Answer
Problem 106 Answer
Problem 106 Answer
c=ab/(a+2*sqr(ab)+b)
Michael Shackleford, A.S.A.
Solution
Problem 106 Solution
Problem 106 Solution
Define u as the center of circle A, v the center of circle B, and w
the center of circle C. Define x as the point tangent to circle A and
the line, y as the point tangent to circle B and the line, and z as
the point tangent to circle C and the line.
uwzx forms a trapazoid with sides of length a,a+c,c,2*sqr(ac) per the
pythagorean formula. Likewise the bottom of trapazoid vyzw is 2*sqr(bc)
and the bottom of trapazoid uvyx is 2*sqr(ab).
The sum of the bottoms of trapazoids uwzx and vyzw is equal to the
bottom of trapazoid uvyx:
2*sqr(ac)+2*sqr(bc)=2*sqr(ab)
sqr(ac)+sqr(bc)=sqr(ab)
sqr(c)*(sqr(a)+sqr(b))=sqr(ab)
sqr(c)=sqr(ab)/(sqr(a)+sqr(b))
c=ab/(a+2*sqr(ab)+b)
Note: Steve Schaefer points out the following general relationship c^{1/2} = a^{1/2} + b^{1/2}
Michael Shackleford, A.S.A.
  236.  Three circles problem #3
   What is the distance from A to B in this diagram: .
Answer
Problem 236 Answer
The answer is 30*sqrt(2) + 10*sqrt(3) = 59.7469.
Michael Shackleford
Solution   15.  Three door problem
   There are three doors numbered 1, 2, and 3. Behind each, placed randomly, are red, blue, and green cars.
There are three contestants one each wearing a red, blue, and green hat.
Each may open two doors, one at a time. If each player sees the car of the same color as his hat, then all three will win cars. If at least one doesn't see the car of his own color, then all three lose.
The three players must go one at a time and may not communicate once they start. Each player must leave all three doors closed after his turn and return to where he started from. They may communicate before the first player's turn.
What is the greatest chance they can achieve of winning the three cars and how should they do it?
For example, if each picked randomly, their chances of winning would (2/3)^3. But can that strategy be improved upon?
This is not a trick question like peeking under the door or anything like that.
Answer
Problem 15 Answer
Problem 15 Answer
The maximum probability that can be achieved of winning is 2/3. See the solution for how to do it.
Michael Shackleford
Solution
Problem 15 Solution
Problem 15 Solution
The maximum probability that can be achieved of winning is 2/3. Here is way to do it.
 Red opens door 1. If first door is blue, he opens 2. If first door is green, he opens 3.
 Blue opens door 2. If first door is red, he opens 1. If first door is green, he opens 3.
 Green opens door 3. If first door is red, he opens 1. If first door is blue, he opens 2.
This should work for these combinations, where the colors are ordered from door 1 first to door 3 last:
 RBG (all three guess correctly on first guess)
 RGB (red is correct first time, blue sees green so opens door 3, green sees blue so opens 2.)
 BRG (red sees blue so opens 2, blue sees red so opens 1, green is correct on first guess).
 GBR (red sees green so opens 3, blue is correct on first guess, green sees red so opens 1).
Michael Shackleford
  79.  Three humans, three monkeys, and a boat problem
   One one side of a river are three humans, one big monkey, two small monkeys, and one boat. Each of the humans and the big monkey are strong enough to row the boat. The boat can fit one or two bodies (regardless of size). If at any time at either side of the river the monkeys outnumber the humans the monkeys will eat the humans. How do you get everyone on the other side of the river alive?
Answer
Problem 79 Answer
Let:
h=human
b=big monkey
s=small monkey
1. Row b and s over ( hhhs / bs )
2. Row b back ( hhhbs / s )
3. Row b and s over ( hhh / bss )
4. Row b back ( hhhb / ss )
5. Row h and h over ( hb / hhss )
6. Row h and s back ( hhbs / hs )
7. Row h and b over ( hs / hhbs )
8. Row h and s back ( hhss / hb )
9. Row h and h over ( ss / hhhb )
10. Row b back ( bss / hhh )
11. Row b and s over ( s / hhhbs )
12. Row b back ( bs / hhhs )
13. Row b and s over ( / hhhbss )
Here is a remark on this problem I received from Rob Pratt,
Department of Mathematics
of The University of North Carolina at Chapel Hill,
"There are exactly three other solutions of the same length, and none
shorter. It probably doesn't make any sense to explicitly list them all,
but you might mention that there are exactly four shortest solutions."
I got this one from a Usenet archive on math problems which can be accessed
here,
look under logic problems, part 2.
Michael Shackleford, A.S.A.
  212.  Three logicans game #3
   Three logicians are on a game show. The host explains that each will be given his own black or white hat with 50% probability. Then each logician will be able to see the other logician's hats, but not his own. After the showing of the hats the host will separate the logicians and ask each of them the color of his own hat. Each logician may choose to not answer. The host will give each logician $1,000 if every answer submitted is correct, and at least one answer is submitted. Communication after the placing of the hats is not allowed. However, they are given time before the hat placing to devise a strategy. What strategy should they follow to maximize their odds of winning the prize, and what would be this probability of winning?
Answer Problem 212 Answer
Answer
If a logitian see both other hats are the same color then he should guess the opposite color.
The probability of winning is 75%.
Michael Shackleford, ASA — June 16, 2012
  198.  Three logicians game #1
   You are playing a game with two perfect logicians. All three must secretly write down a positive integer. The one with the lowest unique integer will win $3. If all three have the same number, each will win $1. The logicians are selfish, and each wishes to maximize his own winnings. Communication is not allowed. What number should you choose?
Answer
Problem 198 Answer
Problem 198 Answer
You should have a random strategy in which you choose x with probability (1/2)^{x}.
Michael Shackleford, A.S.A.
Solution
Problem 198 Solution
Problem 198 Solution
A random strategy is called for, to prevent the other two logicians from exploiting a nonrandom choice. For example, if logician A had a firm strategy to choose the number 1, B and C could gain an advantage by choose 1 with probability 75%, and 2 with probability 25%. The following would be the possible outcomes under that scenario:
B picks 1, C picks 1: probability 0.75×0.75 = 0.5625, A gets $1, B gets $1, C gets $1.
B picks 1, C picks 2: probability 0.75×0.25 = 0.1875, A gets $0, B gets $0, C gets $3.
B picks 2, C picks 1: probability 0.25×0.75 = 0.1875, A gets $0, B gets $3, C gets $1.
B picks 2, C picks 2: probability 0.25×0.25 = 0.0625, A gets $3, B gets $0, C gets $0.
Expected value for A = 0.5625×1 + 0.0625×3 = $0.75.
Expected value for B or C = 0.5625×1 + 0.1875×3 = $1.125.
The optimal strategy is the one that can not be defeated by any other strategy used by both the other logicians. Now, this is where I use some hand waiving logic, but I contend that if two logicians use this same strategy, it won't make any difference what the third logician does, the expected value will be equal for all three.
Let's start by consider the case where the only choices are 1 and 2. Obviously, the only way to win would be to be the singleton in a split game. There is no difference between 1 and 2. So the correct strategy would be to pick 1 or 2 randomly, with 50% each.
Next, let's introduce a third choice. If at least one logician picks 1, the game will be decided for that reason. If none pick number 1, then it reverts to the twonumber game of the last step. Let p be the probability that logicians A or B pick number 1. Logician C should be indifferent to picking 1 or not. If he doesn't, then he should revert to the 50/50 strategy of the 2number game, because it will only come into play if nobody picks 1. The expected value of logician C picking 1 is $1×p^2 + $3×(1p)^{2}. The expected value of C not picking 1 is $1×(1p)^2 + $3×p^2. So, we have $1×p^2 + $3×(1p)^2 = $1×(1p)^2 + $3×p^2. Doing some simply algebra, we get p=0.5. So, in the 3number case, the probability of picking 1 is 50%, 2 is 25%, and 3 is 25%.
Next, let's introduce a fourth choice. Again, if at least one logician picks 1, the game will be decided for that reason. If none pick number 1, then it reverts to the threenumber game of the last step. Let p be the probability that logicians A or B pick number 1. Logician C should be indifferent to picking 1 or not. If he doesn't, then he should revert to the 50/25/25 strategy of the 3number game, because it will only come into play if nobody picks 1. Again, the expected value of logician C picking 1 is $1×p^2 + $3×(1p)^2. The expected value of C not picking 1 is $1×(1p)^2 + $3×p^2. So, we have $1×p^2 + $3×(1p)^2 = $1×(1p)^2 + $3×p^2. The equation is the same as the 3number game, so p will still be 0.5. So, in the 4number case, the probability of picking 1 is 50%, 2 is 25%, 3 is 12.5%, and 4 is 12.5%.
Repeating this logic, the probability of number 1 will always be 50%, and the other numbers will shift down 1, and be divided by 2. Done infinitely, the probability of x (where x>1) will be the probability half of the probability of x1. So, the probability of picking n is (1/2)^n.
Michael Shackleford, A.S.A.
  199.  Three logicians game #2
   Same question as 198, except in the event of a 3way tie, nobody wins anything.
Answer
Problem 199 Answer
Problem 199 Answer
Let p_{n} be the probability you should pick n.
p_{1} = (x1+(1x)^{0.5})/x.
x=1y^{2}.
y=z(2/3). y is also the solution to the equation y^3+2y^22=0.
z=A+B.
A=(b/2+((b^2/4)+(a^3/27))^0.5)^(1/3)
B=(b/2((b^2/4)+(a^3/27))^0.5)^(1/3)
a=4/3.
b=2
After a lot of messy math, p_{1} = 0.456310987.
p_{n} = p_{n1} × (1p_{1}).
Here are the first 34 values of p:
p_{1}=0.456310987308
p_{2}=0.24809127017
p_{3}=0.134884497736
p_{4}=0.073335219402
p_{5}=0.039871553032
p_{6}=0.021677725303
p_{7}=0.011785941067
p_{8}=0.006407886662
p_{9}=0.003483897573
p_{10}=0.001894156832
p_{11}=0.001029832258
p_{12}=0.000559908483
p_{13}=0.000304416091
p_{14}=0.000165507684
p_{15}=0.000089984709
p_{16}=0.000048923698
p_{17}=0.000026599277
p_{18}=0.000014461735
p_{19}=0.000007862686
p_{20}=0.000004274856
p_{21}=0.000002324192
p_{22}=0.000001263638
p_{23}=0.000000687026
p_{24}=0.000000373528
p_{25}=0.000000203083
p_{26}=0.000000110414
p_{27}=0.000000060031
p_{28}=0.000000032638
p_{29}=0.000000017745
p_{30}=0.000000009648
p_{31}=0.000000005245
p_{32}=0.000000002852
p_{33}=0.000000001551
p_{34}=0.000000000843
The probability of winning, assuming any two logicians choose this strategy, is 0.29559774. The probability of a tie is 0.113206772.
Michael Shackleford, A.S.A.
  101.  Three men and three questions problem
   Of three men one man always tells the truth, one always tells lies, and one answers yes or no randomly. Each man knows which man is who. You may ask three yes/no question to determine who is who. If you ask the same question to more than one person you must count it as question used for each person whom you ask. What three questions should you ask?
Answer
Problem 101 Answer
Problem 101 Answer
There are six possible sceanarios:
Truth Lying Random
Guy Guy Guy
  
I A B C
II A C B
III B A C
IV B C A
V C A B
VI C B A
Label the three men A, B, and C and tell each man who is what letter.
 Step 1: Ask A, "Is B more likely to tell the truth than C?" If yes go to step 2, if no go to step 5.
 Step 2: Ask C, "Are you the random guy?" If yes go to step 3, if no go to step 4.
 Step 3: Ask C, "Is A the truth guy?" If yes then scenario IV, if no then scenario II.
 Step 4: Ask C, "Is A the lying guy?" If yes then scenario V, if no then scenario VI.
 Step 5: Ask B, "Are you the random guy?" If yes then step 6, if no then step 7.
 Step 6: Ask B, "Is A the truth guy?" If yes then scenario VI, if no then scenario I.
 Step 7: Ask B, "Is A the lying guy?" If yes then scenario III, if no then scenario IV.
Michael Shackleford, A.S.A.
  12.  Three men, a bell boy, and a motel room problem
   Three men walk into a motel and ask for a room. The desk clerk says a room is $30 so each man pays $10 towards the cost. Later, the clerk realizes he made a mistake, that the room should have been $25. He calls the bell boy over and asks him to refund the other $5 to the three men. The bellboy, not wanting to mess with a lot of change dividing the $5 three ways, decides to lie about the price, refunding each man $1 and keeping the other $2 for himself. Ultimately each man paid $9 towards the room and the bellboy got $2, totaling $29. But the original charge was $30, where did the extra $1 go?
Answer
Problem 12 Answer
Problem 12 Answer
There is no missing dollar, it is a trick question.
Michael Shackleford, A.S.A.
Solution
Problem 12 Solution
Problem 12 Solution
This is a trick question.
The total money paid is $9 * 3 = $27.
The total money received is $25 for the room and $2 to the bellboy = $27.
There simply is no extra dollar. The $2 to the belloy is included in the $27
the three men paid. By adding it to the $27 it is being counted twice.
Michael Shackleford, A.S.A.
  96.  Three people, a bridge, and a bicycle problem
   Three people (A, B, and C) need to cross a bridge. A can cross the bridge in 10 minutes, B can cross in 5 minutes, and C can cross in 2 minutes. There is also a bicycle available and any person can cross the bridge in 1 minute with the bicycle. What is the shortest time that all men can get across the bridge? Each man travels at his own constant rate. Note: Don't let the drawing confuse you, there is just one bicycle.
Answer
Problem 96 Answer
Problem 96 Answer
It takes 73/25 = 2.92 minutes for all men to cross.
Michael Shackleford, A.S.A.
Solution
Problem 96 Solution
A's speed is 1/10 (in bridges per minute), B's speed is 1/5, C's speed is 1/2, and
the bicycle's speed is 1.
It stands to reason that all men should finish at the same time. I can't explain
why this is true, it just makes common sense.
Let me state up front that I made up this problem myself and I am not 100% positive
there is no better answer. If anyone can think of a faster way to cross please
email me. That being said I strongly believe that the fastest way to get everyone
across is for B and C to start out on foot and A to start out with the bicycle. At
a point y A will get off the bicycle and walk the rest of the way. Eventually C will
get to the bicycle abondoned by A, then ride backwards to a point x, leaving the bicycle
there, then turning around and walk until he reaches the end. Person B will walk until
he reaches the bicycle left by C and then ride the rest of the way.
Below are the times that each will take to cross, in terms of x and y:
A: 1*y + 10*(1y)
B: 5*x + 1*(1x)
C: 2*y + (yx) + 2*(1x)
Next equate these equations:
10  9y = 3x + 3y + 2 = 4x + 1.
To solve set up two linear equations:
10  9y = 3x + 3y + 2 > 3x  12y = 8
10  9y = 4x + 1 > 4x + 9y = 9
Then solve for x and y:
x = 12/25, y=59/75.
Given these points it will take each person 73/25 = 2.92 minutes to cross.
Below are other crossing times, given various crossing times of the two fastest people,
assuming the slowest still takes 10 minutes to cross.
Crossing time of fastest

Second
Fastest
Time 1 2 3 4 5 6 7 8 9
         
1 1.00
2 1.62 1.92
3 1.90 2.41 2.74
4 2.06 2.71 3.16 3.48
5 2.16 (2.92) 3.45 3.85 4.16
6 2.23 3.07 3.67 4.13 4.49 4.78
7 2.29 3.18 3.84 4.35 4.75 5.08 5.34
8 2.33 3.27 3.98 4.52 4.96 5.32 5.62 5.87
9 2.36 3.34 4.09 4.67 5.14 5.52 5.85 6.12 6.36
Michael Shackleford, A.S.A., 10/11/1998
  5.  Three wise men and mischievous boy problem
   While three wise men are asleep under a tree a mischievous boy paints their foreheads red. Later they all wake up at the same time and all three start laughing. After several minutes suddenly one stops. Why did he stop?
Answer
Problem 5 Answer
Problem 5 Answer
He stopped laughing because he realized he was also painted.
Michael Shackleford, A.S.A.
Solution
Problem 5 Solution
Problem 5 Solution
The key to this problem is that they are wise men. Assume all
three know that all three are wise. Also, every man can see the
other two but can't see their own forehead. If you still need
the solution scroll down.
Each wise man initially assumes that he isn't painted and can
plainly see the other two are.
However, if only two were painted it would not take long for
either of the painted wise men to realize they were painted. For
either of the painted ones would see the other painted one
laughing and know he is laughing at them, since the other guy is
not painted. In other words if you saw a wise man laughing and
the other guy wasn't he painted he must be laughing at you, thus
you must be painted.
After a sufficient period of time if nobody stops laughing then
it can no longer be assumed that you are not painted, thus you
must be painted as well. When you realize this you stop
laughing.
Michael Shackleford, A.S.A.
  214.  Threeplayer doubleornothing game
   Three players play a game where the loser must double the money of both the other two players. They play this game three times and each player loses once. After three rounds each player has $24. How much did each player start with.
Answer Problem 214 Answer
Answer
The starting bankrolls were 12, 21, and 39.
Michael Shackleford, ASA — Sept. 28, 2012
Solution Problem 214 Solution
Solution
Let's call the players P1, P2, and P3, and the starting bankrolls as follow:
Assume that P1 loses the first round. After he pays the winners the bankrolls we be:
Assume that P2 loses the second round. After he pays the winners the bankrolls we be:
 P1: 2x2y2z
 P2: x+3yz
 P3: 4z
Assume that P3 loses the third round. After he pays the winners the bankrolls we be:
 P1: 4x4y4z
 P2: 2x+6y2z
 P3: xy+7z
All three of these sums must equal 24. So, we have three equations and three unknowns. That is enough to do a some simple matrix algebra to get x=39, Y=21, Z=12.
My thanks to WongBo for this problem.
Michael Shackleford, ASA — Sept. 28, 2012
  197.  Threeplayer random number game
   You are playing a game with three people, you, an opponent, and a referee. Each will pick a real number between 0 and 1. The referee will chose randomly, with each pick equally likely. The player who comes closer, without going over, will win. If both go over, or both pick the same number, then the game will result in a tie. Your oponent is a perfect logician. What strategy should you employ?
Answer
Problem 197 Answer
Problem 197 Answer
Choose a random number (z) between 0 and 3/4, according to the probability density function f(z) = 1/2 × (1z)^{3/2}. This answer coutesy of Rick Percy.
Michael Shackleford, A.S.A.
Solution
Problem 197 Solution
Problem 197 Solution
This problem was solved by Rick Percy of Columbus, Ohio. It is the only problem on the site I have not solved myself. I thought I did, but he proved my answer wrong. Here is a link to Rick Percy's solution (PDF, 26K).
Michael Shackleford, A.S.A.
  131.  Tiger and the lake problem
   A man is standing on a rock in the middle of a circular lake of radius 1. There is a tiger on the shore of the lake that can run four times as fast you can swim, however the tiger can not swim. The tiger is hungry and always attempts to keep the distance between the two of you at a minimum. How can you safely swim to shore?
Answer
Problem 131 Answer
Until you are more than 1/4 of the radius away from the rock you can swim fast enough so that
you can stay 180 degrees away from the tiger. Regardless of which direction
the tiger moves around the circle you swim the other way, always keeping 180 degrees away, and
moving outward with leftover energy.
Before long you will be 1/4 of the radius away from the center and the tiger will be 180
degrees away. At this point swim straight to the point on the shore furthest from the
tiger. You will be able to get there in 3/4 units of time, while it will take the tiger
pi/4 =~ 0.7854 units of time.
Thanks to 'Bob' for this one.
Michael Shackleford, A.S.A., 11/27/1998
  83.  To confess or not confess problem
   You and your partner in crime are both arrested and questioned separately. You are offered a chance to confess, in which you agree to testify against you partner, in exchange for all charges being dropped against you, unless he testifies against you also. Your lawyer, whom you trust, says that the evidence against both of you, if neither confesses, is scant and you could expect to take a plea and each serve 3 years. If one implicates the other, the other can expect to serve 20 years. If both implicate each other you could each expect to serve 10 years. You assume the probability of your partner confessing is p. Your highest priority is to keep yourself out of the pokey, and your secondary motive is to keep you partner out. Specifically you are indifferent to you serving x years and your partner serving 2x years. At what value of p are you indifferent to confessing and not confessing?
Answer
Problem 83 Answer
Problem 83 Answer
The answer is 11/21 =~ 52.4%.
Michael Shackleford, A.S.A.
Solution
Problem 83 Solution
Problem 83 Solution
If you confess there is probability p of two confessions and (1p) probability that
only you implicate your partner. If you don't confess there is probability p
that neither of you implicate each other and a (1p) probability that only you
are implicated. Let s donate one unit of suffering. One
year in jail yourself would cause you 2 units of suffering, and one year of
your partner serving would cause you 1 unit of suffering.
The table below shows the total suffering given the four possibilities:
Your partner

confess not confess
Y confess: 10*2+10=30 0+20=20
o
u not confess: 2*20+0=40 2*3+3=9
So given the probability p of your partner confessing if you confess the
expected, or average, amount of suffering would be 30p + 20(1p)=10p+20.
If you don't confess the expected amount of suffering would be
40p + 9(1p)=31p+9. To find the indifference point equate the two expressions:
10p+20 = 31p+9
11 = 21p
p=11/21 =~ 52.4%.
Thus if you think your partner's probability of confessing is greater than 11/21 you
should confess, otherwise you should keep your mouth shut.
What is a much more interesting question, with no answer that fits
everybody, is what would you do if you assumed your partner had the
same morals and priorities as you, and that he assumes the same
thing of you.
Michael Shackleford, A.S.A.
  140.  Toothpick problem
   Consider a infinite number of parellel lines, spaced one inch apart from each other. If you dropped a one inch toothpick at random on this set of lines what is the probability the toothpick would cross a line?
Answer
Problem 140 Answer
Problem 140 Answer
The answer is 2/ $\pi $ =~ 0.636620
Thanks to
Extra Stuff: Gambling Rambling by Peter Griffin for this problem. See chapter 11.
Michael Shackleford, ASA, August 19 1999
Solution
Problem 140 Solution
Problem 140 Solution
Consider that one end of the toothpick can fall anywhere between two lines with
equal probability. Consider also that the angle formed by the line of the toothpick and the
parellel lines is uniformally distributed. The probability can be set up as:
2/$\pi $ * $\int $(x=0 to 1) cos^{1}(x) dx =
2/$\pi $ * [x*cos^{1}(x)  (1x^{2})^{1/2} for (x=0 to 1)] =
2/$\pi $ * (000+1) =
2/$\pi $ =~ 0.636620
Thanks to
Extra Stuff: Gambling Rambling by Peter Griffin for this problem. See chapter 11.
Michael Shackleford, ASA, August 20 1999
  183.  Train station meeting problem
   Each day a man meets his wife at the train station after work, and then she drives him home. She always arrives exactly on time to pick him up. One day he catches an earlier train and arrives at the station an hour early. He immediately begins walking home along the same route the wife drives. Eventually his wife sees him on her way to the station and drives him the rest of the way home. When they arrive home the man notices that they arrived 20 minutes earlier than usual. How much time did the man spend walking?
Answer
Problem 183 Answer
Problem 183 Answer
The answer is 50 minutes.
I found this problem in Games for the SuperIntelligent by James F. Fixx, which is out of print.
Michael Shackleford, ASA, August 19 1999
Solution
Problem 183 Solution
Problem 183 Solution
Suppose the husband usually took the train that arrived at 5:00, but today took the one that arrived at 4:00. Since the wife would have spent all her time driving at the same speed it stands to reason she saved 10 minutes both directions because her husband walked part of the way.
So, if the wife usually arrived at the station at 5:00 then she would have spotted her husband at 4:50. The husband started walking at 4:00, thus he would have spent 50 minutes walking.
My thanks to MathGent for this elegant solution.
Michael Shackleford, ASA, August 19 1999
  182.  Triangular field problem
   There is a one acre field in the shape of a right triangle, with sides of length x and y. At the midpoint of each side there is a post. Tethered to the posts on each side is a sheep. Thethered to the post on the hypotenuse is a dog. Each animal has a rope just long enough to reach the two adjacent vertices of the triangle. How much area outside of the field do the sheep have to themselves?
Answer
Problem 182 Answer
Problem 182 Answer
The answer is one acre.
Michael Shackleford, A.S.A.
Solution
Problem 182 Solution
Problem 182 Solution
Let the two sides of the triangle have length x and y, and the hypotenuse z. The area outside the field the sheep can cover is pi*(x/2)^{2}/2 + pi*(y/2)^{2}/2.
The area inside the field, which can be covered by at least one sheep, is xy/2.
The part of the sheep's area which the dog can cover is everything on the side of the hypotenuse containing the field, or pi*(z/2)^{2}/2.
So the area the sheep have to themselves is pi*(x/2)^{2}/2 + pi*(y/2)^{2}/2 + xy/2  pi*(z/2)^{2}/2 =
(pi/2)*((x/2)^{2} + (y/2)^{2}  (z/2)^{2}) + xy/2 =
(pi/8)*(x^{2} + y^{2}  z^{2}) + xy/2 =
(pi/8)*0 + xy/2 (because in a right triange x^{2} + y^{2} = z^{2}) =
xy/2
The area of the field is also xy/2, which is one acre, so the answer is one acre.
Here is a simpler solution provided by David Nixon:
Pythagoras really states that the areas of similar shapes on two sides of a
right triangle add to a third similar shape on the hypoteneuse. Here, we
have the semicircle described by the dog on the hypoteneuse, and the two
sheep on the other two sides. The area impinging on the sheep is the dog's
semicircle less the triangle. So the solution is Sheep  shaded areas =
Sheep  (Dog 1) = 1.
Michael Shackleford, A.S.A.
  92.  Trigonometry problem
   In the diagram above AB=3, BC=7, and AC=9. What is BD?
Answer
Problem 92 Solution
Problem 92 Solution
BD = (9(41/18) ^{2}) ^{1/2}
The Wizard of Odds
Solution
Problem 92 Solution
Problem 92 Solution
Define x as angle CAB. Define y as angle ACB.
cos(x)=AD/3, sin(x)=BD/3, cos(y)=CD/7, sin(y)=BD/7.
cos^{2}(x)+sin^{2}(x) = AD^{2}/9 + BD^{2}/9 = 1.
cos^{2}(y)+sin^{2}(y) = CD^{2}/49 + BD^{2}/49 = 1.
BD^{2} = 9  AD^{2} = 49  CD^{2}.
Remember that AD + CD = 9, or CD = 9  AD.
9  AD^{2} = 49  (9AD)^{2}.
9 = 49  81 + 18AD.
AD = 41/18.
(41/18)^{2} + BD^{2} = 9.
BD = (9(41/18)^{2})^{1/2} = 1.95236
Others have pointed out an alternate solution using Heron's Formula, which states the area of a triange is sqr(s*(sa)*(sb)*(sc)), where s=(a+b+c)/2, and a, b, and c are the lengths of the three sides.
In this case a, b, and c are 3, 7, and 9.
s=(3+7+9)/2=9.5
area=sqr(9.5*(9.53)*(9.57)*(9.59)) = sqr(1235/16) = 8.785641695
The area must also equal (1/2)*9*h, where h is the height.
So sqr(1235/16) = (1/2)*9*h
h=sqr(1235/16)*(2/9) = 1.95236
The Wizard of Odds
  76.  Two bags and marble problem
   You choose one of two identical looking bags at random. One bag has three black marbles and one white marble. The other has three white marbles and one black marble. After choosing a bag you draw one marble out at random. You notice it is black. You then put it back and draw another marble out of the same bag at random. What is the probability that the second marble drawn is black?
Answer
Problem 76 Answer
Problem 76 Answer
The answer is 5/8 = 62.5%.
Michael Shackleford, A.S.A.
Solution
Problem 76 solution
Problem 76 Solution
The probability of event A happening given that event B already
happened is the probability of A and B happening divided by the
probability that B happened. This can be pressed as Pr(AB)=Pr(A and B)/Pr(B).
In this case A is drawing a black marble and B is having already drawn a
black marble.
Pr(A and B) = (1/2) * [(3/4)^{2} + (1/4)^{2}] = 5/16.
Pr(B) = 1/2.
Pr(AB) = Pr(A and B)/Pr(B) = (5/16)/(1/2) = 10/16 = 5/8.
Michael Shackleford, A.S.A.
  109.  Two bags of M&Ms problem
   You have two bags of m&m candy. One bag has 99 red candies and 1 blue candy. The other bag has 99 blue candies and 1 red candy. You choose a bag at random, open it, and draw just one candy from the bag. If the candy drawn is red what is the probability what is the probability that it was drawn from the bag with 99 reds and 1 blue?
Answer
Problem 109 Answer
Problem 109 Answer
The answer is 99/100.
Michael Shackleford, A.S.A.
Solution
Problem 109 Solution
Problem 109 Solution
Pr(AB)=Pr(AB)/Pr(B), see problem 16 for further discussion of this theorem.
Pr(AB) = Probability that bag with 99 reds is drawn and 1 red is drawn from that
bag = 1/2 * 99/100 = 99/200.
Pr(B) = Probability that a red is drawn = ((1/2)*(99/100) + (1/2)*(1/100)) = 100/200 = 1/2.
Pr(AB)/Pr(B) = (99/200)/(100/200) = 99/100.
Michael Shackleford, A.S.A.
  238.  Two boards and a moat
   A castle is on a square island, surrounded by a moat 25 feet wide. You have two 24 foot boards, but no nails or any other way to attach them to each other. How can you place the boards to get to the island? For extra credit, how is the shortest length the two boards can be and still reach the island?
Answer
Problem 238 Answer
Place one board across a corner of the moat and the other from the center of that board to the corner of the island, making a T shape with the boards.
The shortest length the boards can be is 50*2^{0.5}/3 =~ 23.5702 feet.
Michael Shackleford
Solution
Problem 238 Solution
 Let the ladder be the length 2x.
 From the diagram above, we can set up the equation: (2x)^{2} + (2x)^{2} = (x/2^{0.5} + 25)^{2}
 8x^{2} = (x*2^{0.5}/2 + 25)^{2}
 8x^{2} = x^{2}/2 + 25x*2^{0.5} + 1250
 15x^{2}  50x*sqrt(2)  1250 = 0
 x = [50*2^{0.5} +/ (5000 + 75000)^{0.5}]/30
 x = 25*2^{0.5}/3
 The whole ladder is length 2x, so the answer is 50*2^{0.5}/3 =~ 23.5702 feet.
Michael Shackleford
  30.  Two boats problem
   There is a boat traveling east at a speed of 20 kilometers per hour. A second boat is 50 kilometers directly north of the first boat at time t=0. The second boat can travel at a speed of 30 kilometers per hour. The direction of the second boat at all times is such that the first boat is directly in front of it. You may assume the earth is flat for the purposes of this problem.
What path does the second boat take and at what time will it intercept the first boat?
In the graph above the speed of the fast boat (green) is 2 kph, and the speed of the slow boat (blue) 1 kph. I changed the parameters because I didn't want to give away the answer in the graph.
Answer
Problem 30 answer
Problem 30 answer
Place the slower boat initially at the origin of a coordinate
plane, and equate "north" with the positive yaxis, and east with
the positive xaxis.
x = 75 * [ 1/5*(y/50)^{5/3}  (y/50)^{1/3} ] + 60
It takes 3 hours for the chasing boat to catch the other boat.
Michael Shackleford, A.S.A.
Solution
Problem 30 solution
Problem 30 Solution
Acknowledgements:
The following solution was provided by Alan Curry. Most of
the words are his, although I have edited his solution in places.
Place the slower boat initially at the origin of a coordinate
plane, and equate "north" with the positive yaxis, and east with
the positive xaxis. The faster boat starts at the point (50,0).
The "curve" shall refer to the path of the faster boat. t shall
refer to time since both boats left their starting positions.
There are two ways to indicate the slope at any point along
the curve. One is dy/dx. The other is (x20t)/y, since the
faster boat is always pointing towards the slower boat. Thus
dy/dx=y/(x20t).
The above equation can also be derived this way (optional):
Pick a point on the curve and label it (x_{1},y_{1}). The x_{1} and the y_{1} are
functions of the current time, t_{1}. The tangent to the curve at point
(x_{1},y_{1}) passes through the point (20t_{1},0), which is the current location
of the 20kph boat. The equation of the line is y=mx+b, where m is the value
of dy/dx at (x_{1},y_{1}) and b is some constant chosen to make sure that
(20_{t1},0) is on the line. b must be 20*t_{1}*y'(x_{1}), where I've used y'(x_{1})
to mean the value of dy/dx evaluated at the point (x_{1},y_{1}). The equation of
the tangent line is therefore:
y = y'(x_{1})*x  20*t_{1}*y'(x_{1})
Rearranging that, and phrasing it as a general statement rather than a
particular point (the point x_{1} was arbitrary, so this applies to every point
on the curve).
dy/dx * (x20t) = y
dy/dx = y/(x20t)
Something interesting can be found if you treat x as a function of y,
instead of the other way around:
dx/dy = (x20t)/y
This is justifiable, though it may look like abuse of Leibniz, the
derivative of the inverse of a function is the reciprocal of the derivative
of the function.
Then multiply both sides by y:
y * dx/dy = x  20t
And here's the killer... differentiate with respect to y, using the product
rule on the left hand side:
(1 * dx/dy) + (y * d^{2}x/dy^{2} ) =
dx/dy  (20 * dt/dy)
There's a dx/dy term on each side of the equation, and they now happily
disappear.
[1] y * d^{2}x/dy^{2} = 20 * dt/dy
Now it's time to turn dt/dy into something we can work with. I split
it up as follows:
dt/dy = (dt/ds) * (ds/dy)
Where s is the arc length, the distance traveled so far by the chasing
boat.
dt/ds is known (it's the reciprocal of the chasing boat's speed ds/dt=30kph),
and ds/dy can be expressed in terms of dx and dy by applying the arc length
formula:
The formula for arc length is:
s=Integral of [ (1+(dx/dy)^{2})^{1/2} ] dy
Take the derivative of both sides:
ds/dy=(1+(dx/dy)^{2})^{1/2}, remember that
the square root can be positive or negative.
Here is another way to derive this:
ds^{2} = dx^{2} + dy^{2}
(ds^{2})/(dy^{2}) = (dx^{2})/(dy^{2}) + 1
(ds/dy)^{2} = (dx/dy)^{2} + 1
sqrt both sides...
ds/dy = sqrt((dx/dy)^{2} + 1)
ds/dy = +/sqrt((dx/dy)^{2} + 1)
We need to choose a sign here. Look which way the curve goes. As t increases,
y decreases. At the same time, s (the distance traveled by the chasing boat)
increases. Therefore, ds/dy must be negative.
ds/dy = sqrt((dx/dy)^{2} + 1)
Substituting all this back into the equation marked [1], you get:
y * d^{2}x/dy^{2} = 20 * (1/30) * sqrt((dx/dy)^{2} + 1)
y * d^{2}x/dy^{2} = 2/3 * sqrt((dx/dy)^{2} + 1)
That's a secondorder differential equation, but not one of the extremely
difficult kind. Rewrite it with a new variable u=dx/dy, and it becomes a
firstorder diffeq:
y * du/dy = 2/3 * sqrt(u^{2}+1)
Breathe a sigh of relief, because that's separable. (I spent most of my
solution time needlessly hacking away at this with more advanced diffeq
techniques before I noticed it was separable. D'oh!)
du 2 dy
 =  * 
sqrt(u^2+1) 3 y
Integrate them... (I had to use a table for the left one)
ln(u + sqrt(u^{2}+1)) = 2/3 * ln y + K
I've dispensed with the absolute value stuff, since u+sqrt(u^{2}+1) can't be
negative, and y isn't negative either, for the part of the curve that's
relevant to the original question.
exp() both sides...
u + sqrt(u^{2}+1) = C*y^{2/3}
K was an integration constant; C is exp(K).
To find C, use the point (0,50) on the curve. This is at t=0, when the
target boat is due south of the chasing boat. The curve's tangent at that
point is vertical, so u=dx/dy=0. Plug in u=0 and y=50:
0+1 = C*50^{2/3}
C = 1/50^{2/3}
Put that back in...
u + sqrt(u^{2}+1) = (y/50)^{2/3}
We've still got some differential equations work to do.. u is a
derivative! argh, a derivative in a radical. But it works itself out,
just rationalize it:
sqrt(u^{2}+1) = (y/50)^{2/3}  u
u^{2}+1 = (y/50)^{4/3}  2u(y/50)^{2/3} + u^{2}
Lucked out again. Cancel u^{2}'s...
1 = (y/50)^{4/3}  2u(y/50)^{2/3}
2u(y/50)^{2/3} = (y/50)^{4/3}  1
2u = (y/50)^{2/3}  (y/50)^{2/3}
u = 1/2*[ (y/50)^{2/3}  (y/50)^{2/3} ]
dx/dy = 1/2*[ (y/50)^{2/3}  (y/50)^{2/3} ]
A simple integration:
x = 1/2 * [ 3/5*(y/50)^{5/3}*50  3*(y/50)^{1/3}*50 ] + K
Factor out those 3's and 50's
x = 75 * [ 1/5*(y/50)^{5/3}  (y/50)^{1/3} ] + K
Again look at the starting point. When x is 0, y is 50.
0 = 75 * [ 1/5*1^{5/3}  1^{1/3} ] + K
0 = 75 * [ 1/5  1 ] + K
0 = 60 + K
60 = K
x = 75 * [ 1/5*(y/50)^{5/3}  (y/50)^{1/3} ] + 60
That's the path the chasing boat follows. If that curve has a name, I don't
know what it is.
The chasing boat intercepts the other boat when y=0.
x = 75 * [ 1/5*0  0 ] + 60
x = 60
The boats meet at the point (60,0). Since the slower boat is traveling due
east at 20kph, it must have taken 3 hours to get there.
It takes 3 hours for the chasing boat to catch the other boat.
Below is another solution provided by Tristan Simbulan:
let t = s/30, distance traveled by first boat is 20 * s/30 = 2/3 * s dy/dx=y/( x  2/3 * s), at t = 0 dy/dx is 50/0 is undefinable.
At t=0 dx/dy=0/50=0 is definable.
dx/dy = (x2/3*s)/y, y * dx/dy = x  2/3 * s, take second derivative.
Let first derivative dx/dy = p and second derivative dp/dy, ds/dy = sqrt(1 + (dx/dy)^2) = sqrt( 1+ p^2 ) dy/y = 3/2 * dp/sqrt(1+ p^2), lny + C = 3/2 * ln( p + sqrt( 1 + p^2 )), substitute initial condition y = 50, p = 0.
C = ln 50. p = dx/dy = 0 y = { p + sqrt(1 + p^2 ) }^(3/2) * 50 = { dx/dy + sqrt( 1 + (dx/dy)^2) }^(3/2) * 50
Integration result: 2 * 50^(2/3)x =50^(4/3) * y^(1/3) * 3  { y^(5/3) * 3/5 } + C C = 12/5 * 50^(5/3) at initial x = 0 y = 50 substitute value of C and you get equation of the curve.
At y = 0, x = 60 . divide distance by rate = time = 60/20 = 3 hours
Note: For the general case in which the slow boat
travels at a k.p.h, the fast boat travels at b k.p.h., and the
initial distance is d kilometers the formula of the curve is:
x = 1/2*[ db/(a+b)*(y/d)^{(a+b)/b}  db/(ba)*(y/d)^{(ba)/b}
+ db/(ba)  db/(a+b) ]
The time to interception is [db/(ba)  db/(a+b)]/2a.
Michael Shackleford, A.S.A.
  157.  Two boats problem
   Two boats on opposite sides of a river head towards each other at different speeds. When they pass each other the first time they are 700 yards from one shoreline. They continue to the opposite shoreline, turn around, and move towards each other again. When they pass the second time they are 300 yards from the other shoreline. How wide is the river? Assume both boats travel at a constant speed and ignore factors such as turnaround time and the current of the river.
Answer
Problem 157 Answer
Problem 157 Answer
The answer is 1800 yards.
Michael Shackleford, ASA  November 18, 2000
Solution
Problem 157 Solution
Problem 157 Solution
New Solution
Let's call:
w = width of river
r = rate of the boat which was 700 yards from the destination shore at the time of the first meeting
t = duration of time from start to first meeting
 t=(w700)*r
 3t=(2w300)*r
This step is the humdinger. If you draw the pathes of both ships you'll see that after the first meeting point the combined distance covered was the width of the river. At the second meeting the combined distance traveled was
three times the width of the river. So, at the second meeting time, both boats traveled three times as far compared to the first meeting.
 Equate 3t in equations (1) and (2): 3*(w700)*r = (2w300)*r
 3*(w700) = (2w300)
 3w2100 = 2w300
 w=1800
Old Solution
Define the following variables:
w=width of river
s_{1}=speed of first boat
s_{2}=speed of second boat
t_{1}=time until boats meet the first time
t_{2}=time until boats meet the second time
I shall arbitrarily say that both boats are 700 yards from the shore of original of the first boat at time t_{1}.
The following equations can be inferred from the information given:
(1) w=t_{1}*(s_{1}+s_{2})
(2) s_{1}*t_{1}=700
(3) w+300=s_{1}*t_{2}
(4) 2w300=s_{2}*t_{2}
Adding equations (1) and (2) we get:
(5) 3w=t_{2}*(s_{1}+s_{2})
Substituting s_{1}+s_{2}=w/t_{1} from equation (1) we get:
3w=t_{2}*r/t_{1}
3w*t_{1}=r*t_{2}
(6) 3*t_{1}=t_{2}
Combining equations (3) and (6):
(7) w+300=3*s_{1}*t_{1}
Combining equations (2) and (7):
w+300=2100
w=1800
This problem was asked in the 'Ask Marilyn' column in the November 19, 2000, issue of Parade magazine.
Michael Shackleford, ASA  November 18, 2000
  22.  Two brothers and a flock of x sheep
   Two brothers share a flock of x sheep. They take the sheep to the market and sell each sheep for $x. At the end of the day they put the money from the sales on the table to divide it equally. All money is in $10 bills, except for less than ten excess $1 bills. One at a time they take out $10 bills. The brother who draws first also draws last. The second brother complains about getting one less $10 bill so the first brother offers him all the $1 bills. The second brother still received a total less than the first brother so he asks the first brother to write him a check to balance the things out. How much was the check?
Answer Problem 22 AnswerThe answer is $2. Michael Shackleford, A.S.A. Solution Problem 22 SolutionLet the total number of sheep be 10x+y, where y<10. The total money raised is (10x+y)^{2} = 100x^{2} + 20xy + y^{2}. Regardless of the values of x and y 100x^{2} + 20xy will be divisible by 20. Because the number of $10 bills is odd y^{2} mod 20 must be greater than 10 and less than 20. The only values of y where this is true is 4 and 6, where y^{2} is either 16 or 36. Either way there will be 6 $1 bills left over. Before the check the first brother will have $4 more than the second brother. A $2 check will balance things out.
Thanks to Bill Feldman for suggesting this problem. Michael Shackleford, A.S.A.   52.  Two children problem
   A woman is chosen at random among all women that have two children. She is asked do you have at least one boy, and she answers 'yes.' What is the probability her other child is a boy? Assume every pregnancy has a 50/50 chance to be a boy or a girl.
Answer
Problem 52 Answer
Problem 52 Answer
The answer is 1/3.
Michael Shackleford, A.S.A.
Solution
Problem 52 solution
Problem 52 Solution
Since the probability is 50/50 for a boy or girl in each pregnancy
there are four possibilities for a two child pregnancy, each with
equal probability:
 Boy then boy
 Boy then girl
 Girl then boy
 Girl then girl
We know that for this particular mother the last possibility is
not the case. The other three are still equally likely. In two
of the other possibilities the other child is a girl, in one
it is a boy. Thus the probability that the other child is a boy
is 1/3.
Note: This is similar to the question in
problem 16 which approaches the solution another way.
Michael Shackleford, A.S.A.
  16.  Two coins problem
   A box contains two coins. One coin is heads on both sides and the other is heads on one side and tails on the other. One coin is selected from the box at random and the face of one side is observed. If the face is heads what is the probability that the other side is heads? Note: I have had more email on this problem than any other.
Answer
Problem 16 Answer
Problem 16 Answer
The answer is 2/3.
Michael Shackleford, ASA
Solution
Problem 16 Solution
Problem 16 Solution
Yes, the answer is 2/3. Numerous people have tried to explain why
they think the answer is 1/2, arguing that since both coins have a
head then seeing a head doesn't rule out anything and thus it could
be either coin with equal probability. While it is true that seeing a head does not
rule out anything it still provides valuable information that can not be
ignorred.
To illustrate why the 1/2 answer is wrong let propose a similar
problem. Suppose a soup company canned 2,000,000 cans of tomato
soup. Suppose further that they put the wrong label on half of the
cans. The company then sent 999,999
correct cans and 1 incorrect can to one store and 999,999 incorrect
cans and 1 correct can to another store. To determine which store
has the correct cans you go to one store at random and open one
can. You see it does indeed have tomato soup inside. What is the
probability you went to the store with 999,999 correct cans? It is
not reasonable to assume the probability is 50/50 that you are in
either store. The odds that you would pick the one correct can in the
store with 999,999 incorrect cans are extremely remote. It is true that
before opening a can the odds were 50/50 but once you do open a can that
information gives you a clue that can not be ignorred. In fact the
probability that you are in the store
with the correctly labled cans is 99.9999%.
Now let me introduce Bayes' theorem on conditional probability to explain why the
answer is 2/3:
Let the events A_{1}, ...,A_{k} for a partition of the space
S such that Pr(A_{j} > 0 for j=1,...,k and let B be any event such that
Pr(B) > 0. Then, for i=1,...,k,
Pr(A_{i}  B) =
Pr(A_{i}) * Pr(BA_{i}) /
[ $\Sigma $ (for j=1 to k) Pr(A_{j})*Pr(BA_{j}) ] .
For the problem in question we can reword this to:
Pr(two headed coinhead chosen) = Pr(choosing two headed coin) * Pr(choosing heads given that two headed coin was chosen) /
[Pr(choosing two headed coin) * Pr(choosing heads given that two headed coin was chosen) +
Pr(choosing one headed coin) * Pr(choosing heads given that one headed coin was chosen)] =
1/2 * 1 / [ (1/2 * 1) + (1/2 * 1/2) ] =
1/2 / (1/2 + 1/4) = (1/2)/(3/4) = (1/2)*(4/3) = 2/3.
If this all went over your head think if it another way. There were 3 heads to begin with. 2 of
them have another head on the other side. You could have chosen any head with equal probability
and since 2 out of 3 have a head on the reverse side the answer is 2/3. Of course this answer
would probably not merit very much in partial credit in probability class.
Some other people have questioned my exact wording of this problem, mainly objecting over
the tense of the verb I use for when the coin is chosen. What appears now was taken
almost exactly from a similar problem in Probability and Statistics (second
edition) by Morris H. Degroot on page 63, problem number 5. I changed cards to coins and
eliminated what would be the coin with two tails. The book is a commonly used college
text on the subject and should be above reproach. Here is how they stated a similar problem:
A box contains three cards. One card is red on both sides, one card is green
on both sides, and one card is red on one side and green on the other. One
card is selected from the box at random, and the color on one side is
observed. If this side is green, what is the probability taht the other side of
the card is also green?
Essentially the same problem, but worded using sunken boats, secret compartments, and
bars of gold, is asked on page 603 of Statistics for Business and Economics by
Edwin Mansfield.
A reader sent in the following explanation that he thought might
help:
You have probably received similar emails to the one I am now sending, but
just in case, ...
Perhaps the best way to demonstrate that 2/3 is the correct answer to this
problem is to approach it in a simple intuitive manner, such as:
Denote the double headed coin A and the regular coin B. Further, denote
the sides of the double headed coin A1 (a head) and A2 (another head), and
the sides of the regular coin B1 (a head) and B2 (a tail).
We then have the following possible outcomes:
coin selected side shown other side
A A1 (H) A2 (H)
A A2 (H) A1 (H)
B B1 (H) B2 (T)
B B2 (T) B1 (H)
We exclude the last possibility from consideration since it is not a head.
Three possibilities remain, all equally likely, two of which have a head on
the other side. The required probability is thus 2/3.
Gary.
Here is my favotite email I received on this problem. I x'd out
some private information.
Dear Michael,
I am a biology teacher at Xxxxxxx High School in Xxxxxxx, MA. I
recently was searching for an extra credit question to put on a test
for tenth graders. I used the three coins in a bag problem (#16).
Unfortunately, none of my students got the correct answer. They all
answered 1/2. I have done my best to explain the problem and some of
the students now understand why the correct answer is 2/3. The debate
has spread throughout the whole school. Most teachers and students
beleive the answer to be 1/2. So I started an organization called
Team 2/3  The Probability Masters. Students take an oath to become
official members of the team. The mission of the team is to recruit
as many new members as possible. Team members get a membership card
that has your original question on the back which helps them recruit
new team members. Many people think I have gone off the deep end, but
I find it amazing that so many people including most teachers and the
principal still do not see the truth. If fact, the day after I
started the Team 2/3, the principal got on the morning announcements
and stated the the answer is 1/2 and the group that thinks otherwise
should rethink the problem.
One student created a computer program that simulates 10,000 coin
picks in 1 second. The data clearly show the 2/3 answer. But that
same student insists that the probability is 2/3 only if many trials
are conducted. He doesn't understand that the probability is the same
in the first and last trial. People are stuck on the idea that you
must have either the HH or HT coin and say it is 1 out of two. I
have tried to simplify your solution without the TT coin to help
people understand.
If two coins: HH, HT
Conditional Prob = 1/2 divided by 3/4 = 2/3
Team 2/3 will recruit for two more days and then reveal the truth. We
will try to get the principal to announce that Team 2/3 was correct
all along. All team members will hold their heads high if that
happens. I only have one math teacher on my side at this time. One
of my student team members had a hat made for me that says "Team 2/3
Captain". I wore it on Friday and will wear it this week. We have
almost everyone in the school discussing your coin problem. Many of
the students are giving it a great deal of thought.
We are very confident that the answer is 2/3, however most people
still disagree.
It would be most helpful if you would place a call to the school and
talk to Mr. Xxxxxxx Xxxxxxxxx, the principal. This would be greatly
appreciated by the team, your team, Team 2/3.
Xxxxxxx High School 1(xxx)xxxxxxx
Sincerely,
Troy X. Xxxxxxx
Biology Teacher
P.S. Please feel free to call me at home: 1(xxx)xxxxxxx
I did call the principal and tried to explain why the
answer was 2/3. He seemed pretty entertained that somebody
would call all the way from Baltimore about this. After
going over the problem he seemed to partially understand but
added that "he didn't want to be confused with facts." Later
the Biology teacher wrote me back, here is his second email:
I really appreciate you talking to the principal.
He did talk to me afterward and he seemed very unsure of the correct
answer. Team 2/3 has 42 oathtaking official members including an
English teacher and the guidance secretary. We allowed people to join
unofficially by signing the unofficial membership list. That way they
still get a membership card without taking the oath; it's our "chicken
list". The Latin teacher, the US history techer, and the alternative
education teacher are on that list. Our membership card states"TEAM
2/3 the Probability Masters" and there is a picture of a man holding two
coins. On the back of the card is your original problem. Another team
formed (not as organized as our team) called Team 1/2. They also have
membership cards. Just this past Thursday, a good week after your call,
I reminded the principal that we needed closure to the debate. I just
assumed he believe the answer was 2/3. And I told him it would be nice
to recognize the courage of team 2/3 and congratulate them. He got on
the announcements and stated "After consulting with several mathematical
experts" (and other people heard about the controversy and contacted him
with explanations for 2/3) "the answer to the coin probability problem
is 'probably'
2/3, so congratulations to Team 2/3, you are probably correct. While we
would have liked a stronger statement(without the probablies), we have
accepted the principal's announcement as victory and feel that we have
been successful in our mission.
Thanks for your help,
Troy
Michael Shackleford, A.S.A.
  164.  Two courier pigeons problem
   A courier pigeon departs Las Vegas for Reno at the same time as another courier pigeon departs Reno for Las Vegas. Both pigeons fly at constant speeds, although different from each other. They cross paths 2x miles from Las Vegas. After each arrives at their destination they immediately turn around, going back and forth without breaks. They cross paths the second time x miles from Reno. Where will they cross paths the third time?
Answer
Problem 164 Answer
Problem 164 Answer
They will meet exactly in Las Vegas (where as some have pointed out pigeons always come to meet).
Michael Shackleford, ASA  June 28, 2001
Solution
Problem 164 Solution
Problem 164 Solution
First solve for x. Assume that the distance between Las Vegas and Reno is 1. Call r the pigeon that starts in Reno and v the pigeon that starts in Las Vegas. For any given amount of time the ratio of the distance traveled by r to v will be the same. This ratio at the first meeting is (12x)/2x. At the second meeting this ratio is (2x)/(1+x). Equating these two ratios:
(12x)/2x = (2x)/(1+x).
Then cross multiply...
(12x)(1+x) = 2x(2x)
1x2x^{2} = 4x  2x^{2}
1x = 4x
5x = 1
x = 1/5
So the first time they meet is 2/5 of the way from Vegas and the second time is 1/5 of the way to Reno.
I find the easiest approach at this point is to just follow the two paths as the birds continue to fly and see where they meet. When v has traveled 2/5 of the distance between Vegas and Reno to a point 3x the distance from Reno r will have traveled 3/5 of this distance to 2x the distance from Reno. When v has traveled another 2/5 he will be exactly in Vegas. Conventiently this is also where r will be.
Thanks to Alan Goldberg and Contingencies magazine for this problem.
Michael Shackleford, ASA  June 28, 2001
  201.  Two dice question, twelve before two sixes.
   Two dice are rolled until either a total of 12 or two consecutive totals of 7. What is the probability the 12 is rolled first?
Answer
Problem 201 Solution
Problem 201 Answer
The answer is 7/13.
Michael Shackleford, ASA — Mar. 23, 2009
Solution
Problem 201 Solution
Problem 201 Solution
Question
Two dice are rolled until either a total of 12 or two consecutive totals of 7. What is the probability the 12 is rolled first?
Let's call p the probability that a 12 is rolled first from the first roll, or at any time when the last roll was not a 7.
Let's call q the probability that a 12 is rolled first when the last roll was a 7.
The probability of rolling a 12 first from the first roll can be expressed as:
p = (1/36) + (1/6)*q + (29/36)*p
36p = 1 + 6q + 29p
7p = 1 + 6q
The probability of rolling a 12 first when the last roll was a 7 can be expressed as:
q = (1/36) + (6/36)*0 + (29/36)*p
36q = 1 + 29p
29p = 36q  1
So, we have two equations and two unknowns:
(1) 7p = 1 + 6q
(2) 29p = 36q  1
Multiplying equation (1) by 6:
(3) 42p = 6 + 36q
Multiplying equation (2) by 1
(4) 29p = 36q + 1
Adding equations (3) and (4):
13p = 7
p = 7/13
Substituting 7/13 for p in equation (1):
49/13 = 1 + 6q
49 = 13 + 78 q
78q = 36
q = 36/78
q = 6/13
So, from the first roll the probability that a 12 will be rolled first is 7/13. If any point a 7 is rolled, the probability of a 12 being rolled before two sevens is 6/13.
Michael Shackleford, ASA — Mar. 23, 2009
  6.  Two envelope problem
   There are two envelopes in front of you each with a nonzero number. You will receive an amount of money equal to the final envelope you choose. You are informed one has twice as much money as the other. You are then allowed to select either envelope. After you select one and before opening it you are given the option to change your mind and switch to the other one? You think to yourself that if your envelope has x dollars there is a 50% chance the other one has x/2 dollars and a 50% chance it has 2x dollars. The expected return, you compute, is .5[.5x + 2x]=1.25x which seems like a favorable gamble. Do you switch and why? Assume you are neither risk averse nor risk prone, in other words you will take any good gamble and avoid any bad one.
Answer
Problem 6 Answer
Problem 6 Answer
There is no advantage or disadvantage to switching.
Michael Shackleford, A.S.A.
Solution
Problem 6 Solution
Problem 6 Solution
I get a lot of Emails and forum posts about this problem. Based on feedback given, I've changed my "solution" to this problem numerous times. Suffice it to say, I've never been entirely satisfied with my explanation of why there isn't a 25% expected gain in switching. My readers have never been satisfied with my explanations either.
As of this writing, I feel comfortable that I understand the answer, but I don't know the proper way to explain it. I think there are two ways you can attack the 25% argument but I think one of them is a red herring and there needs be only one.
Suffice it to say, that I recently took down my latest explanation because it was getting too much negative feedback. So, come back later and maybe I'll have another solution to present.
Michael Shackleford, A.S.A.
  227.  Two ferries problem
   Two ferries depart at right angles from opposite shores of a river. Each ferry travels at a consistent speed, although one is faster than the other. The pass each other 700 meters from the nearest shore. Upon reaching the other side, each spends 10 minutes docked. On the return trip they pass each other 400 meters from the opposite shore.
How wide is the river?
Answer Problem 227 Answer
The answer is 1700 meters.
Michael Shackleford, Nov. 28, 2017
Solution Problem 227 Solution
Solution #1
Let's call x the distance between the two meetings points.
We can ignore the 10 minutes each spends turning around and assume the turn around instantly. This is simply obvious.
At the first meeting, the two boats combined would have traveled the width of the river.
At the second meeting, the two boats combined would have traveled three times the width of the river.
Since both boats travel at a consistent speed, both boats would have traveled three times as far at the second meeting as the first.
Let's solve for x looking at the fast boat.
3 × (time traveled until first meeting) = (time traveled until second meeting).
3 × (400+x) = 1800 + 2x.
x = 600.
So the total width of the river is 400+600+700 = 1700 meters.
Solution #2
Let's call:
 x = the distance between the two meetings points.
 f = speed of the fast ferry.
 s = speed of the slow ferry.
 t_{1} = time of first time ferries cross in the river.
 t_{2} = time of second time ferries cross in the river.
We can ignore the 10 minutes each spends turning around and assume the turn around instantly. This is simply obvious.
We all know that distance = rate × time. Let's express this equality at the first meeting for both ferries.
 Fast ferry: x + 400 = f * t_{1}
 Slow ferry: 700 = s * t_{1}
To reorder the above two equations:
 Fast ferry: t_{1} = (x+400)/f
 Slow ferry: t_{1} = 700/s
Those two expressions are equal:
t_{1} = (x+400)/f = t_{1} = 700/s
Let's express the distance=rate*time equality at the second meeting for both ferries.
 Fast ferry: 2x + 1800 = f * t_{2}
 Slow ferry: x + 1500 = s * t_{2}
To reorder the above two equations:
 Fast ferry: t_{2} = (2x+1800)/f
 Slow ferry: t_{2} = (x+1500)/s
Those two expressions are equal:
t_{2} = (2x+1800)/f = t_{1} = (x+1500)/s
It would seem we have two equations and three unknowns. However, we don't have to actually solve for f and s. We can express the the ratio f/s two ways, as follows:
f/s = (x+400)/700 = (1800+2x)/(1500+x).
Cross multiply to get:
x^{2} + 500x  660000.
The quadratic formula easily gives us x=600. So the whole river is 700+600+400 = 1700 meters wide.
My thanks to Marvin Gardener, which can be found as problem 11 (with slightly different distances) in his book My Best Mathematical and Logic Puzzles.
Michael Shackleford, Nov. 28, 2017
  117.  Two incense sticks problem
   A Buddhist monk is told to meditate for exactly 45 minutes. He is given two incense sticks, each of which will burn for exactly an hour, but at variable rates over the length of the stick. In other words, the length it takes a section to burn is not necessarily proportional to its length. You are given plenty of matches and scissors. How do you measure off 45 minutes?
Answer
Problem 117 Answer
 Use the scissors, if necessary, to cut off the nonburning handle of each stick, so that you may light both ends.
 Light three of the four ends at the same time.
 The stick with both ends lit will burn out in 30 minutes. At that time, light the other end of the other stick.
 In 15 more minutes, the other stick will burn out, for a total of 45 minutes.
Michael Shackleford
  136.  Two marble collections problem
   Jack and Jill each have marble collections. The number in Jack's collection in a square number (1,4,9,16, etc). Jack says to Jill, "If you give me all your marbles I'll still have a square number." Jill replies, "If you gave me the number in my collection you would still left left with an even square." What is the least number of marbles Jack has?
Answer
Problem 136 Answer
Problem 136 Answer
Posted May 27, 1999
Jack has 25 and Jill has 24.
I would like to thank Recreations in the Theory of Numbers by Albert H.
Beiler for this problem (#20 on page 296).
Michael Shackleford, ASA
  59.  Two primes problem
   Show that any prime number other than 2 can be expressed as the difference of two squares, where each square is an integer squared.
Answer
Problem 59 Answer
Let p be the prime number.
Let a=(p+1)/2, b=(p1)/2.
Unless p is 2, p is the difference between a^{2} and
b^{2}:
a^{2}  b^{2} = (p^{2}+2p+1)/4  (p^{2}2p+1)/4 = 4p/4 = p.
Thanks to Guy de Kindler
for this one, to Nick Hobson for showing that this property to true
not only of primes but odd numbers and numbers evenly divisible by
four, and to Terry Ryder for pointing out some previous errors.
Michael Shackleford, A.S.A., 10/21/1998
Solution
Problem 59 Solution
Let p be the odd number.
Let p=a^{2}  b^{2}.
a^{2}  b^{2} = (a+b)*(ab).
Let ab=1 and a+b=p.
Adding the two equations yields 2a=p+1, so a=(p+1)/2.
Since b=a1, b=(p1)/1.
Thanks to Guy de Kindler
for this one and to Nick Hobson for showing that this property to true
not only of primes but odd numbers and numbers evenly divisible by
four.
Michael Shackleford, A.S.A.
  180.  Two switches problem #1
   You are trapped in a small phone booth shaped room. In the middle of each side of the room there is a hole. In each hole there is a light switch. You can't see in the holes but you can reach your hands in them and flip the switches. You may stick your hands in any two holes at the same time and flip none, either, or both of the switches as you please. Nothing will happen until you remove both hands from the holes. You succeed if you get all the switches into the same position, after which time you will immediately be released from the room. Unless you escape, after removing your hands the room will spin around, disorienting you so you can't tell which side is which. How can you escape? The fewest possible turns that I know if is five.
Solution
Problem 180 Solution
Problem 180 Solution
Here is the solution. At the end of any step you may win, otherwise proceed to the next step.
 Pick two adjacent holes and turn both switches to on.
 Pick two opposite holes and turn both switches to on. Assuming I didn't win then I must have a DUUU configuration.
 Pick two opposite holes. If one switch is down turn up and win. Otherwise turn one down for a DDUU configuration.
 Pick two adjacent holes. If the switches are in the same position then switch both (and win). If different switch them for a DUDU configuration.
 Pick ossosite holes and switch both switches (and win).
Thanks to Stacy Friedman for this puzzle.
Michael Shackleford, A.S.A.  December 14, 2000
  181.  Two switches problem #2
   You are trapped in a small phone booth shaped room. In the middle of each side of the room there is a hole. In each hole there is a push button that can be in either an off or on setting. You can't see in the holes but you can reach your hands in them and push the buttons. You can't tell by feel whether they are in the on or off position. You may stick your hands in any two holes at the same time and push neither, either, or both of the buttons as you please. Nothing will happen until you remove both hands from the holes. You succeed if you get all the buttons into the same position, after which time you will immediately be released from the room. Unless you escape, after removing your hands the room will spin around, disorienting you so you can't tell which side is which. How can you escape? The fewest possible turns that I know if is seven.
Solution
Problem 181 Solution
Problem 181 Solution
Here is the solution. At the end of any step you may win, otherwise proceed to the next step.
 push opposite buttons
 push adjacent. buttons
 push opposite buttons (if you didn't win by this step then you know
you have a 3/1 situation)
 push any one button (either winning or resulting in a 2/2 situation)
 push opposites (either winning or resulting in an UUDD situation)
 push adj. buttons
 push opposites
Thanks to Stacy Friedman for this puzzle.
Michael Shackleford, A.S.A.  December 14, 2000
  196.  Twoplayer random number game
   You are playing a game with three people, you, an opponent, and a referee. Each will pick a real number between 0 and 1. The referee will chose randomly, with each pick equally likely. The player who comes closer, without going over, will win. If both go over, or both pick the same number, then the game will result in a tie. You know your opponent will chose a number randomly between 0 and 1, each equally likely. What strategy should you employ?
Answer
Problem 196 Answer
Problem 196 Answer
Pick 1/3.
Michael Shackleford, A.S.A.
Solution
Problem 196 Solution
Problem 196 Solution
Link to Rick Percy's solution (PDF, 6K).
Michael Shackleford, A.S.A.
  48.  Volume of dodecahedrom problem
   What is the volume of a dodecahedron with side (of pentagon) equal to 1? Hint: you may wish to solve problem 123 first.
Answer
Problem 48 Answer
Problem 48 Answer
The answer is (15+7*sqr(5))/4 =~ 7.663119
Thanks to Mark Hoyle for some help with this problem.
Michael Shackleford, A.S.A.
Solution
Problem 48 Solution
Problem 48 Solution
From problem 123 we have the area of a pentagon of side 1 to be sqr(25+10*sqr(5))/4. The critical question to be answered is the distance from the center of a face to the center of the dodecahedron.
This solution is going to be hard to explain without pictures so bear with me. Call any corner of the dodecahedron a. Call the three corners surrounding a to be b, c, and d. Call q the point at the center of triangle bcd. Call p the point at the center of the dodecahedron.
From problem 123 we know the distance from two nonadjacent corners in a pentagon to be (1+sqr(5))/2. Knowing this and that cos(30)=sqr(3)/2 we can determine that BQ=(sqr(3)+sqr(15))/6.
It is given that AB=1, using the pythagorian formula we can deduce AQ from AB and BQ, which we know. After going through the math we find AQ to be sqr((3sqr(5))/6).
Lets review what we know so far:
AQ + QP = AP
BP = AP
BQ^{2} + QP^{2} = BP^{2}
BQ = (sqr(3)+sqr(15))/6
AQ = sqr((3sqr(5))/6)
So:
PQ^{2} + BQ^{2} = BP^{2}, or
PQ^{2} + ((sqr(3)+sqr(15))/6)^{2} = BP^{2}, or
(BP  sqr((3sqr(5))/6))^{2} + ((sqr(3)+sqr(15))/6)^{2} = BP^{2} (substituting sqr((3sqr(5))/6) + PQ = BP)
After working through the math we get BP = sqr(6*(3+sqr(5)))/4
From problem 123 we know the distance from the corner of a pentagon to the center of the pentagon is 2*sqr(5+2*sqr(5))/(5+sqr(5)) .
Lets call the center of a pentagon E, where B is one of the corners of the pentagon. We now know the distance from B to P and we know the distance from E to B from problem 123. Next we must work through the pythagorean formula again to solve for EP, which turns out to be sqr((25+11*sqr(5))/40) =~ 1.113516.
The volume of the dodecahedron is composed of 12 pyramids. Each pyramid has base sqr(25+10*sqr(5))/4 and height (25+11*sqr(5))/40. The volume of each of these pyramids is
(1/3) * sqr(25+10*sqr(5))/4 * sqr((25+11*sqr(5))/40)
The volume of the entire dodecahedron is 4 * sqr(25+10*sqr(5))/4 * sqr((25+11*sqr(5))/40) =
sqr(25+10*sqr(5)) * sqr((25+11*sqr(5))/40) =
(1/sqr(40)) * sqr(625+275*sqr(5)+250*sqr(5)+550) =
(1/sqr(40)) * sqr(1175+525*sqr(5)) =
(5/sqr(40)) * sqr(47 + 21*sqr(5)) =
(sqr(10)/4) * sqr(47 + 21*sqr(5)) =
(sqr(5)/4) * sqr(94 + 42*sqr(5)) =
(sqr(5)/4) * sqr(49 + 42*sqr(5) + 45) =
(sqr(5)/4) * sqr(((7+3*sqr(5))^{2}) =
(sqr(5)/4) * (7+3*sqr(5)) =
(15+7*sqr(5))/4 =~ 7.663119
Thanks to Mark Hoyle for some help with this problem.
Michael Shackleford, A.S.A.
  8.  Volume of octahedrom problem
   What is the volume of an octahedron with edge length of 1?
Answer
Problem 8 Answer
Problem 8 Answer
The answer is 2 ^{.5}/3 = 0.471404521
Michael Shackleford, A.S.A.
Solution
Problem 8 Solution
Problem 8 Solution
Michael Shackleford, A.S.A.
  7.  Volume of tetrahedron problem
   What is the volume of an tetrahedron with edge length of 1?
Answer
Problem 7 Answer
Problem 7 Answer
The answer is sqr(2)/12 = 0.11785113
Michael Shackleford, A.S.A.
Solution
Problem 7 Solution
Problem 7 Solution
Solution 1
The following can be derived using the Pythagorean theorem:
The height of a face is 3^{.5}/2.
The area of any face is 3^{.5}/4.
The distance from any corner to the center of a joining face is 1/3^{.5}.
The height of the tetrahedron is (2/3)^{.5}
The area is 1/3*base*height =
(1/3) * 3^{.5}/4 * (2/3)^{.5} = 2^{1/2}/12.
Solution 2
Provided by
Mircea Petrache, of "Liceo scientifico Varano" in Camerino, Italy
The above cube has sides of length 1/sqr(2). The area formed by the diagonals along four of the faces as shown in the diagram is a tetrahedrom. Left over are four pyramids. The base of each pyramid is half of the side of the cube and the height is the height of the cube. Thus the area of each pyramid is:
1/2*(1/sqr(2))^{2} * 1/sqr(2) * 1/3 = (12*sqr(2))^{1}
The total area of the cube is (1/sqr(2))^{3} = (2*sqr(2))^{1}
The area of the tetrahedrom is thus (2*sqr(2))^{1}  4*(12*sqr(2))^{1} =
3/(6*sqr(2))  2/(6*sqr(2)) = 1/(6*sqr(2)) = sqr(2)/12
Michael Shackleford, A.S.A.
  167.  Wagon wheel problem
   The front wheels of a wagon measure 3.5 feet in diameter. The real wheels measure 4.25 feet in diameter. While the wagon is stopped somebody makes a chalk mark on both a front and rear wheel. How far must the wagon travel before both chalks marks return to their initial position at the same time?
Answer
Problem 167 Answer
Problem 167 Answer
59.5*pi =~ 186.92 feet.
Michael Shackleford, ASA  July 27, 2001
Solution
Problem 167 Solution
Problem 167 Solution
The circumference of the front wheel is 3.5*pi and the circumference of the rear wheel is 4.25*pi. We must find the least common multiple of both these numbers. Both these numbers are divisible by .25*pi, so the least common multiple is .25*pi*LCM(14,17) = .25*pi*14*17 = 59.5*pi =~ 186.92.
Michael Shackleford, ASA  July 26, 2001
  121.  Wait for a 6 problem
   You are to roll a die over and over until you get a six. What is the expected total of all throws before you throw a six? By the "total" I mean the sum of the die faces, for example the sequence 1356 would have a total of 9.
Answer
Problem 121 Answer
The answer is 15
Michael Shackleford, A.S.A.
Solution
Problem 121 Solution
Let x denote the answer on any given trial of this experiment.
Let E(x) denote the expectation of x, in other words the answer to this problem.
E(x)=(1+E(x))/6 + (2+E(x))/6 + (3+E(x))/6 + (4+E(x))/6 + (5+E(x))/6
E(x)=(15+5*E(x))/6
E(x)=15
Thanks to Nick Hobson for suggesting a similar problem.
Michael Shackleford, A.S.A.
  43.  Wall and two houses problem
   There is a castle surrounded by a circular wall. The wall has a north gate and a south gate. A yellow house lays 3 miles north of the north gate. A blue house lays 9 miles east of the south gate. What is the radius of the circle formed by the wall?
Answer
Problem 43 Answer
Problem 43 Answer
The answer is 4.5
Michael Shackleford, ASA
Solution
Problem 43 Solution
Problem 43 Solution
Let a be the angle between the line from the yellow house to the north gate
and the line from the yellow house to the blue house.
sin(a) = r/(r+3)
tan(a) = 9/(2r+3)
cos(a) = sqr(1sin^{2}(a)) = sqr(1sin^{2}(r/(r+3))) = sqr(6r+9)/(r+3)
tan(a) = sin(a)/cos(a) = (r/(r+3)) * ((r+3)/sqr(3*(2r+3))) = r/sqr(3*(2r+3))
Equating the two expressions for tan(a):
9/(2r+3) = r/sqr(3*(2r+3))
9 * sqr(3) * sqr(2r+3) = r*(2r+3)
9 * sqr(3) = r * sqr(2r+3)
243 = r^{2}*(2r+3)
2r^{3} + 3r^{2}  243 = 0
A visit to
The MathServ Calculus Toolkit will factor this to (2r9)*(r^{2} + 6r + 27) = 0.
The real solution is r=4.5
Here are two more solutions presented by Heng Cheng Suang:
I would like to suggest another soln to problem 43.
 just by using pythagoras thm (and/or similar triangles)
the solution may be solved too.
Let the triangles be labelled as
A
\
 \E It is noted that angle AED is 90 deg,
 /\ and EC is 9 units (properties of tangents)
/ \
D \
 \

B C
USING ONLY PYTHAGORAS THM

Based on triangle AED (By pyt. thm.), STEP 1
AE = sqrt[(3+r)^2  r^2)] = sqrt(9+6r)
Based on triangle ABC (By pyt. thm.), STEP 2
(3+2r)^2 + 9^2 = [sqrt(9+6r) + 9]^2
Solving STEP 3
4r^4 + 12r^3 + 9r^2  486r  729 = 0
r = 4.5
OR

USING SIM TRIANGLES PROPERTIES

STEP 1 same
STEP 2 is replaced by
9 / (3 + 2r) = r / sqrt (9 + 6r)
STEP 3 same
I would like to thank Elliot Hunter for suggesting this problem which originally appeared
the the November 1937 issue of Popular Science Monthly. The solution is by Dr. Anthony of
the math forum.
Michael Shackleford, ASA
  41.  Water flow and tank problem
   Water flows into a tank at a rate of 1 gallon per second. Water leaves the tank at a rate of 1 gallon per second for each 100 gallons in the tank. The tank is initially empty. How long will it take for the tank to fill with 50 gallons of water?
Answer
Problem 41 Answer
Problem 41 Answer
The answer is 100*ln(2) seconds =~ 69.3 seconds.
Michael Shackleford, A.S.A.
Solution
Problem 41 Solution
Problem 41 Solution
You will need to understand the basics of differential equations
for this one.
Let V_{t} = V = The volume in the tank at time t.
Let K_{x} = Some constant of integration.
V_{t+Δt} = V_{t} + 1*Δt  (Δt/100)*V_{t}.
dv/dt = (V_{t}  V_{t+Δt}) / Δt = 1  V/100.
dv = (1  V/100) dt
100/(100V) dv = dt
Integrate each side:
100*ln(100V)=t + K_{1}
exponentiate each side:
(100V)^{100} = K_{2} * e^{t}
100V = e^{t/100} * K_{3}
V = 100  K_{3}*e^{t/100}
We know V=0 when t=0, thus K_{3}=100
V = 100*(1e^{t/100})
Substitute 50 for V and it is not difficult to solve for t=100*ln(2).
Tristan Simbulan sent in the following solution you may find better:
Let v be the volume at time t and let dv/dt the volume rate of change at time t.
Volume rate of change = rate in  rate out at time t. rate in = 1 gal. per second,
rate out = v/100 * 1 gal. per sec.
Therefore dv/dt = 1  v/100 * 1 ====> dv = 1/100 * (100  v ) dt ======>
dv/(100  v ) = dt/100  ln ( 100  v ) = t/100 + C.
At t = 0 and v = 0, C =  ln 100. v = 100 * { 1  e^(t/100) } , let v = 50 .
Find value of t: e^(t/100) = 2, t = 100 * ln 2
Michael Shackleford, A.S.A.
  210.  What holds more water, a cup or cone?
   Given an equal amount of paper, which paper cup would hold more water, a cylinder or cone shape? Assume the cyliner is open on one end and optimal dimensions in both cases.
Answer Problem 210 Answer
Question
Given an equal amount of paper, which paper cup would hold more water, a cylinder or cone shape? Assume optimal dimensions in both cases.
The answer is the cone holds 7.4569932% more water than the cylinder.
Michael Shackleford, ASA — Feb. 7, 2012
Solution Problem 210 Solution
Question
Given an equal amount of paper, which paper cup would hold more water, a cylinder or cone shape? Assume optimal dimensions in both cases.
We can see from problem 208 that a cylinder of surface area 1 can hold at most 0.108578.
We can see from problem 209 that a cone of surface area 1 can hold at most 0.116675.
The ratio of the volume of the cone to cylinder is 0.116675/0.108578 = 1.074570
The answer is the cone holds 7.4569932% more water than the cylinder.
Michael Shackleford, ASA — Feb. 7, 2012
  232.  Which is more e^pi or pi^e?
   Which is more π^{e} or e^{π}?
Answer
Problem 232 Answer
e^{π} is greater.
Michael Shackleford
Solution   13.  Wine and cigar problem
   Joe's happiness is proportional to the function w^{2}*c (w squared * c), where w stands for daily consumption of glasses of wine, and c stands of daily consumption of number of cigars. Wine costs $3 a glass and cigars cost $2 each. His daily budget for both is $100. How should he divide his money between wine and cigars to maximize his happiness? Assume he does not have to buy integer numbers of either cigars or wine.
Answer
Problem 13 Answer
Problem 13 Answer
He should spend $66.67 on wine and $33.33 on cigars.
Michael Shackleford, ASA
Solution
Problem 13 Solution
Problem 13 Solution
Let U=happiness.
U=w^{2}*c.
3*w + 2*c = 100.
c=(1003w)/2.
U=w^{2}*(1003w)/2.
dU/dw=2w*(1003w)/2 + w^{2}*3/2=0.
Solving for w yields w=200/9.
With wine costing $3 a glass this will cost 3*200/9=$66.67, leaving $33.33 for cigars,
for which he can buy 16.67.
Michael Shackleford, A.S.A.
  54.  Work or pleasure problem
   Your job allows you to work any number of hours per week you desire. Your take home pay is proportional to the number of hours worked (no overtime). After subtracing time for sleeping and routine daily tasks you have 80 hours per week left for work and pleasure. You wish to maximize your income multiplied by the amount of pleasure time you have to enjoy it. How many hours per week should you work?
Answer
Problem 54 Answer
Problem 54 Answer
The answer is 40 hours. Isn't this an
interesting coincidence, or is it?
Michael Shackleford, A.S.A.
Solution
Problem 54 Solution
Problem 54 Solution
Le U be your function the how much you get to enjoy your money per week.
Let h be the number of hours worked per week.
U = h*(80h).
U^{'} = 802h.
Set the derivative equal to 0:
802h=0, thus h=40.
Michael Shackleford, A.S.A.
  50.  World series problem
   The World Series consists of up to seven games, the first team to win four wins the series. Team A will play in their home field games 1, 2, 6, and 7. Games 3, 4, and 5 will be played at team B's home field. Assuming that each time has a 50% chance of winning every game, which team will is more likely have have the home field advantage the majority of the time? For extra credit, what is the expected number of games played in each home field?
Answer
Problem 50 Answer
Problem 50 Answer
The majority of the games are likely to be played
at team A's home field. The expected number of games
at team A's home filed is 188/64, and 184/64 at
team B's home field.
Michael Shackleford, A.S.A.
Solution
Problem 50 solution
Problem 50 Solution
In general the probability of either team winning exactly 3 out of n (n>3) games
is (n1)*(n2)*(n3)/(3*2 ^{n}).
For a series to last exactly n games at least one team must have won exactly
3 out of the last (n1). Thus:
The probability of the series lasting 4 games is 1/8.
The probability of the series lasting 5 games is 1/4.
The probability of the series lasting 6 games is 5/16.
The probability of the series lasting 7 games is 5/16.
The expected number of games played in team A's home field is:
2*(1/8) + 2*(1/4) + 3*(5/16) + 4*(5/16) = 188/64.
The expected number of games played in team B's home field is:
2*(1/8) + 3*(1/4) + 3*(5/16) + 3*(5/16) = 184/64.
Michael Shackleford, A.S.A.
  120.  Yahtzee problem
   What is the expected number of turns needed to form a Yahtzee.
Answer
Problem 120 Answer
Problem 120 Answer
The answer is 151496136/13660416 =~ 11.09015538
Michael Shackleford, ASA
Solution
Problem 120 Solution
Problem 120 Solution
Let E _{x} = expected number of more rolls to obtain a yahtzee given the player
already has an x of a kind.
The problem is to find E_{1}.
E_{5}=0 since the player already has a yahtzee and needs no further rolls.
E_{4}=1+(5/6)*E_{4}+(1/6)*E_{5}
6*E_{4}=6+5*E_{4}+E_{5}
E_{4}=6+E_{5}
E_{4}=6
Next solve for E_{3}
E_{3} = 1 + (25/36)*E_{3} + (10/36)*E_{4} + (1/36)*E_{5}
36*E_{3} = 36 + 25*E_{3} + 10*E_{4} + E_{5}
11*E_{3} = 36 + 10*E_{4} + E_{5}
11*E_{3}  10*E_{4} + E_{5} = 36
Next solve for E_{2}
At this point it starts to get less obvious what the probabilities are to advance closer
to a yahtzee. Lets assume the two of a kind is of the number a. Below are the number of
ways to advance closer and the number of corresponding combinations.
abc: 3!*combin(5,2)=60
abb: 3*5=15
bbb: 5
So there are 60+15+5=80 ways to progress to a three of a kind.
aab: 3*5=15
So there are 15 ways to progress to a four of a kind.
There is obviously only 1 way to progress to a five of a kind.
cde: 3!*combin(5,3)=60
bbc: 5*4*3=60
So there are 60+60=120 ways to stay at a pair.
There are 6^{3}=216 ways to roll 3 dice.
E_{2} = 1 + (120/216)*E_{2} + (80/216)*E_{3} + (15/216)*E_{4} + (1/216)*E_{5}
216*E_{2} = 216 + 120*E_{2} + 80*E_{3} + 15*E_{4} + 1*E_{5}
96*E_{2} = 216 + 80*E_{3} + 15*E_{4} + 1*E_{5}
96*E_{2} = 216 + 80*E_{3} + 15*E_{4}
96*E_{2}  80*E_{3}  15*E_{4} = 216
Next solve for E_{1}. Lets assume that the one of a kind is already an a.
There is obviously 1 way to get to a yahtzee.
Going to a four of a kind:
aaab: 5*4=20
bbbb: 5
20+5=25
Going to a three of a kind:
aabc: combin(5,2)*4!/2! = 10*24/2 = 120
aabb: 5*combin(4,2) = 5*6 = 30
abbb: 5*4 = 20
bccc: 5*4*4 = 80
120+30+20+80 = 250
Going to a pair:
abcd: combin(5,3)*4! = 240
abbc: 5*4*4!/2! = 240
bbcc: combin(5,2)*combin(4,2) = 60
bbcd: 5*combin(4,2)*4!/2! = 360
240+240+60+360 = 900
Staying at one of a kind:
bcde: combin(5,4)*4! = 120
E_{1} = 1 + (120/1296)*E_{1} + (900/1296)*E_{2} + (250/1296)*E_{3} + (25/1296)*E_{4} + (1/1296)*E_{5}
1296*E_{1} = 1296 + 120*E_{1} + 900*E_{2} + 250*E_{3} + 25*E_{4} + 1*E_{5}
1176*E_{1} = 1296 + 900*E_{2} + 250*E_{3} + 25*E_{4} + 1*E_{5}
1176*E_{1} = 1296 + 900*E_{2} + 250*E_{3} + 25*E_{4}
1176*E_{1}  900*E_{2}  250*E_{3}  25*E_{4} = 1296
So now we have four equations and four unknowns and can solve with matrix algebra. To summarize
here is what we know thus far:
1176*E_{1}  900*E_{2}  250*E_{3}  25*E_{4} = 1296
96*E_{2}  80*E_{3}  15*E_{4} = 216
11*E_{3}  10*E_{4} + E_{5} = 36
E_{4}=6
Here is the matrix that needs to be solved.
1176 
900 
250 
25 
1296 
0 
96 
80 
15 
216 
0 
0 
11 
10 
36 
0 
0 
0 
1 
6 
I'll leave the solving to the matrix to you. Here are the final answers.
E_{1} = 151496136/13660416 =~ 11.09015538
E_{2} = 11046/1056 =~ 10.46022727
E_{3} = 96/11 =~ 8.727272727
E_{4} = 6
Thanks to Nick Hobson for suggesting a similar problem.
Michael Shackleford, ASA
  39.  Zeros and ones problem
   What is the smallest integer greater than 0 that can be written entirely with zeros and ones and is evenly divisible by 225?
Answer
Problem 38 Answer
Problem 38 Answer
The answer is 11111111100.
Michael Shackleford, A.S.A.
Solution
Problem 39 Solution
Problem 39 Solution
The prime factorization of 225 is 5*5*3*3. So the answer will be both a multiple of 25 and of 9.
All multiples of 25 end in either 00, 25, 50, or 75. The only one of these composed of 0's and 1's is obviously 00, so the answer must end in 00. The hard part is finding a series of 0's and 1's preceeding the 00 that will make the entire number divisible by 9.
If you didn't already know the following trick then this problem would be very hard. If you did know it then the problem was was likely very easy. The trick is that if the sum of digits of a number is divisible by 9 then the number itself is also divisible by 9. Note that this is true for 3 also. For example the number 17685 is divisible by 9 because 1+7+6+8+5=27, and 27 is divisible by 9.
To prove this let's consider any five digit number, abcde. This number can be expressed as follows.
a*10000 + b*1000 + c*100 + d*10 + e =
a*(9999+1) + b*(999+1) + c*(99+1) + d*(9+1) + e*1 =
a*9999 + b*999 + c*99 + d*9 + a + b + c + d + e =
9*(a*1111 + b*111 + c*11 + d*1) + a + b + c + d + e
Thus 9*(a*1111 + b*111 + c*11 + d*1) is a multiple of 9. So if a+b+c+d+e is also a multiple of 9 then the entire number must be a multiple of 9. Note also that the remainder of abcde/9 is the same as the remainder of (a+b+c+d+e)/9.
The smallest number consisting of all 1's and divisible by 9 is thus 111,111,111. Adding the two zeros at the end results in the answer to the problem: 11,111,111,100.
Michael Shackleford, A.S.A.
 
