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.
  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.
  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.
  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.
  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.
  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
This is a well known problem known as the "two envelope paradox." Sometimes it is asked where the player may open the first envelope before switching. The argument in favor of switching is that the other envelope has either half, or twice, as much, with 50% probability each. If there is $x in the first envelops, then the expected value of the other envelope is ($2x+$0.5x)/2 = $1.25x. However, this argument could be used to go back and forth infinitely, if the first envelope is not opened. If you know the option will be granted in advance, you could use it to save the trouble of switching and pick the other envelope in the first place, but then you could use it again to switch back again, infinitely.
Whether or not the first envelope is opened, you can not blindly say the other envelope might have twice as much. To make an extreme example, if the total amount of money in the world is t, and the first envelope has more than t/2, then it is impossible for the other envelope to have twice as much. More practially, if the person offering the money must have a finite amount to give. So we can't blindly say there is a 50% chance of doubling if the first envelope. For example, if the person offering the envelopes has exactly $1,000,000 dollars, then the most he can offer is $500,000 and $1,000,000. So if the first envelope had $1,000,000, then it would be impossible for the other envelope to have $2,000,000.
However let's take the case where a billionaire is hosting the game, and the first envelope has less than $100. Any amount under $10,000 would mean nothing to the host. So bumping up against the maximum is apparently not an issue. A million players could use the expected value to switch, but would they be better off? Indeed, on average half would end with 200% of the first envelope and half 50%, for an average amount in the second envelope of 125% the first.
However, what is important is not the rate of increase, but the amount of the increase. As a group, switching will not have helped. This is the hard part to explain but when the switch was the right decision, the player is doubling off of the small amount, and when it is wrong the player is halving from a larger amount. Either way, the decision to switch will result a win or loss of the amount in the smaller envelope. So, the expected rate of increase is 25%, but the expected amount of increase is $0. So although the expected rate of increase is 25%, that does not mean the other envelope will have 25% more. The second envelope has a 100% chance of having 100% more, if the first envelope is the smaller amount, and a 100% chance of having 50% less if the first envelope is the larger amount. So, once the first envelope is chosen the second envelope is no longer a random variable.
To backtrack, in the interests of making another point, we can't blindly say the other envelope might have twice as much. As mentioned, if the chosen envelope had more than half the wealth of the host, then the other couldn't have twice as much. However, let's suppose the host promised to put in small amounts compared to his weatlth, to nix this argument. However, it would still aplly. Suppose the wealth of the host was $1,000,000. If he guarantees the larger envelope will have $500,000 or less, then any amount between $250,000 and $500,000 must be the larger envelope and the 50/50 argument wouldn't apply. We could keep going down and down in amounts recursively to show that no amount is safe to using the 50/50 argument.
Suffice it to say I still don't sleep well at night if this problems comes to mind. I'm not sure what is the correct reason the 50/50 argument doesn't hold or if multiple reasons hold water. Sorry if this solution was less than satisfactory.
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.
  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.
  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.
  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
 
