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   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
  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
  209.  Maximum volume of a cone problem
   A cone with an open top, has a surface area of S. What is the maximum volume it can hold?
Answer Problem 209 Answer
Question
A cone with an open top, has a surface area of S. What is the maximum volume it can hold?
The answer is 3^^{5/4} × ((2×S^{3})/(3×pi))^{1/2}
You may also wish to know that:
height=(S^{2}pi^{2}×r^{4})^{1/2}/(pi×r)
radius=3^{1/4}×(S/pi)^{1/2}
Michael Shackleford, ASA — Feb. 7, 2012
Solution Problem 209 Solution
Question
A cone with an open top, has a surface area of S. What is the maximum volume it can hold?
Let h = height of cone.
Let r = radius of cone.
Let S = surface area of cone.
Let V = volume of cone.
The formula for the Volume of a cone is V=pi×r ^{2}×h/3.
The formula for the surface area of a cone is S=pi×r×(h ^{2}+r ^{2}) ^{1/2}.
The above can be found by cutting the cone from the tip to a point on the edge and then flattening it out. I'll leave that part to you.
Solving for h in terms of S...
h = (S ^{2}pi ^{2}×r ^{4}) ^{1/2}/(pi×r)
Then plug that expression for h in into the V formula...
V = (1/3)×r×(S ^{2}pi ^{2}×r ^{4}) ^{1/2}
Next, take the derivative of V and set equal to 0. Don't forget the product rule: (uv)' = u×v' + u'×v
V' = (1/3)×r×(1/2)×(S ^{2}pi ^{2}×r ^{4})^(1/2) × 4×pi ^{2}×r ^{3} + (1/3)×(S ^{2}pi ^{2}×r ^{4}) ^{1/2} = 0
V = (2/3)×pi ^{2}×r ^{4}×(S ^{2}pi ^{2}×r ^{4})^(1/2) + (1/3)×(S ^{2}pi ^{2}×r ^{4}) ^{1/2} = 0
((4×pi ^{2}×r ^{4})/(6×(S ^{2}pi ^{2}×r ^{4}) ^{1/2}) = (S ^{2}pi ^{2}×r ^{4}) ^{1/2}/3
12×pi ^{2}×r ^{4} = 6×(S ^{2}pi ^{2}×r ^{4})
2×pi ^{2}×r ^{4} = S ^{2}  pi ^{2}×r ^{4}
S ^{2} = 2×pi ^{2}×r ^{4} + pi ^{2}×r ^{4}
S ^{2} = 3×pi ^{2}×r ^{4}
r ^{4} = (1/3)×(S/pi) ^{2}
r = 3 ^{1/4} × (S/pi) ^{1/2}
Next, put those values back into the equation for V...
V = (1/3)×r×(S ^{2}  pi ^{2}×(1/3)×(S/pi) ^{2}) ^{1/2}
V = (1/3)×r×(S ^{2}  S ^{2}/3) ^{1/2}
V = (1/3)×r×(2×S ^{2}/3) ^{1/2}
V = (1/3)×3 ^{1/4}×(S/pi) ^{1/2}×(2×S ^{2}/3) ^{1/2}
V = 3 ^{5/4} × (2×S ^{3}/(3×pi)) ^{1/2}
For example, here are the values for a surface area of 1...
S = 1.000000
r = 0.428691
distance from tip to edge = 0.742515
h = 0.606261
V = 0.116675
My thanks to Ed in Connecticut for his help with this solution.
Michael Shackleford, ASA — Feb. 7, 2012
  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
  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
  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
  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
  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
  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 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.
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, ASA — 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
 
