Doubling Dice Game
Probability puzzles require you to weigh all the possibilities and pick the most likely outcome.
You are playing a game of dice with two friends, Tom and Harry. Each person has a cup with six dice. Each round the players ante $1 each and then all slam their dice cup openend down on the table. Each player lifts the edge of their cup to look at the roll of their six dice, but keeps their roll concealed from the other players.
On their turn each player may either pass or double the antes for all the players. After the third player's turn all the dice are revealed to determine the winner.
The winner is the person with the most dice with the same value. If two or more players have the same number of matching dice, then the higher value of the dice wins. For example, four ones beat three sixes, but three sixes beat three fives. If two or more players have the same number and value for the winning roll, then they split the antes. The starting person rotates so that the person who went first in a round goes last in the next round.
On the first round you roll a pair of sixes. Harry passed on his turn and now it's your turn. You know that Harry would have doubled the ante if he had three of a kind or better, but that he wouldn't double the ante with a pair when having to go first. So Harry could have the same roll as you, but you're not worried about Harry beating you. You also know that Tom will also double the ante on his turn if he has three of a kind or better, but that he won't double it with a pair.
Should you pass or double the antes?
Bonus question: Does your action change if instead of splitting the antes on a tie, the antes instead stay in the pot for the next round?
HintYou need to first determine the probability of each outcome (win/lose/tie) and then determine the expected value for each outcome. The expected value of an outcome is determined by multiplying the probability of the outcome by the net gain/loss associated with that outcome. Add up all the expected probabilities for each outcome to get the expected value for each action.
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Answer
You should double the antes.
You want to take the action that maximizes your expected value. The expected value is the sum of the products of the probability of each outcome multiplied by the gain or loss for the outcome.
First you need to determine the probability that Tom has a higher roll, and the probability that Tom or Harry or both have the same roll as you.
There are 6^6 = 46656 possible rolls of six dice.
To roll six of a kind, all the dice must be one of the six values, so there are six ways to get six of a kind.
To roll five of a kind, five of the dice must be one of the six values and then one of the six dice must be one of the five remaining values, so there are:
6*6*5 = 180 ways to get five of a kind.
To roll four of a kind, choose four of the six dice to be one of the six values and each of the remaining two dice is any of the remaining five values, so there are:
C(6,4)*6*5^2 = 15*6*25 = 2250 ways to get four of a kind.
(The notation C(6,4) reads as "six choose four". The binomial coefficient C(n,k) is computed as n! / (k!*(nk)!). C(6,4) = 6! / (4!*(64)!) = (6*5*4*3*2*1) / (4*3*2*1*(2*1)) The 4*3*2*1 in the numerator and denominator cancel out to leave C(6,4) = (6*5) / (2*1) = 30/2 = 15.)
Three of a kind can be formed either as two three of a kinds or as three of a kind and three other dice from one of the other five values that aren't all three the same value. So there are:
C(6,2)*C(6,3) + 6*C(6,3)*(5^35)
= 15*20 + 6*20*120 = 14700 ways to roll three of a kind.
This gives:
6 + 180 + 2250 + 14700 = 17136
ways to roll three of a kind or better for a probability of
17136 / 46656 = 119 / 324 ~ .3673
A pair of sixes can be formed as a pair of sixes plus either two pairs, or one pair and two other values, or four different values. So first choose two of the six dice as the sixes and then either:
a) choose two of the remaining five values for the two other pair and two of the remaining four dice for one of the values; or
b) choose one of the remaining five values for the other pair and two of the remaining four dice for the pair and one of the four remaining values for the first remaining die and one of the three remaining values for the last die; or
c) choose four of the remaining five values in any of the 4! arrangements (i.e. the permutation of four of the remaining five values).
a: C(6,2)*C(5,2)*C(4,2) = 15*10*6 = 900
b: C(6,2)*5*C(4,2)*4*3 = 15*5*6*4*3 = 5400
c: C(6,2)*C(5,4)*4! = 15*5*24 = 1800
900 + 5400 + 1800 = 8100 ways to roll a pair of sixes for a probability of
8100 / 46656 = 25 / 144 ~ .1736
So the probability of the outcomes for a pair of sixes versus a random roll is:
Lose: .3673
Tie: .1736
Win: 1(.3673+.1736) = .4591
Now calculate the expected values for each action: either pass or double. The possible outcomes are win, lose, or tie with one and beat the other, or all three tie. Since you assume your roll either beats or ties Harry, the probabilities of each of these is:
Tie Harry: .1736 / (.1736 + .3673) = .2744
Beat Harry: .3673 / (.1736 + .3673) = .7256
The probability of each outcome is:
Tie both: .1736 * .2744 = .0476
Tie one, beat the other: .1736*.7256 + .2744*.4591 = .2519
Win both: .4591*.7256 = .3331
Lose: .3673
If you pass, then you will lose $2 if Tom has a better roll since Tom will double the antes; you will win $2 if neither Tom nor Harry has a pair of sixes or better since nobody doubles the antes; and you will win $0.50 if you beat one and tie the other. You just get your money back if all three tie. So the expected value of passing is:
($2)*.3673 + $2*.3331 + $0.50*.2519 = $0.0577 (less than 6 cents positive expectation)
If you double, then you will lose $4 if Tom has a better roll since Tom will double the antes; you will win $4 if neither Tom nor Harry has a pair of sixes or better since nobody doubles the antes; and you will win $1 if you beat one and tie the other. You just get your money back if all three tie. All the win/loss values are doubled, so the expected value is also doubled to $0.115.
It may seem counterintuitive, but since Tom will only double when he wins, his action has no impact on the expected value. Since you have a positive expected value, even though it is a small positive, you should double the antes to double your expectation.
Bonus question:
You should pass if the money stays in the pot on any tie.
The reason is twofold.
First, your expectation when passing is now:
($2)*.3673 + $2*.3331 = $0.0683 (almost 7 cents negative expectation)
Second, you will be going first next round. First is a disadvantages position since you must act without any knowledge about your opponent's rolls. For this reason, you would not want to increase the size of the starting pot when you have to go first, so even if you had a zero or small positive expectation in this case, you might still not want to double the antes because of the significant probability of a tie.
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Comments
dreamndesire
Feb 11, 2009
 i understood half of the question and very little of the answer, but nice! 
Zag24
Feb 14, 2009
 It wasn't quite clear to me what you meant by "doubling the antes." I assume the other players have to match your increase (or it would just be stupid) but do they have to match it or can they fold? 
javaguru
Feb 14, 2009
 I tried to make it clear: "double the antes for all the players". I called them antes instead of bets, since that's what you have to put in just to participate in the wager, whereas a bet implies something you can either accept or not.
So no, they can't drop. All players will have to double the money they have at risk. 
(user deleted)
May 10, 2009
 I think there is an error in the calculation of the dice probabilities. Specifically with regards to making three of a kind, the number of ways you can make 2 threes of a kind is C(6,3) * 5 * 6, where C(6,3) is the first three dice assigned to some value, 5 is the number of feasible solutions for the other 3 dice, and the 6 represents the number of ways for each of the dice. 
javaguru
May 12, 2009
 Nate, the problem with C(6,3) * 5 * 6 is that you count each combination of triples twice: once as three ones plus three twos, and then again as three twos plus three ones.
Here are all the combinations of two different numbers from 1 to 6: (1,2), (1,3), (1,4), (1,5), (1,6), (2,3), (2,4), (2,5), (2,6), (3,4), (3,5), (3,6), (4,5), (4,6), (5,6).
Your approach also counts (2,1), (3,1), (4,1), ... 
azbycx
Jun 09, 2010
 That was a long expanation. I got part of it, but not all. 
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