# Fair Game and Dice

Two players put a dollar in a pot. They decide to throw a pair of dice alternatively. The first one who throws a total of $5$ on both dice wins the pot. How much should the player who starts add to the pot to make this a fair game?

So my interpretation of this problem is that I first throw a dice and then the other person throws a dice afterwards. E.g. if I throw a 1, and if the other person throws a 4 wins the pot. So we are trying to find the expected payout?

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"Throws a total of $5 on both dice"? Please clarify. Also 0% accept rate? – Sev Oct 20 '10 at 18:32 Should just be 5. – PEV Oct 20 '10 at 18:40 Trevor, you should tell us how the rules of the game are. The answer to the question of course depends on that. – Rasmus Oct 20 '10 at 18:42 This is all you are given. – PEV Oct 20 '10 at 18:46 -1 for not defining the problem, even when asked – Ross Millikan Oct 21 '10 at 14:13 show 2 more comments ## 3 Answers First, find the probability p that a 5 is thrown and define q = 1 - p. The chance That the first player wins is C=p+pq^2+pq^4+\cdots; calculate this value. Now set C=(1-C)(1+x) and solve for the extra contribution x. - So player$1$throws on odd throws? On the first throw, the probability that he wins is$p$. On the third throw, the probability that he wins is$pq^2$(e.g 2 failures and 1 success)? But he has only thrown the dice once right? So how can he have 2 failures? – PEV Oct 20 '10 at 19:22 For player 1 to win on turn 3, player 1 must have failed on turn 1, player 2 must have failed on turn 2, and player 1 must have won on turn 3. – Charles Oct 20 '10 at 19:34 add comment Building on your interpretation, let us define the rules as follows: A and B alternately throw a single die. If the sum of the last two throws is 5, the one who just threw wins. Note that as stated, an initial 5 could win, or a series of throws of a different length than 2. The analysis would change, but follow the same route. The below requires precisely two throws to add to 5. Let p be the probability that the first player wins. Let q be the probability that the next player to throw wins given that he has received a chance of winning, that is that the last throw is less than 5. Then if you receive a throw of 5 or 6 your chance of winning is p. So q=1/6 (that you win on this throw) + (1-p)/3 (that you throw 5 or 6 and then win) +(1-q)/2 (that you throw <5, don't win this throw, but finally win. p=(1-p)/3 (that you throw 5 or 6 and win)+2(1-q)/3 (that you throw <5 and win).$q=\frac{1}{6}+\frac{1-p}{3}+\frac{1-q}{2}p=\frac{1-p}{3}+\frac{2(1-q)}{3}$If I have the algebra right, p=15/32 and q=9/16, so the first player should withdraw$2/17.

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Player 1 throws on odd throws and player 2 throws on even throws. This is a geometric distribution. So his distribution is $\left(\frac{8}{9} \right)^{2k} \cdot \frac{1}{9}$ and player 2's distribution is $\left(\frac{8}{9} \right)^{2k-1} \cdot \frac{1}{9}$. So player 1 must give $1/8$.