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Assume player $A$, $B$ throw a pair of fair dices in turn. Player $A$ is the winner if the sum of two dices is $5$ at the round he throw and player $B$ is the winner if the sum of the dices is $9$ at his turn. Once any one of the player achieve their number, the game end and he will become the winner. Suppose $A$ throw first, what's the probability $A$ will be the winner.

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3 Answers

up vote 2 down vote accepted

Player A may win at the first leg, which happens with probability p = P(sum of two dice is 5). Or, later on, but only if A's first leg and B's first leg are both unsuccessful, and then everything starts again.

Hence the probability w that A wins is w = p + (1-p)(1-q)w, where q = P(sum of two dice is 9). Solving for w yields w = p/(p+q-pq).

Thus, when p = q (as in the question), the first player wins with probability w = 1/(2-p) and the second player wins with probability 1 - w = (1-p)/(2-p) (in the question, p = q = 1/9 hence w = 9/17).

On the contrary, to get equal probabilities for both players, one should require that p = q(1-p), that is, p = 1/(n+1) and q = 1/n for some n > 1.

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I've been reluctantly convinced that the use of "dice" as a singular has become quite usual, but that doesn't imply the same for the use of "dices" as a plural :-) –  joriki Nov 5 '12 at 13:10
    
Now one of us screw up, probably me on this early morning... gonna re check after my class –  Jean-Sébastien Nov 5 '12 at 13:11
    
@did So I was right :P. Your shortcut is quicker though –  Jean-Sébastien Nov 5 '12 at 13:23
    
yup, quicker, btw, it is a method to solve certain type or probability question? –  Mathematics Nov 5 '12 at 13:30
    
Many problems in probability cna be solved with symmetry. You just need to see it. Sometimes the long road gives other insight though –  Jean-Sébastien Nov 5 '12 at 13:37
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In general, let $p$ be the probability that $A$ wins the game at any given (odd) round and $q$ the probability that $B$ wins at any given (even) round. Note that $p+q\neq 1$ . We condition on $A$ winning the game his turn number $k$, $k=1,2,\ldots$

  • $A$ wins on his first turn with probability $p$
  • $A$ wins on his second turn with probability $(1-p)(1-q)p$. This is because $A$ lost his first turn, as did $B$.
  • $A$ wins on his third turn with probability $(1-p)^2(1-q)^2p$
  • $A$ in general, $A$ wins on his $k$th turn with probability $(1-p)^{k-1}(1-q)^{k-1}p$.

Summing all these give $$ \mathbb{P}(A \text{wins})=\sum_{k=0}^{\infty}(1-p)^{k}(1-q)^{k}p=\frac{p}{q+p-qp} $$ Taking $q=p=\frac{4}{36}$, this gives that $A$ wins with probability $9/17$.

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Name the dice outcomes at each turn X and Y. $p(X+Y = 5) = p(X+Y = 9) = \frac{4}{36} = \frac{1}{9}$. You can compute these by simply counting the cases when the sum is 5 and 9.

A will win if he wins on the first roll or if he fails and B also fails and then he wins, etc.

Therefore:

$$p(A_{win}) = \frac{1}{9}\sum_{i=0}^{\infty} \left(\frac{8}{9}\right)^{i} \left(\frac{8}{9}\right)^{i} = \left(\frac{1}{9}\right) \sum_{i=0}^{\infty} \left(\frac{64}{81}\right)^{i}$$

Now you can apply the formula for the sum of the geometric series:

$$p(A_{win}) = \frac{1}{9} \cdot \lim_{n \to +\infty} \frac{1 - \left(\frac{64}{81}\right)^{n}}{1 - \left(\frac{64}{81}\right)} = \frac{1}{9} \cdot \frac{1}{1 - \left(\frac{64}{81}\right)} = \frac{81}{153} = \frac{9}{17}$$

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This is wrong because p(X+Y = 5) = 4/36, sorry. –  pedrosorio Nov 5 '12 at 13:17
    
And it is corrected! –  pedrosorio Nov 5 '12 at 21:52
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