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Players $A$ and $B$ decide to play chess until one of them wins. Assume games are independent with $P(A\text{ wins})=0.3$, $P(B\text{ wins})=0.25$, $P(\text{draw})=0.45$ on each game. If the game ends in a draw another game will be played. Find the probability $A$ wins before $B$.

Since the games are independent, I can simply calculate $P(A \text{ wins} \mid \text{somebody wins})$ right? The textbook does not have a solution.

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up vote 2 down vote accepted

Yes. This is because we have:

$$\Pr(A \text{ wins first}) = \sum_{n \ge 0} 0.3 \cdot 0.45^n$$

which by the geometric series can be evaluated to:

$$\sum_{n \ge 0} 0.3\cdot 0.45^n = 0.3\cdot \sum_{n \ge 0} 0.45^n = 0.3 \cdot \frac1{1-0.45} =\frac{0.3}{1-0.45}$$

and the latter expression equals $\Pr(A \text{ wins}\mid \text{someone wins})$ because "$A$ wins" and "nobody wins" are mutually exclusive.

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By independence, the probability the series goes on more than $n$ games is $(0.45)^n$, so the probability the game goes on forever is $0$.

Because of this, draws are irrelevant, so effectively we could assume that we are playing a game in which the probability A wins is $p=\frac{0.3}{0.55}$, and B wins with probability $1-p$.

In the modified game, A wins if and only if she wins the first game. This has probability $p$.

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By the law of total Probability you have $$P(A\mathrm{\,wins})=P(\mathrm{someone\,wins})P(A\mathrm{\,wins}|\mathrm{someone\,wins})+P(\mathrm{nobody\,wins}|)P(A\mathrm{\,wins}|\mathrm{nobody\,wins})$$ $P(\mathrm{nobody\,wins})$ means that all the games are draw so $P(\mathrm{nobody \, wins})\leq P(\mathrm{The \,first \,n\,games\,are\,draw})=(0,45)^n$ since this holds for every $n$ it implies $P(\mathrm{nobody\,wins})=0$. Which makes your assertion correct.

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Probability of A winning(before B) is =$0.3\sum_{i=0}^{\infty}(0.45)^i=0.3\times \frac{1}{1-0.45}=\frac{0.3}{0.55}=30/55$

Reason: If A wins after the first game it must be with prob $0.3$, if he wins after the 2nd game then the first game must be a draw and A must win the 2nd game, this happens with prob. $0.45\times 0.3$in this way it goes on. Ultimately all these prob. must be added to get the required prob.

Yes it is always possible to calculate the prob. of A winning given that somebody wins.

In the same way you can calculate the prob. of B winning before A.Adding both these prob. you will get the prob. of somebody winning.Then using the formula for conditional prob. you can find the above cond. prob.

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The OP does not benefit from an ASAP braindump as you seem to have produced as much as from a proper, well-written and clear answer. Especially with the consideration that there is no limited amount of space on which the answer is to fit, there is absolutely no reason to make it appear as rushed as the above. – Lord_Farin Apr 5 '13 at 15:34

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