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$A$ and $B$ are drawing marbles with replacement where there is one white and one black marble. When the individual draws the white ball, they win. Assume $A$ draws first, what is the probability that A wins the game?

Since we only want to look at $P$($A$ wins), we need to look at the (2i + 1)th draw. We also have 1/2 chances of drawing a white marble each trial. From this I have $\frac{1}{2}^{2i+1}$ but I don't understand how to turn this into something I can evaluate as a general probability for the question.

goldenlinx
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2 Answers2

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Set up a tree where $A$ has a $\frac{1}{2}$ chance of winning on the first draw and an equal chance of not winning. Along the non-winning branch, there are again two possibilities: $B$ wins or $B$ doesn't in which case the probability of $A$ winning is the same as the original probability.

This should give $$P(A)=\frac{1}{2} +\frac{1}{4}P(A)\implies P(A)=\frac{2}{3}$$

John Douma
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Let $p$ be the probability that $A$ wins. Then $1-p = \frac p2$, since if $A$ doesn't win on the first draw, which will happen $\frac 12$ of the time, then the probability ($1-p$) that $B$ wins is $p$ (because $B$ is now in the position $A$ was in). Thus, $\frac {3p}{2}=1$, so $p= \frac 23$.

More generally, if the probability that a player wins on a specific draw is a constant $c$, then the probability $1-p$ that $B$ will win is a weighted average of $0 \cdot c+p (1-c)$, so $1-p=(1-c)p=p-cp$, so $1=p(2-c)$ and $p=\frac{1}{2-c}$.

Robert Shore
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