$A$ and $B$ are playing a coin tossing game. $A$ starts the game and if he gets a head the game is over and he wins $\$100$ cash. If he gets a tail $B$ will toss the coin and if he gets a head the game is over and $B$ will get $\$100$ cash. The game continuous until one gets a head. Find the probabilities that $B$ will get $\$100$ cash and the probability that $A$ will win $\$100$ cash.
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Let $p$ be the probability that A (ultimately) wins. We condition on the result of the first toss. If it is head, then A wins. If it is a tail, the roles are reversed, and the probability A ultimately wins is $1-p$. Thus $$p=\frac{1}{2}+\frac{1}{2}(1-p).$$
Remark: Alternately, by tracing the conditions under which A wins, namely H, HTH, HTTTH, and so on, we find that the probability A wins is $$\frac{1}{2}\left(1+\frac{1}{4}+\frac{1}{4^2}+\cdots\right).$$ Sum the infinite geometric series.
André Nicolas
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is it for both? – anı Oct 18 '14 at 22:14
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1Solving for $p$ we get $2/3$. Thus the probability B wins is $1/3$. One can also calculate using an infinite series. – André Nicolas Oct 18 '14 at 22:18
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ı get it thank you :) – anı Oct 18 '14 at 22:23
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1You are welcome. The same methods work also with a biased coin. Conditioning on the first result is a useful tool, both for probabilities and for expectations. – André Nicolas Oct 18 '14 at 22:25