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A biased coin yields heads with probability $\frac{1}{3}$ and tails with probability $\frac{2}{3}$. Adam and Bob use this coin to play a game, in which I flip the coin twice. If both flips are tails, Adam wins. If the flips differ, then Bob wins. Otherwise, this process is immediately repeated.

How many flips are expected in a game (until either player wins)?

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    Thanks! This actually isn't homework. I'm trying to teach myself probability with some problems with no solutions :/ The chance of terminating per round is $\frac{8}{9}$ since either player has a $\frac{4}{9}$ chance of winning. I think the probability of winning after$k$times is $\frac{8}{9}\frac{1}{9^{k-1}}$?2012-04-30

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Let $N$ be the number of flips needed, and let $E(N)$ be the expected number of flips needed. You always need a flip, but with probability $\frac{1}{9}$ you will not make progress and need to "start over" and flip again, with another $E(N)$ expected flips needed to finish. So

$E(N) = 1 + \frac{1}{9} E(N).$

Solving for $E(N)$ gives $\color{blue}{E(N) = \displaystyle\frac{9}{8}}$.

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    @DavidFaux: because it is the first round you are counting. Then the 1/9 is the chance you have to keep going. This is a general property of geometric distributions (think of summing a geometric series)-the expected number of rounds is the inverse of the probability of stopping on one round.2012-04-30
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Hint: what is the chance the game terminates on the first round? If it doesn't, the chance of terminating on the next round is the same.

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    @DavidFaux: that is true. Then let n be the ex$p$ected number of rounds. You either end the first round (p=8/9), or (p=1/9)you have one round in the bank and the same expected number of rounds left.2012-04-30