# How does one calculate the expected number of coin flips for this game to last?

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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This looks a lot like homework, and if so, please add the homework tag. Hint: What is the probability that neither wins when you toss the coin twice? What is the probability that the game is won (don't care by whom) when you toss the coin twice? If you think of your double-flip as a single trial of a new experiment, what kond of probability distribution are you looking at? That is, what is the probability that you double-flip exactly $k$ times at which point either Adam or Bob has won the game? – Dilip Sarwate Apr 30 '12 at 21:11
Let $X$ have geometric distribution. Probably you have information about $E(X)$. Double because a "trial" consists of two flips. – André Nicolas Apr 30 '12 at 21:19
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 $\frac{8}{9}\frac{1}{9^{k-1}}$? – David Faux Apr 30 '12 at 21:19

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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Thanks, but why is the first expected value 1, not $\frac{8}{9}$? – David Faux Apr 30 '12 at 21:21
@David: You always need a flip, you never terminate with no flips. Equivalently you could say $$E(N) = P(\text{terminate}) \cdot (1) + P(\text{continue}) \cdot (1 + E(N)) \\= \frac{8}{9} \cdot 1 + \frac{1}{9} \cdot (1 + E(N)) = 1 + \frac{1}{9} E(N).$$ – TMM Apr 30 '12 at 21:25
@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. – Ross Millikan Apr 30 '12 at 21:27

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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Hmm, well the chance of terminating per round is $\frac{8}{9}$, right? Each player has a $\frac{4}{9}$ chance of winning per round. The game only proceeds to a new round if both flips yield heads. – David Faux Apr 30 '12 at 21:07
@DavidFaux: that is true. Then let n be the expected 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. – Ross Millikan Apr 30 '12 at 21:11