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A game has probability $\frac13$ of winning. Someone would like to play this game and continue to play until he loses two in a row. What is the expected number of playing the game?

Darío A. Gutiérrez
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Interception
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    Any thoughts? Hint: consider the early tosses. Either you get your two consecutive losses quickly, or you restart. – lulu Oct 22 '18 at 14:54
  • theres a $\frac 49$ chance of losing twice, given two arbitrary tosses. – Alexander Day Oct 22 '18 at 14:55
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    Try and write out equations for $\mathbb{E}[T\vert L=0]$ and $\mathbb{E}[T\vert L=1]$ where $T$ is time and $L$ is current number of losses – kcborys Oct 22 '18 at 14:59

1 Answers1

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This is a question on the law of total expectation(which works just like the law of total probability).

EDIT: I just realized I answered the question for winning twice in a row: to apply it to the OP's problem, just change probabilities around.

Let $n =$ the expected length of the game (i.e., the answer you are after).

If the first game is a loss (with probability $2/3$), you have made no progress, and then your expected length is 1+n.

If the first game is a win (with probability $1/3$), you have made some progress: with another win (probability $1/3$), your expected length is 1+1; but with another loss (probability $2/3$), your expected length is $1+(1+n)$ since you are now in the previous case.

So $$ n=(2/3)(1+n) + (1/3)[(1/3)2 + (2/3)(2+n)] $$ and you can solve for $n$.

Matthew C
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  • Does this equation apply for a statement "until he wins 2 times in a row" instead of "until he loses 2 times in a row"? But I got your idea thanks – Interception Oct 22 '18 at 15:37
  • Yes @GerryDunda sorry I thought the question was asking about two wins in a row, however just make the obvious changes for the other case. – Matthew C Oct 22 '18 at 15:55