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In this question this game is presented:

Suppose in an independent game which has 2 players, player 1 and player 2, the probability of player 1 to win each game is r. To be the overall winner of the game, one of the players needs to win 2 more games than the other.

Now, instead of asking for the probability of player 1 to win, I want to know what is the expected number played to end the game?

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    Since there are already answers to your earlier question, it is good to ask a new one instead of retroactively changing the old one. But please make your new question self-contained so it contains all the necessary information rather than just referring back to the earlier one. (It is good _also_ to link to the old question, but it should not be necessary to follow the link in order to figure out what you're asking).2012-10-09

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Let $a$ be the expected number of additional games that need to be played if the two players are tied in wins. In particular, $a$ is our required number, since the players start off tied.

Let $b$ be the expected number of additional games if Player $1$ is leading by $1$. And let $c$ be expected number of additional games if Player $1$ is trailing by $1$. We have the equations $\begin{align}a&=1+rb+(1-r)c,\\ b&=1+(1-r)a, \\ c&=1+ra.\end{align}$

Use the above system of linear equations to find $a$.

Remarks: $1.$ To justify the first equation, note that for sure we will be playing one more game. That's the $1$ in the equation. And we will not be finished. With probability $r$ our expected number of games beyond that will be $b$, and with probability $1-r$ our expected number of games beyond that will be $c$. That yields the first equation. It may be prettier and clearer to write the equation as $a=r(1+b)+(1-r)(1+c)$.

The justifications for the other two equations are similar. Again, one might like to write the second equation as $b=r(1)+(1-r)(a+1)$, and do a similar rewrite of the third equation.

$2.$ As pointed out by Marc van Leeuwen, the argument tacitly assumes that the expectations exist. To show that they do is not difficult. Whatever $r$ is, the probability of two opposite results in a row is $2r(1-r)$, which is $\le 1/2$. So the probability that a game length $2n$ or more is $\le (1/2)^n$, and therefore the expected length is finite.

$3.$ We used a strategy broadly similar to the one used in the related question about the probability that Player $1$ wins. This kind of strategy tends to work more widely for expectations than for probabilities, because of the linearity of expectation.

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    The theory of stochastic processes.2012-10-10