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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. What is the probability that player 1 will be the overall winner?

My sketch to solve the question: Note that to be the overall winner, one player should have won 2 games consecutively. So if player 1 is the winner, the outcome should either a draw, i.e. each player wins a game consecutively or player one won 2 games consecutively. But I am not sure how to start calculating.

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    What is the tie probability?2012-09-30

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Let $a$ be the probability that the first player (ultimately) wins if the two players are tied in wins. Let $b$ be the probability that she wins if she is $1$ ahead. And let $c$ be the probability she wins if she is $1$ behind. We have the equations $\begin{align}a&=rb+(1-r)c,\\ b&=r+(1-r)a, \\ c&=ra.\end{align}$

Solve the system of linear equations.

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    i see, at the very begining, the circumstances is equivalent to tie so a is the answer2012-10-09
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The probability player 1 wins the first two games is $r^2$ while the probability player 2 wins the first two is $(1-r)^2$; otherwise they start again.

So the probability player 1 wins the first two games given that either of the players does is $\dfrac{r^2}{r^2 + (1-r)^2}$ and this is therefore the probability overall that player 1 wins overall.

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    @abc: Conditional probability gives $\frac{r^2}{r^2 + (1-r)^2}$, which then needs to be multiplied by the probability that the overall match finishes at a particular time and then added up, i.e multiplied by $1$.2012-09-30