# Finding the winning probability of the game

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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Just to clarify, they play a game, say head or tail, until one of them has a $2$ games lead onto the other? – Jean-Sébastien Sep 30 '12 at 0:54
I am not sure as in a head tail game, it is possible to have 2 winner at the same time, i think may be consider a simplier case first, i.e.each game always have one winner – abc Sep 30 '12 at 1:01
What is the tie probability? – Jean-Sébastien Sep 30 '12 at 1:07

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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Situation is not that clear, as mentionned in a comment, the games can tie... – Jean-Sébastien Sep 30 '12 at 1:04
@Jean-Sébastien: if that was a possibility then the question could not be answered without knowing the probability of player 2 winning an individual game. abc's comment says "each game always has one winner" – Henry Sep 30 '12 at 1:08
yes right, he said that he could consider a simpler case. I asked him to specify what is the Tie probability, or player 2's – Jean-Sébastien Sep 30 '12 at 1:11
@Jean-Sébastien: The probability of an overall tie is $0$. – Brian M. Scott Sep 30 '12 at 1:11
how do you get that form? – abc Sep 30 '12 at 1:23

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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Would you mind briefly explain how you get the equations – Mathematics Oct 9 '12 at 14:45
We justify the first equation. Suppose the two are tied. Player $1$ can ultimately win if (i) she wins next game and ultimately wins or (ii) she loses next game but ultimately wins. For (i), the probability she wins next game is $r$, and given that, she will be $1$ ahead, and the probability she ultimately wins is by definition $b$. So the probability of (i) is $rb$. For (ii), the probability she loses the next game is $1-r$, and given that, she will be $1$ behind, so by definition the probability she ultimately wins is $c$. So the probability of (ii) is $(1-r)c$. (Continued $\dots$) – André Nicolas Oct 9 '12 at 14:56
Now we add the probabilities of (i) and (ii). We get $a=rb+(1-r)c$. The other two equations are obtained the same way, except that for the third, if Player $1$ is $1$ behind and wins (probability $r$, they are tied, but if she loses the game, then it's all over, Player $1$ has lost. So probability Player $1$ ultimately wins if she is $1$ behind is $ra$. – André Nicolas Oct 9 '12 at 15:01
I did the problem this way because it is a general approach to similar problems. A system can be in any one of several states. In our case there are $3$ states (tied, $1$ ahead, $1$ behind). There are various transition probabilities between states. We form a matrix with these transition probabilities, and multiplying by that matrix lets us trace out the evolution of the system. If you are familiar with Linear Algebra, you will note we found an eigenvector of the matrix with eigenvalue $1$. – André Nicolas Oct 9 '12 at 15:12
What a gorgeous solution, Thank you very much. – Mathematics Oct 9 '12 at 15:23