# expectation of chess game

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#### Guest

##### Guest
Joe and Bob are playing chess. Joe wins with probability p and Bob wins with probability q and finally the game is a tie with probablity 1-p-q.
I got the Probability Mass Function of the number of games they would play as:
P(X=x) = (1-p-q)^(x-1) (p+q)

I need the expectation of the number of games. So far I have:
(p+q) Summation from 1 to infinity of x (1-p-q)^(x-1)

I was told that I need to take the intergral of this to form a ore managable equation but I'm still stuck. Please help.

#### royhaas

##### Full Member
A derivative would be easier. From $$\displaystyle \sum_{x=1}^\infty \theta^x = \theta/(1-\theta)$$ (sum of a geometric series), take the derivative with respect to $$\displaystyle \theta$$.

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