expectation of chess game

G

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
Joined
Dec 14, 2005
Messages
832
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 \(\displaystyle \theta\).
 
G

Guest

Guest
Thanks for the reply.
If you're still there, the final answer I got is:
(p+q)/(1-x)^2
Hope this is fight
 
G

Guest

Guest
also, the answer is in the form 1/(1-x)^2. Doesn't this have a problem with x=1? (or someone winning the first game which is totally possible)
 
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