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        riddles >> easy >> Chess Players' Career
        
(Message started by: Johno-G on Jan 17^th, 2003, 9:02am)

Title: Chess Players' Career
Post by Johno-G on Jan 17^th, 2003, 9:02am

At the start of his profesional career, a chess player decides that he will retire after his very first defeat. He has a probability p of winning any given game.
x is the number of games of chess the player plays in his career.
k = 1,2,3,4,...

Prove that P[x=k] = p(1-p)^(k-1) and verify that the probabilities for all k sum to 1.

Assume each game to be independant of each other.

Title: Re: Chess Players' Career
Post by BNC on Jan 17^th, 2003, 9:28am

Is P[x=k] the probability of the player retiring after the k'th game? Then p should be the probability of loosing each game.

Probability of retiring after game 1: P[x=1]=p
Probability of retiring after game 2: winning game 1, then loosing game 2: P[x=2]=(1-p)*p

Probability of retiring after game k: winning games 1-(k-1), then loosing game k: P[x=k]=(1-p)^k*p (the probability of winning n consecutive games is (1-p)^n

Title: Re: Chess Players' Career
Post by BNC on Jan 17^th, 2003, 9:34am

As for probability sum:

Sigma(k=1,k->inf, P[x=k]) = Sigma(k=1,k->inf, p*(1-p)^(k-1)) = p* Sigma(k=1,k->inf, (1-p)^(k-1)) = p* [1/(1-(1-p))] = p* [1/(1-1+p)] = p* (1/p) = 1

Title: Re: Chess Players' Career
Post by wowbagger on Jan 17^th, 2003, 9:44am

For enhanced legibility:
http://tcw2.ppsw.rug.nl/~towr/PHP/formula.php?formula=%5Csum_%7Bk%3D1%7D%5E%7B%5Cinfty%7D+P%28x%3Dk%29+%3D%0D%0A%5Csum_%7Bk%3D1%7D%5E%7B%5Cinfty%7D+p+%281-p%29%5E%7B%28k-1%29%7D+%3D%0D%0Ap+%5Csum_%7Bk%3D1%7D%5E%7B%5Cinfty%7D+%281-p%29%5E%7B%28k-1%29%7D+%3D%0D%0Ap+%5Cfrac%7B1%7D%7B1-%281-p%29%7D+%3D%0D%0Ap+%5Cfrac%7B1%7D%7Bp%7D+%3D+1