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Topic: Chess Players' Career (Read 920 times) |
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Johno-G
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Chess Players' Career
« on: Jan 17th, 2003, 9:02am » |
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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.
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BNC
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Re: Chess Players' Career
« Reply #1 on: Jan 17th, 2003, 9:28am » |
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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
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| « Last Edit: Jan 17th, 2003, 9:29am by BNC » |
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How about supercalifragilisticexpialidociouspuzzler [Towr, 2007]
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BNC
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Re: Chess Players' Career
« Reply #2 on: Jan 17th, 2003, 9:34am » |
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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
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How about supercalifragilisticexpialidociouspuzzler [Towr, 2007]
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wowbagger
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Re: Chess Players' Career
« Reply #3 on: Jan 17th, 2003, 9:44am » |
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For enhanced legibility:
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