The Gambler's Ruin
At the beginning of play, gamblers A and B have m and n chips, respectively. Let their probabilities of winning be p for A, and q=1-p for B. After each play, the winner gets a chip from the loser, and play continues until one of the players is ruined.
What is the probability of A being ruined?
What is the probability of A being ruined?
Labels: mathschallenge, Probability, SharedPuzzle
posted by Zaux at
2:11 PM
4 Comments:
Probability of A being ruined is the probability of losing n times before winning m times.
Prob of losing m straight is (p-1)^m
Prob of winning 1 while losing m (i.e. m+1 games played, but losing the last) is mp*(p-1)^n.
Prob of winning 2 while losing m is m(m+1)p^2*(p-1)^n
Prob of winning n-1 and losing m is m(m+1)..(m+n-2)p^(n-1)*(p-1)^n
These all add up to:
(p-1)^n(1+mp(1+(m+1)p(1+(m+2)p(1+...(m+n-2)p^(n-1)))..)
. . . . whatever. Easy!
First sentence above should read the other way, i.e. losing m times before winning n times.
Make that last formula line
(p-1)^n(1+mp(1+(m+1)p(1+(m+2)p(1+...(m+n-2)p))..)
Not that it matters - probably wrong anyway. I suppose this expression can be simplified, but I don't know how.
Pretty high. The house always wins.
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