GamblersRuinExample

GamblersRuinExample - Pi = Example Alice and Bob are...

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Gambler’s Ruin Derivation and Example Suppose a gambler starts with $i and wins or loses $1 on each game. P(win) = p, P(lose) = 1 – p = q. They will stop if they go bankrupt or if they reach $k. Let P i = P(gambler reaches $k before going bankrupt, given they started with $i) We can clearly see that P 0 = 0 < P 1 < P 2 < … < P k–1 < P k = 1 Condition on the outcome of the first game. P i = P(reach goal | win first)*P(win) + P(reach goal | lose first)*P(lose) = P i+1 *p + P i–1 *q Rearranging, we get the following set of recursive equations for the P i ’s: P i+1 – P i = (q/p)(P i – P i–1 ) P 0 = 0 so P 2 – P 1 = (q/p)P 1 P 3 – P 2 = (q/p)(P 2 – P 1 ) = (q/p) 2 P 1 P i+1 – P i = (q/p) i P 1 P k – P k–1 = (q/p) k–1 P 1 But P k = 1 You can check (or derive) that the solution to these relations is
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Unformatted text preview: Pi = Example Alice and Bob are playing a game. Each game, the loser pays the winner $5. Bob has $50 and Alice has $25 to start. a) If the game is fair, find the probability that Bob loses b) If the game is biased so that Alice has a 60% chance of winning each game, again find the probability that Bob loses. Solution Let 1 unit be $5. Then Bob has 10 units and Alice has 5. The total bankroll is 15 units so k = 15. a) p = q = 0.5 here. P(Bob loses) = P(Alice wins) = i/k = 5/15 = 0.33 b) p = 0.6, q = 0.4, and P(Bob loses) = P(Alice wins) = [1 (0.4/0.6) 5 ] / [1 (0.4/0.6) 15 ] = 0.87...
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This note was uploaded on 12/12/2010 for the course STAT 4670 taught by Professor Skrzydlo,dianakatherine during the Spring '10 term at Waterloo.

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