My Math Forum 'Winning' at roulette

 May 19th, 2009, 11:05 AM #1 Global Moderator     Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms 'Winning' at roulette I have a question that may be of interest. Suppose a standard European roulette with simple rules: 37 slots, payoffs (2, 3, 36) with probabilities (18/37, 12/37, 1/37) for a unit bet. (These include the value of the returned bet; payoffs are (1, 2, 35) otherwise.) Now any strategy for this game loses, on average, 1/37th of the amount wagered. But suppose instead that the interest was making a profit of at least p chips, regardless of the magnitude of the potential loss. In particular, the game consists of n ? 0 rounds, during a bet of one of the three types above must be placed. The minimum bet is 1 chip and the maximum bet is m ? 1 chips. You start with a bankroll of c chips. You lose if you cannot make a bet, or if n rounds have passed and you have less than c + p chips. You win if n rounds have passed and you have at least c + p chips. The goal is to design a strategy that maximizes the probability of winning. Let P(n, m, c, p) be the probability of winning with an optimal strategy and n rounds, a max bet of m, c chips, and a target profit of p. Clearly for small values the probability can be calculated with recurrence: P(n, m, c, p) = max(P(n - 1, m, c - 1, p) * 19/37 + P(n - 1, m, c + 1, p) * 18/37, P(n - 1, m, c - 1, p) * 25/37 + P(n - 1, m, c + 2, p) * 12/37, P(n - 1, m, c - 1, p) * 36/37 + P(n - 1, m, c + 35, p) * 1/37) P(0, m, c, p) = 1 for c ? p P(0, m, c, p) = 0 for c < p But this requires 4^n steps to calculate, so it's not workable for large n. What are good strategies for large n? Say, n > 100,000 and |p-c| small.
 May 19th, 2009, 05:37 PM #2 Global Moderator     Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms Re: 'Winning' at roulette Oops, sorry. The formula I gave only works for m = 1. The correct one would loop over all possible values of the bet from 1 to m, taking the max as appropriate. So the true effort is something like (4m)^n, which is far worse.

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