User Name Remember Me? Password

 Advanced Statistics Advanced Probability and Statistics Math Forum

 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. Tags roulette, winning Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Bucephalus Probability and Statistics 2 January 25th, 2012 03:47 PM mdenham2 Algebra 1 February 3rd, 2010 06:00 PM bogazichili Advanced Statistics 1 December 21st, 2009 03:50 PM Stephy7878 Advanced Statistics 1 February 18th, 2009 08:56 PM Stephy7878 Algebra 1 February 18th, 2009 12:56 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top      