A gambler playing roulette has a superstition that he must place bets for exactly 10 spins of a roulette wheel (American wheel with 00). He has a bankroll of $300 and the table limits are minimum $1 and maximum $150, with $1 increments.

Can you find a strategy which guarantees he will be able to place ten bets and will have at least a 96% chance of making a profit (finishing bankroll of at least $301)?

(In reply to Solution by hoodat)

Hoodat has the right idea, but if his first 4 bets come out wins, then he can start right then with a 2 dollar bet. If he makes it, he is covered. If not, bet 5, 11, 23, 47, 95. for each successive loss.

The probability of this alternate scenario is 1/16, and has chance of losing equal to (20/38)^6, so the probability of winning using this strategy for 4 wins and the other for anything else is 1/16*(20/38)^6+(20/38)^5=.3919

Posted by Gamer on 2007-04-05 17:58:07