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 re: Solution by Charlie)

Less than 4/100ths of a percent?  Sheesh!

Consider then if he wins four in a row of his first five bets (all considered losses in the first trial).  If he doubles up after each win, then he will show a profit of $8 which is more than enough to cover the final 5 bets.  The odds of winning 4 in a row on the first 5 are 2 * (18/38)^4 = 10.069%, so the odds of this not happening is 89.931%.  Multiply this by the probability of losing the next five to find the overall probability of losing (3.632%), which leaves the probability of turning a profit at 96.368%.

Edited on April 6, 2007, 12:57 am

Posted by hoodat on 2007-04-05 20:57:02