Can you think of a method of playing standard casino roulette that gives the player a better than 90% chance of winning each time he plays a game?

A game is considered to be at least 10 bets on consecutive spins of the roulette wheel. A game is considered won if the gambler stops with more money than he started. The player starts with $1100 and table limits are min: $1 max: $1000).

(In reply to Classic martingale by Old Original Oskar!)

(19/37)^10 is 0.13%, not 1.3%.

Considering wheels with 00 in addition to 0, 20/38 is the appropriate chance of losing an individual bet, leading to 0.16% probability of ultimately failing in the martingale.

In that small chance of losing the martingale, the player will have lost $1023, for the chance of winning $1. He will be left with $77.

The martingale need have gone only for 4 bets: $1, $2, $4 and $8 (or at the other extreme 73, 146, 292 and 584, leaving the player $1095 in the hole if he failed to gain $73). It's 7.67% likely that he would lose overall, so his chance of winning $1 (or $73) is over 90%.

Posted by Charlie on 2004-08-23 10:20:16