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 Trouble with Martingale by Larry)

The problem didn't specify exactly 10 spins, it said at least 10 spins.

I will only add that you don't need to win on the 10th spin.  If you have a negative balance after the 10th spin, you need only be in a series of bets that ends without you running into gambler's ruin (running out of money to double).

Also, if you have a positive balance after a losing 10th spin, you can stop after that loss, having won the "game".

For example, if I win the first eight spins (winning $1 each spin for a total of $8, then lose the ninth and tenth spins, $1 and 2$, respectively), I'm still up $5, and I can stop.

Or, for the first case, if I win the first six spins (winning $1 each spin for a total of $6, then lose the next four spins, losing $1, $2, $4, and $8, I am now down $9), but I can continue doubling for another six spins, and if I win any of those, I will come out having won the game.

Posted by SilverKnight on 2004-08-23 18:02:08