# Fool proof way of winning at roulette?

Imagine you are playing a game of roulette... There are 2 blocks with the names "odd" or "even". If you choose odd and the random number selected is odd, you win double your money. If it isn't the correct selection, you lose what you bet.

You decide to bet based on a "fool proof" system:

At first you bet 1 on odd. If you lose, you bet 2 on odd... If you lose again you bet 4 on odd... Etc... You continue until you win: and when you win you reset your system... Starting at 1 and you continue the method. ($1,2,4,8,16,32,...$)

Let's follow a few example rounds:

• Win (gain 1) $+ 1$
• Lose, Win (lose 1, gain 4) $+ 3$
• Lose, Lose, Win (lose 3, gain 8) $+ 5$

I don't see any "fault" in this system? Is there an error, and if there is... What is it? Any answers/comments will be highly appreciated :)

Kind Regards

Joshua Lochner

• The problem is that your wallet is finite – H. H. Rugh Aug 27 '16 at 20:21
• Your strategy is known as the martingale – pi66 Aug 27 '16 at 20:23
• One, eventually you run out of money and/or hit the house betting limit. Two, there's a 00 on the roulette wheel that makes all bets losers. So even given unbounded bankroll and betting limit, you would slowly lose with this system. – user4894 Aug 27 '16 at 20:24
• Along similar lines, you might want to read about the St. Petersburg Paradox – lulu Aug 27 '16 at 20:27
• Your calculations of winnings are not correct. If you lose and then win, you lose 1, then just gain 2 for a net of 1. Any sequence that ends in a win will net you 1. The problem comes when you cannot double any more, either from a lack of resources or from the house limit. – Ross Millikan Aug 27 '16 at 21:12