I think that I have figured out a way to beat the game of roulette with simple probability and a very useful equality...(adsbygoogle = window.adsbygoogle || []).push({});

The bet is either red, black, odd, even, low, or high.

Each of these bets have a probability of 18/38 and the payout is double the bet.

The strategy is to increase the bet such that the payout is more than the additive losses and then after each win cycle back to the first bet.

For instance: first bet is a dollar, second bet is 2 dollars, third bet is 4 dollars, fourth bet is 8 dollars. In mathematical terms, we have

[tex]\sum_{0}^{n}2^{n}=2^{n+1}-1[\tex]

This way, with each winning, bet the gain is more than the loss.

Now, say you lose 4 times in a row betting red, the fifth bet will be 32 dollars probability that you will win the fifth time is already

1-(18/38)^5-2/38=.923

In which the 2/38 represents the 0 and double 0 squares.

Once you win then you start back with bidding a dollar.

Granted this particular choice in cycle would take a long amount of time to earn money since the net gain is only a dollar. But there are obviously an infinite amount of other choices in which the net is larger.

The fall back is that long losing streaks would amount to ridiculously large sums of money with 2^20 being just over a million dollars.

Thoughts?

**Physics Forums | Science Articles, Homework Help, Discussion**

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Strategy for beating roulette using simple probability

Loading...

Similar Threads for Strategy beating roulette |
---|

B Strategy for a Lottery-Style Draw Application |

I Average profits and losses on a roulette table |

About the strategy of reducing the total suffering in a queue |

I Are these alleged equal probabilities really equal? |

**Physics Forums | Science Articles, Homework Help, Discussion**