Strategy for beating roulette using simple probability

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The discussion centers on a proposed strategy for beating roulette using probability, specifically through a betting system that doubles the wager after each loss, known as a martingale strategy. The strategy relies on betting on outcomes with an 18/38 probability, with the aim of recouping losses and making a profit after a win. However, the approach is criticized for its inherent risk, as long losing streaks can lead to substantial financial losses, potentially exceeding available funds. Critics emphasize that the house edge, due to the presence of 0 and 00, ultimately favors the casino, making consistent winning unlikely. Overall, while the strategy may seem appealing, it is fundamentally flawed due to the risks involved and the mathematical advantage held by the house.
cardicorona
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I think that I have figured out a way to beat the game of roulette with simple probability and a very useful equality...

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

\sum_{0}^{n}2^{n}=2^{n+1}-1[\tex]<br /> <br /> This way, with each winning, bet the gain is more than the loss.<br /> <br /> 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<br /> <br /> 1-(18/38)^5-2/38=.923<br /> <br /> In which the 2/38 represents the 0 and double 0 squares. <br /> <br /> Once you win then you start back with bidding a dollar. <br /> <br /> 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.<br /> <br /> 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.<br /> <br /> Thoughts?
 
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The strategy you describe is called a "martingale." You'll eventually go broke unless you start with an infinite amount of cash, in which case you probably wouldn't be bothering with roulette to begin with.
 
The amount of cash you start with determines how many consecutive losses you can sustain without going broke. Once you know how many consecutive losses you can survive, you can calculate the odds of that happening. I think you will find that you will win remarkably little money compared to the amount you start with before you go broke.
 
There is one ridiculously simple way to win at roulette, be the house. 0 and 00 give you a real mathematical advantage. As far as I am aware that is the only way.
 
If there are an infinite number of natural numbers, and an infinite number of fractions in between any two natural numbers, and an infinite number of fractions in between any two of those fractions, and an infinite number of fractions in between any two of those fractions, and an infinite number of fractions in between any two of those fractions, and... then that must mean that there are not only infinite infinities, but an infinite number of those infinities. and an infinite number of those...

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