Strategy for beating roulette using simple probability

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    Probability Strategy
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Discussion Overview

The discussion revolves around strategies for beating roulette using probability, specifically focusing on a betting system that involves increasing bets after losses. Participants explore the viability of this approach and its implications for long-term play.

Discussion Character

  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant proposes a betting strategy that involves increasing bets after losses, suggesting that this could lead to a net gain despite the risks involved.
  • Another participant identifies the proposed strategy as a "martingale" system, cautioning that it is likely to lead to bankruptcy without infinite funds.
  • A different participant emphasizes that the starting cash amount limits how many losses one can endure before going broke, suggesting that the potential winnings are minimal compared to the initial investment.
  • One participant asserts that the only reliable way to win at roulette is to be the house, highlighting the mathematical advantage provided by the presence of 0 and 00 on the wheel.

Areas of Agreement / Disagreement

Participants express differing views on the effectiveness and sustainability of the proposed betting strategy, with some arguing against its viability and others exploring its mechanics. There is no consensus on the best approach to winning at roulette.

Contextual Notes

The discussion does not resolve the uncertainties surrounding the long-term outcomes of the proposed betting strategy, nor does it address the mathematical implications of the house edge in detail.

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.
 

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