[Math] How to place bets to get > 50% chance of winning

gamblingprobability

Assuming we have a simple game of fair-coin throwing, there's a 50% chance to win and a 50% chance to lose.

Let's assume that we have a large amount of tokens and there is no cap to the amount placed in the bet, I have this strategy to sure-win:

  1. Bet x amount.

  2. If win, pocket x and go to 1. If lose, bet 2x.

  3. If win, pocket x and go to 1. If lose, bet 4x.

  4. If win, pocket x and go to 1. If lose, bet 8x.

  5. If win, pocket x and go to 1. If lose, bet 16x.

  6. If win, pocket x and go to 1. If lose, bet 32x.

  7. If win, pocket x and go to 1. If lose, bet 64x.

  8. If win, pocket x and go to 1. If lose, bet 128x.

  9. If win, pocket x and go to 1. If lose, bet 256x.

  10. etc…

Since there is no cap to the bet amount, using this strategy would guarantee us > 50% chance of winning.

Now if there is a cap at 50000x, I believe this strategy doesn't work any more (chance of winning no longer > 50%).

When there is a cap at 50000x, How should we modify the strategy to guarantee us > 50% chance of winning?

Best Answer

This betting strategy is called a martingale.

In fact, providing you stop playing once you have won, the probability of winning is more than a half, both with and without a cap. In reality, there is always a cap, as most people will not let you bet an amount higher than you could possibly repay.

The key point is that if you lose (i.e. hit the cap), then you lose a much larger amount than when you win. This can be disastrous, which is why most casinos ban it.

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