How to place bets to get > 50% chance of winning?

Question

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?

Answer 1

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.

Answer 2

Even just considering the first two rounds, your chance of winning is $75\%$. Later rounds make it very close to $100\%$. But as it is a fair game, the expectation is zero. On average, you will neither win nor lose. Even in an unfair game, like roulette, your chance of winning after two spins will be well greater than $50\%$, but then your expectation will be negative.