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

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?

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.

• You said that if we stop playing once we win, probability of winning is more than half. Can you demonstrate it? If we stop playing once we win, isn't probability of winning exactly half? Commented Oct 20, 2013 at 15:23
• @Pacerier: say the cap were $2$, so you could only play two rounds. You win the first round $50\%$ and win half the remaining, or $25\%$ on the second round. That makes $75\%$. More rounds only makes it higher. Commented Oct 20, 2013 at 15:27
• @RossMillikan, but after factoring in the losses, we have 75% chance to win x and 25% chance to lose 3x... so final probability to win is still exactly 50% right? Commented Oct 20, 2013 at 16:37
• You are confusing the probability to win with the expected value. The probability to win is the chance you are ahead, by however much or little. The expected value is the average profit, which takes account of the size of your winnings/losses. Your last two calculations are correct, but the final claim that the probability to win is 50% is wrong. It contradicts the 75% in the previous part of the sentence. Commented Oct 20, 2013 at 17:06
• You may end up with $\$1000$or not. The expectation will be$\$1000$, which is the average over many runs. But each run could well end up somewhere else. Commented Nov 20, 2013 at 14:07

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.

• Yes I meant in the question what strategy should we use to get an above-zero expectation? Commented Oct 20, 2013 at 16:40
• There is no strategy that gives positive expectation. Each bet is fair, so has zero expectation. Add up as many zeros as you want, you still get zero. You can make a series of bets to get essentially any probability distribution you want, as long as the expectation is zero. Commented Oct 20, 2013 at 20:43

One thing to note is that each time you "win" and reset back to your starting bet of $x$, you are only an additional $x$ above where you started. So you have a very high probability of gaining a relatively small amount of money each time you initiate this strategy.

The reason that your expected gain remains at zero is that the very high probability of making a small profit, $x$, is balanced out by the very small probability of incurring a huge loss, $512x$ or whatever your cap happens to be.