# On Martingale betting system

I was reading the Martingale betting system on wikipedia and there is something not completely clear to me. The article says at some point

However, the gambler's expected value does indeed remain zero because the small probability that he will suffer a catastrophic loss exactly balances with his expected gain. It is widely believed that casinos instituted betting limits specifically to stop Martingale players, but in reality the assumptions behind the strategy are unsound. Players using the Martingale system do not have any long-term mathematical advantage over any other betting system or even randomly placed bets.

I don't quite understand this. If you keep doubling, it is true that you might encounter catastrophic losses, but the probability that you get 5 heads in a row is 1/32, quite low, and it decreases to zero exponentially. Thus, to me it seems that it should be rare , assuming that the probability of head is .5, to have to play more than 6/7 rounds in real life.

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Do you doubt the mathematical fact that the expected value is zero, or do you doubt that using this system is a bad idea? – Stefan Nov 20 '11 at 10:30

## 3 Answers

The probability that you get e.g. five heads in a row is low, but nonzero. It'll happen to you eventually.

I wrote a quick script in Matlab to simulate players following the Martingale strategy. I started 100 players with a fortune of 100 each, and had them all play the martingale redoubling strategy. Here are the results of one run:

As you can see, there are only 9 players who make it past 2,000 time steps, only 5 who make it past 4,000 and only 3 who make it past 6,000. By timestep 10,000 everyone has gone broke. "In the long run, we're all dead."

It's a common psychological bias to discount low-probability events when making decisions (even when you know about the effect!) which is why the martingale strategy seems like such a good deal on first examination.

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And the lifetimes would be shorter in a real casino where the even money bets are not fair. – Henry Nov 20 '11 at 13:06

Imagine a huge number of martingale players all starting to bet at the same time. After $n$ rounds, an overwhelming proportion of players, say $1-\varepsilon_n$, is broke but the tiny proportion $\varepsilon_n$ not broke yet has some good chances of being rich.

To see this, note that statistically the total amount of money of the players does not change hence, if $N$ players start with a given fortune $F$, the total amount of money at any time is about $NF$. After $n$ rounds, about $(1-\varepsilon_n)N$ of the players are broke, their fortune is $0$, hence the players not broke yet won about $(1-\varepsilon_n)NF$ globally. There are about $\varepsilon_nN$ of these hence the mean gain of each is about $(1-\varepsilon_n)F/\varepsilon_n$. Since $\varepsilon_n\to0$, this is about $F/\varepsilon_n$, which is huge if $n$ is large.

The trouble is that, when you start playing, nothing guarantees that you will be one of the around $\varepsilon_n N$ lucky players after $n$ rounds, and that the chances that you will be one of the unlucky players whose fortune is $0$ become huge when $n$ becomes large.

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The gambler's expected value does indeed remain zero. Assuming that an individual bet is 50/50 probability that means each bet has a zero expected value. Whether the gambler makes 1 bet, or 100 bets doesn't affect this.

However what the Martingale betting system does change is the probability distribution of how much you win or lose. If you bet the same amount every time you are more likely to win or lose a little bit.

If you keep increasing the stakes, you are likely to win or lose a lot.

If you place a $1 bet, 64 times you expect your winnings to follow a binomial distribution, and you are unlikely to win or lose more then 20 dollars. However if you follow the martingale system, and place 64 strings of bets starting at$1, you end up with a ~35% chance of winning 64 dollars, a ~27% chance of losing 64 dollars, and the remaining odds of losing some amount of money less than 64 dollars, or breaking nearly even.

This is the probability distribution if you instead use the martingale betting system, and make 64 strings of bets

both of those have zero expected value, but in the martingale system you are more likely to win a lot of money, or lose a lot

I have done a more extensive write-up of the martingale system here http://www.fairlynerdy.com/gamblers-ruin/

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