Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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 Walter Nov 20 '11 at 10:30
up vote 5 down vote accepted

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.

share|cite|improve this answer
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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.