Martingale Betting System

Ever since the very beginning when people started to gamble (which was probably man's second hobby, the first was masturbation) people tried to devise methods of gambling, various systems, that they believed would make them win at gambling games. All of these systems are flawed. None of them work. The basic idea is that if the casino has a small advantage over you, then you will lose money, not matter how you play.

We will demonstrate the failure of betting systems by focusing on a popular system known as the "Martingale". The idea is the following. Suppose you are playing blackjack. You bet 10. If you win, then you won 10. If you lose then you bet again, but this time 20. If you win, you win 20 minus the 10 you lost earlier, which is a net gain of 10. If you lose then your net lost if 30. But if you lose then the next game you bet 40. If you win then your net gain would be 40 minus the 30 you lost earlier which works out to be 10. If you lose your next lose would be 70. But then next game you bet 80. And this process continues. The general rule is that if you lose you double your bet. The idea of the Martingale is based on that eventually you will win. And when you win you will get back all your lost money. If you start win 10 then your bets will be: 10,20,40,80,160,320,640, .... It is highly unlikely for a casino to consecutively to win all those games. Thus, you expect, with extreme likelihood of winning back your lost money.

It is often useful to consider extreme examples when working on a problem. Here we will consider the example when the player has an infinite amount of money and the other case when the casino has an infinite amount of money.

If the player has an infinite money of money the Martingale always works. This follows by the infinite monkey theorem. The idea is simple enough. A monkey hitting a typewriter randomly will eventually type out a book. Because no matter how unlikely it is for a monkey to randomly type out a book it still has a chance happening. So eventually is must happen. So this means the Martingale system will always work, provided the probability of winning a game is positive. Even if the casino has a 99% advantage over the player the Martinage system will eventually, by the theorem, make the player win all his money back.

But there are two obvious problems with such an extreme example. The first one is that no one has an infinite amount of money (not even the Federal Reserve, the counterfeiting that they do has to come from paper, there is a limited amount of paper). The second one is that what would be the point of gambling if you have an infinite amount of money? The point of gambling is to try to make some profit. If you have an infinite amount of money and you gain a thousand dollars you still have an infinite amount of money. The feeling of danger and risk is also absent from gambling. If you have an infinite amount of money and you lose a million, you have not really lost anything. Gambling with an infinite money supply is pointless because the person no longer values his money, in general the more a person has of something the less he values it. However, this thought experiment was just suggested to see what is going on with the Martingale system.

So let us then suppose that the player does not have an infinite money supply, but that he has a very large money supply. Let us assume you have 10 quadrillion dollars. And you want to buy up the entire casino, which is worth, say 100 million dollars. So you bet 100 million dollars. If you lose you keep doubling your bet. The infinite monkey theorem is no longer applicable here because you have a finite supply of money. But because it is so unlikely for the casino to win every single game your winning chances are overwhelmingly on your side that you would win up the entire casino. In this case the casino would not bet against you. It is unfavorable for them to bet. Because they are in a losing position.

Now consider the extreme case when the casino has an infinite supply of money. Ignoring the issue that no casino would have an incentive to be a casino given that it had an infinite amount of money, let us just pretend that there was such a casino. Such a casino would always play against a player with a finite amount of money. The player, even if he has 10 quadrillion dollars, will probably win billions or trillions of blackjack hands with the Martingale. However, the infinite monkey theorem is now applicable on the side of the casino this time. So eventually the player would have to disastrously lose every single game that he has played. So the casino would lose trillions of dollars, but eventually they will get lucky and win, at once, more trillions of dollars back. Why this happens? Well we will explain the reasoning for this below.

So now we need to consider actual situations when both the player and the casino always have a finite amount of money. Let $b$ be the bet a player bets in a game (say the red/black game on the roulette wheel, to make it as simply as possible). For most people $b$ will be little money, like 10. Not many people in the world have enough money to bet 1000 dollars and double it many times along the way. Let $n$ represent the number of games the player is able to play. Because it is assumed that the player has a finite amount of money there is a maximum number of games he can play before he loses all his money by being dreadfully unlucky. For example, if he has 1000 dollars with him and he bets b=10, then he can only play games with bets: 10,20,40,80,160,320. Therefore, n=6, he can only play six games. He cannot play seven because he does not have enough money for the seventh game, that would be 640, but that is too much. Let $p$ be the probability that the player wins. On the classical roulette wheel there are 26 red slots, 26 black slots, and 2 green slots. The chances of guessing the right color between red and black is almost 50%, except the casino has a slight advantage, because your chances are at 26/54 = 48%.

What are the possible ways to win on the roulette wheel under the Martingale? You can win your first game (W). You can lose your first game but win on your second game (LW). You can lose your first game, lose the second game, but win on your third game (LLW). Or (LLLW). And this keeps on going (LL...LW) until you are at your n-th game, the last possible game you can play. What are the possible ways to lose on the roulette wheel under the Martingale? There is only one way. When you lose all your possible money by doubling bets, i.e. the situation is (LL...L) where you played n-games.

So your winning events are: (W),(LW),(LLW), ... (LL...LW). Where the last one contains n total games. Your losing event is just: (LL...L) where the number of times L appears is n-times.

What is the probability that you would win? It is any one of the possibilities (W),(LW),.... happening. To compute this probability it is a lot easier to look at the reverse problem. What is the probability of losing everything? It is just (LL...L). The probability of this is $(1-p)^n$. The reason for this is because you have probability p of winning, so that your probability of losing is (1-p). Therefore, losing n-times in a row is just $(1-p)^n$ by just multiplying (1-p) by itself n times. This means the probability of winning, in any one of these winning events, is $1-(1-p)^n$.

The probability that you would win under Martingale on the roulette wheel with probability p=.48 is therefore given by $1-(.52)^n$. Notice that if $n$ is a large number, say n=10. This quantity works out to be .998, your chances of winning the bet b are at 99.8%. The higher $n$ the smaller $(.52)^n$, the smaller $(.52)^n$ the closer and closer $1-(.52)^n$ is to 1. Therefore, as $n$ goes to infinity the probability of winning $1-(.52)^n$ goes to 1. However, in our finite case the number $1-(.52)^n$ is very close to one, almost one, but not quite.

A gambler would look at these numbers and scream "success, the Martingale works!". But he is missing one very important point. When you win, you win b. But if you are to lose the game, you are to lose disastrouly. It means you would lost b in the first game, 2b at the second game, $2^2b$ at the third game, and so on until you would lose $2^{n-1}b$ on the last n-th game. In total you would lose $(1+2+2^2+...+2^{n-1})b$. Now using the identity that $1+2+2^2+...+2^{n-1}=2^n - 1$ we see that if you were to lose you lose $(2^n-1)b$. This is an exponential lose. You win little, but you lose exponentially.

A better way of understading how you win or lose is by looking at the expected value. Say that you are playing a game where your chances of winning at 1/10. This means if you bet a dollar you essentially get back 10 cents, and lose the other 90 cents. Your expected gain/lose in such a game, if you bet b, will be $-(9/10)b$. If you were playing a game where your chances of winning at 1/2, then your expected gain/loss would be 0. The expected gain/lose can be calculated by: (win on bet) TIMES (probability of winning) MINUS (loss on bet) TIMES (probability of losing).

Let us try to figure out the expected value for our Martingale system. When you win, you win $b$, at probability $1-(1-p)^n$. When you lose, you disastrously lose $(2^n-1)b$, at probability $(1-p)^n$. The expected gain/loss is therefore, $b(1-(1-p)^n) - (2^n-1)b(1-p)^n$.

It is interesting to consider the case when p=1/2, i.e. the game is fair. In this case, $b(1-(1-p)^n-(2^n-1)b(1-p)^n$ works out to be $b(1 - (1/2)^n) - b(2^n-1)(1/2)^n$. Now expand everything out. To get, $b-b(1/2)^n - b + b(1/2)^n = 0$. The expected return under the Martingale system in a fair game is 0. All this means is that what you win, but eventually after many many games you will be lose, and you would be back to no overall winnings.

Let us consider when the expected gain is positive. So we need to solve the inequality $b(1-(1-p)^n) - (2^n-1)b(1-p)^n >0$. Dividing by b we get $1-(1-p)^n-(2^n-1)(1-p)^n>0$. Substract 1 from both sides and multiply by -1, but remember when you multiply by a negative number the inequality is switched. So we get $(1-p)^n+(2^n-1)(1-p)^n<1$. Divide by $(1-p)^n$ to get $1+(2^n-1)<(1/(1-p))^n$. This works out to be $2^n<(1/(1-p))^n$. For the left hand side to be less than the right hand size it means the base of the exponent must be smaller, so that $2<1/(1-p)$. Multiply to get $2(1-p)<1$. Divide to get $(1-p)<1/2$. Multiply by -1 and switch inequality to get $p-1>-1/2$. Add 1 to both sides to finally get $p>1/2$. So as long as the probability is greater than 1/2, i.e. the game is favored for you, you are expected to earn a gain by the Martingale system. Of course, if you have an advantage there is no need to to use this system at all. If however, $p<1/2$ then your expected gain would be negative. Which means you would expect to lose money with this system.

These simple calculations show that the expectation of using the Martingale system is negative in casinos (since they all have an advantage over the player). And so the Martingale system is doomed to fail.

The Martingale is a short-term system. If you just want to win 100 briefly in a casino then go ahead and use this system. But do not expect on being able to make a living doing this. Since eventually you will lose it all in one unfavorable game. Long-term effectiveness of the Martingale is just as effective as standard betting. Do not use it, unless you want a short-term win once in Las Vegas or Atlantic City (assuming you are not banned from entering the casinos).