# Martingale betting strategy in fair casino

## Problem

Martingale strategy is a rather famous gambling strategy which has been recently discussed in the Numberphile video.

In the video, it has been stated that if one decides to follow the strategy the probability of doubling the initial amount of money approaches 1/e as the initial amount approaches infinity. A simpler way to say it is that you get only a 36-37% chance of doubling your money if you start with more than about 25$ and you put 1$ as a first bet.

I believe that the claim is false and actual probability is equal to 50%. The goal of the article is to show that the produced result leads to strange consequences and afterwards I want to prove that the actual probability is 50%. At the very end, I will try to point to the actual mistake in the calculations shown in the video.

## Assumptions

It’s important to remember that calculations assume that a person that follows the Martingale strategy plays in the casino on the fair roulette wheel. It means that the probability of observing red equals to 50% and the other 50% of observing black. In addition, the assumption of the fair casino assumes that outcomes of all games are i.i.d.

## Initial indicator of the problem

The most obvious problem with the formula shown in the video can be noticed if you look at the probability of doubling the money if one enters the casino with 1$. The formula says that

\[P(2N\text{ | }N) = \left(1 - \frac{1}{N}\right)^N\]So if we plug \(N=1\) the formula says that the probability of doubling the initial amount of money is equal to 0 which is nonsense, since you either win on the first try and get 2$ or lose everything with no other money to play with.

One might argue that the formula just gets more accurate when \(N\) increases, but even that’s not the case as will be shown later.

## Martingale strategy as two person zero-sum game

Rephrasing the problem can help to see why it’s strange when the strategy produces a probability of winning less than 50%.

Let’s assume that person A enters the casino with \(N\) dollars and wants to leave it with \(2N\) dollars. The person A is determined to follow the Martingale strategy and play the fair roulette wheel until loses all of the money in which case person A will have to leave the casino with nothing (0 dollars). Imagine that instead of going to a casino that person plays with person B that also has \(N\) dollars and the game ends when person A or B ends up with no money (and the other player will get \(2N\) dollars). It’s easy to see that if player A plays with another player or casino it has no effect on the game since it doesn’t change the final goal of the game or intermediate outcomes. So each turn person A makes a bet by putting a certain amount of money on red or balck and person B puts exactly the same amount of money on the opposite color. The following set up exactly reflects how the casino works for our specific problem and shows it from the perspective of the two person zero- sum game. It’s very easy to see why this is a zerosum game, because only one person wins and takes all of the money on the table, so the loss for one person is a gain for another. In addition, since our casino is fair and all outcomes are i.i.d. It absolutely doesn’t matter on which color we bet as long as the person follows the specified strategy. We can also say that the flip of a coin determines on which color person A has to bet, so that none of them actually picks a color on which to bet.

The reason why probability below 50% is a strange result because it indicates that person B has a larger probability of winning (above 50%) meaning that strategy which person B follows is actually better. Notice that both players start with the exact amount of money, they both have the exact random chance of winning the game, but the strategy that they follow is the only thing that makes them different. Person B actually follows a slightly different strategy compared to person A. Recall that in Martingale strategy each time person A loses the player has to bet double of what was lost and since person B bets the same amount it means that it happens every time person B wins. Person B actually follows quite the opposite strategy. Since the selected color doesn’t have any effect on the outcome, the conclusion actually indicates that strategy used by person B is better. In order to see why the previous conclusion has to be true (assuming that the Martingale strategy has probability of winning below 50%) for the given set up we can try switching the strategy. Person A starts the first bet with 1$ as before and doubles it each time person A wins. It means that as before Person B has to bet on the opposite color, but this time double the bet each time person A wins or we can say that it happens each time person B loses which is exactly the Martingale strategy.

Does it mean that by following a “bad” Martingale strategy we actually discovered another strategy which can be profitable? The answer is NO, since as you remember we started with the assumption that probability of winning is below 50% which as we will see later is false.

## Expected revenue for unknown winning probability

It can also be shown that probability below 50% produces negative expected revenue, meaning that on average a person that follows Martingale strategy will lose money. For example, if a person goes every day to the casino with \(N\) dollars then more often than not that person will come out with no money and losses from this daily “job” won’t cover the expenses. The expected daily profit can be calculated very easily, but notice that on successful days revenue will be \(N\) dollars and on the other days it will be \(-N\).

\[\begin{align} \mathbb{E}[R] &= Np + (-N) (1 - p) \\ &= (2p - 1)N \end{align}\]where \(R\) is a daily profit and \(p\) is a probability of doubling the initial amount of money on a specific day. It’s easy to see that for any \(N\) and \(p \lt 0.5\) expectation is negative which means that overtime a player will be losing money. And from the “zero-sum game” point of view we can easily see that casinos have positive expected revenue even with the fair roulette wheel.

As has been stated before the main assumption is false and it can be shown that both players have expected revenue equal to zero which at the same time proves that for \(N > 0\) probability has to be equal to \(0.5\).

## Expected revenue

There are multiple ways to show that the actual probability of winning is 50%, but I would like to continue with the most intuitive proof (in my opinion) which is based on the previous findings and addresses the problem a bit less directly. Specifically, I want to show that the expected revenue obtained by following the Martingale strategy is equal to zero which will at the same time prove that the probability of doubling the initial amount of money is 50% (see previous section in order to understand how one conclusion follows from the other).

We can rewrite expectation using the law of the total expectation

\[\begin{align} \mathbb{E}[R\text{ | }N,D] &= \mathbb{E}[\mathbb{E}[R\text{ | }M,N,D]\text{ | }N,D] \\ &= \mathbb{E}[\mathbb{E}[R\text{ | }M]\text{ | }N,D] \end{align}\]where \(R\) is a revenue generated by the strategy, \(N\) is an initial amount of money that player has, \(D\) is a desired amount of money that player wants to get and \(M\) is a maximum number of steps that player can play with the given amount of money (\(M \geq 1\)) one cycle of the Martingale strategy.

Equation above implies that revenue is conditionally independent from \(N\) and \(D\) given \(M\). It’s easy to see that inner expectation doesn’t depend on \(N\) or \(D\) when \(M\) is known, but first we’ll need to slightly modify the problem in a way that wouldn’t effect our expectation. We can imagine the following process. Each time a player makes one dollar bet we can say that the player starts a cycle of the Martingale strategy. Before starting the cycle the player can calculate what is the maximum number of steps the player can play in one cycle of the strategy given the amount of money that the player has and we can call this value \(M\). It’s perfectly fine that a player can play and win at least once in less than \(M\) steps, it’s just when the player loses all games within a cycle then the cycle has to be terminated and a new one begins in case the player has enough money. But we can imagine an alternative process. The player can always play \(M\) games within a cycle. It clearly happens anyway when player loses all \(M\) games, but when player wins in \(K\) (\(K \leq M\)) games we can imagine that the player can just wait near the table table the remaining \(M-K\) games without making any bets. Waiting near the table has no effect on the expectation since the player doesn’t gain or lose money by doing that. So now in \(M\) games (including those where player doesn’t play the game) player can lose \((-2^M + 1)\) dollars with \(1/2^M\) probability or win 1 dollar in which case conditional expectation will be

\[\begin{align} \mathbb{E}[R\text{ | }M] &= \left(1 - \frac{1}{2^M}\right) \, + \frac{-2^M + 1}{2^M} \\ &= 0 \end{align}\]And since conditional expectation is equal to zero it means that distribution of \(M\) is completely irrelevant to the final expectation and overall expected revenue is \(\mathbb{E}[R\text{ | }N, D] = 0\).

If player starts with \(N\) dollars and wants to finish the game with \(D\) dollars (for \(D > N\)) then the expected revenue will be

\[\begin{align} \mathbb{E}[R\text{ | }N, D] &= (D - N)p + (-N) (1 - p) \\ &= Dp - N \end{align}\]but since expectation is equal to 0 we get

\[p = \frac{N}{D}\]and if \(D=2N\) than \(p=0.5\)

## Simulation

Very basic simulation can show similar conclusions (they can’t be exactly the same) and a large number of simulations can produce probabilities very close to 0.5. In each simulation player enters the casino with 50$ and plays on the fair roulette wheel by following the Martingale strategy until wins 100$ or loses everything. When a player doesn’t have enough money to double the previous bet, the player gives up and starts another round by betting 1$ and doubling each time the person loses the game. We can run simulation 100,000 times and observe that roughly 50% of the time game ends up with double the initial amount (and it works for any positive \(N\))

Python code with the simulation can be found here.

## Problem with the conclusion in the video

I believe that the main problem in the video happens by ignoring the possibility of players recovering large losses from money that couldn’t cover doubled the amount of losses in the previous bet. Specifically, we can rewrite the initial amount of money in the following way.

\[N = 2^k - 1 + m\]where \( 0 \leq m \lt 2^k\) and \(k \geq 1\)

The new way of writing \(N\) allows us to derive the actual probability of winning an extra dollar by following the Martingale strategy with a finite budget. For any \(N\) we basically have a budget that allows us to continue the Martingale strategy, but we will terminate it as soon as we run out of budget (after \(k\) subsequent losses). The remaining \(m\) dollars can help us to recover previous losses (which gives us another chance and increases probability of winning). Notice also that \(m\) changes overtime (because \(N\) changes) which means that the more successful games a player had before the more likely the player will recover larger losses produced by the Martingale strategy. The correct way to specify probability of winning can be shown to be

\[\begin{align} P(N+1\text{ | }N) = &P(N+1\text{ | }W,N)\,P(W\text{ | }N) + \\ &P(N+1\text{ | }L,N)\,P(L\text{ | }N) \end{align}\]where \(W\) indicates that a player managed to win 1$ by following the Martingale strategy and \(L\) means that we ran out of money that would have allowed us to continue following the strategy (\(W\) and \(L\) are the only possible outcomes). In this case, it’s obvious that \(P(N+1\text{ | }W,N)=1\), because we started with \(N\) dollars and got 1$ after following the Martingale strategy. Also we can show that \(P(N+1\text{ | }L,N) = P(N+1\text{ | }m)\) which means that we lost \(2^k-1\) dollars and now we’re left with \(m\) dollars, which gives us a small probability of still recovering from our initial losses and earning back \(N+1\) dollars. And at last, we can show that

\[P(L\text{ | }N) = \frac{1}{2^k}\]and since \(P(L\text{ | }N) = 1 - P(W\text{ | }N)\) we get the following result

\[P(N+1\text{ | }N) = 1 - \frac{1}{2^k} + \frac{1}{2^k} P(N+1\text{ | }m)\]In addition, it’s important to specify that \(P(X\text{ | }0) = 0 \, \forall \, X \gt 0 \), because with no money there won’t be an option to earn some.