# Martingale betting strategy in fair casino

## Problem

Martingale strategy is a rather famous gambling strategy that 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 more straightforward 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.

The claim is false, and the actual probability equals 50%. The article aims to show that the produced result leads to strange consequences, and afterward, I want to prove that the true probability is 50%. Ultimately, I will point to the actual mistake in the calculations shown in the video.

## Assumptions

It’s important to remember that calculations assume a person following the martingale betting strategy plays in the casino on the fair roulette wheel. It means that the probability of observing red equals 50% and the other 50% of observing black. In addition, the assumption of the fair casino implies 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 equals 0, which is nonsense, since you either win on the first try and get 2$ or lose everything.

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

## Optional stopping theorem

Martingale is also a name for the general family of stochastic processes, and the game described in the video is one of them. The most famous result associated with martingales is the optional stopping theorem. According to the theorem, no strategy can modify your initial expectation, meaning that the martingale betting strategy and “all-in” betting strategy would have precisely the same expectation, namely 50%.

Even without any knowledge about martingales, there are ways to convince ourselves that the answer is wrong.

## Martingale strategy as two person zero-sum game

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

Let’s assume that Alice enters a casino with \(N\) dollars and wants to play the game until she wins \(2N\) dollars. Alice is determined to follow the martingale betting strategy and play the fair roulette wheel until she loses all of the money, in which case Alice will have to leave the casino with no money. Imagine that instead of going to a casino, Alice plays with her friend Bob that also has \(N\) dollars, and the game ends when Alice or Bob ends up with no money (and the other player gets \(2N\) dollars). It’s easy to see that if Alice plays with Bob or a casino, it has no effect on the game since it doesn’t change the game’s final goal or intermediate outcomes. So each turn, Alice makes a bet by putting a certain amount of money on red or black, and Bob puts the same amount on the opposite color. The following setup 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 easy to see why this is a zero-sum game because only one person wins and takes all 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 coin flip determines which color Alice has to bet, so neither Bob nor Alice pick a color on which to bet.

Now we can go back to our original problem and understand why the probability below 50% is strange. A probability below 50% indicates that Bob has a larger probability of winning (above 50%), meaning that strategy that Bob follows is better. Notice that both players start with the exact amount of money and have the exact random chance of winning the game, but the strategy they follow is the only thing that makes them different. Bob actually follows a slightly different strategy compared to Alice. Recall that in the martingale betting strategy, each time Alice loses, she has to bet double of what was lost, and since Bob bets the same amount, it means that it happens every time Bob wins. Bob actually follows quite the opposite strategy. Since the selected color doesn’t affect the outcome, the conclusion indicates that the strategy used by Bob is better.

Does it mean that by following the martingale betting strategy, we discovered another strategy that can be profitable? The answer is no, since, as you remember, we started with the assumption that the probability of winning is below 50%, which, as we will see later, is false.

## Expected revenue for unknown winning probability

It can be shown that a probability below 50% produces negative expected revenue, meaning that, on average, a person that follows the martingale betting 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 the probability of doubling the initial amount of money on a specific day and the expectation is implicitly conditions on \(N\) and \(p\). It’s easy to see that for any \(N \gt 0\) and \(p \lt 0.5\), the expectation is negative, meaning that a player will lose money over time. And from the “zero-sum game” point of view, we can see that casinos have positive expected revenue even with the fair roulette wheel.

As stated before, the main assumption is false, and it can be shown that both players have expected revenue equal to zero, which simultaneously proves that for positive \(N\), probability has to be equal to 50%.

## 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 betting 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 the previous section to understand how one conclusion follows from the other).

We can rewrite expectation \(\mathbb{E}[R]\) using the law of the total expectation

\[\begin{align} \mathbb{E}[R] &= \mathbb{E}[\mathbb{E}[R\text{ | }T]] \\ &= \mathbb{E}[\mathbb{E}[\sum_{t=1}^{T}R_t\text{ | }T]] \\ &= \mathbb{E}[\sum_{t=1}^{T}\mathbb{E}[R_t\text{ | }T]] \\ \end{align}\]where \(R\) is the total revenue generated by the strategy, \(T\) is the total number of martingale cycles encountered in the game, and \(R_t\) is a revenue on \(t\)-th cycle. It’s important to clarify at this point the meaning of the term “martingale cycle”. The strategy follows “cycles” where each cycle begins with a fixed bet (e.g., 1$) and continues until we either win or can no longer double the previous amount.

We can apply the law of total expectation one more time to the inner most expectation by implicitly defining a distribution of steps \(K\) in the \(t\)-th cycle.

\[\mathbb{E}[R_t\text{ | }T] = \mathbb{E}[\mathbb{E}[R_t\text{ | }K,T]\text{ | }T]\]In \(K\) steps of a cycle a player can lose \((2^K - 1)\) dollars with \(1/2^K\) probability or win 1$ in which case conditional expectation will be

\[\mathbb{E}[R_t\text{ | }K,T]= \left(1 - \frac{1}{2^K}\right) \, - (2^K - 1)\frac{1}{2^K} = 0\]And since the conditional expectation is equal to zero, it means that distributions over \(T\) and \(K\) are completely irrelevant to the final expectation and overall expected revenue is zero, \(\mathbb{E}[R] = 0\). So if player starts with \(N\) dollars and wants to finish the game with \(D\) dollars (for \(D > N\)) then the expected revenue will be

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

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

## Simulation

Simple 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. Each simulation player enters the casino with 50$ and plays on the fair roulette wheel by following the martingale betting strategy until the player 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 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 because it was ignored that a player can recover 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 probability of winning a dollar by following the martingale betting strategy with a finite budget. For any \(N\), we have a budget that allows us to continue the 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 the probability of winning). Notice also that \(m\) changes over time (because \(N\) changes), which means that the more successful games a player had before, the more likely that the player will recover larger losses. The correct way to specify the 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 strategy for one cycle, and \(L\) means that we ran out of money and can no longer continue with the cycle (\(W\) and \(L\) are the only possible outcomes). In this case, it’s evident that \(P(N+1\text{ | }W,N)=1\) because we started with \(N\) dollars and got 1$ after following the martingale betting 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 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

\[P(N+1\text{ | }N) = 1 - \frac{1}{2^k} + \frac{1}{2^k} P(N+1\text{ | }m)\]The formula above slightly differs from the one produced in the video and highlights the mistake.