Gambling Systems and Strong Convergence

Question

Say you're betting on "red" at roulette. From one trial to the next, you vary your bet size between the house minimum and maximum based on the outcomes of previous trials. Can you prove that you go broke with probability 1, or that if you have an unlimited bankroll to start, your long-term return rate (winnings per unit wagered) converges to the expectation of a single unit wager? Well, with probability 1, at least. That seems plausible.

Answer 1

Your question is sometimes called the "gambler's fallacy". In fact, you need not even assume that there is a house maximum. The issue is a failure of integrability to apply the optional sampling theorem.

Let $X_n$ denote your winnings at time $n$, and let $\mathcal{F}_n$ be the associated $\sigma$-algebra. Simply speaking, $\mathcal{F}_n$ encodes the information associated to $X_1, \dots, X_n$. The house will never allow you a strategy that favors you, so regardless of what the game is or what you do, your expected winnings at the next step of the game given all the past information can be no greater than your current winnings: $$ \mathbb{E}[X_{n+1} | \mathcal{F}_n] \leq X_n.\tag{1} $$ If you suppose the house offers you a fair game, a known strategy is "double-or-nothing", which I think you describe. Let $\xi_i, i \geq$ be i.i.d. mean zero $\{\pm 1\}$ random variables, which represent whether we won (+1) or lost (-1). Let $B_n$ denote our wager associated to the outcome of $\xi_n$. Note that $B_n$ must be $\mathcal{F}_{n-1}$-measurable (we need to bet $B_n$ before we know the outcome $\xi_n$). Our winnings at step $n$ are given by $$X_n = X_0 +\sum_{i=1}^n B_i \xi_i.$$

Take our initial budget to be $X_0 = 1$ for simplicity. Double or nothing means $$B_n = \begin{cases} 2B_{n-1} & \xi_{n-1} = -1 \\ 1 & \xi_{n-1} = 1 \end{cases}.$$ Observe that $\mathbb{E}[X_{n+1} | \mathcal{F}_n] = X_n$, so equality is achieved in (1) and in this sense the strategy is the best we can hope for. $X_n$ is called a martingale and in particular $\mathbb{E} X_n = \mathbb{E} X_0$ for any deterministic time $n \geq 1$.

The optional sampling theorem allows you to make the claim $\mathbb{E} X_\tau = \mathbb{E} X_0$ for random times $\tau$. Let $\tau = \inf\{n \geq 1 : X_n = 2\}$, the first time we hit 2 (when our strategy succeeds). The probability that we succeed is $$\mathbb{P}(\tau < \infty).$$ In fact, $\mathbb{P}(\tau < \infty) = 1$. However, the strategy is not valid, because the expected time to hit is infinite: $\mathbb{E} \tau = \infty$. In particular, one cannot apply the optional sampling theorem, $$ 1 = \mathbb{E} X_0 = \mathbb{E} X_{\min(\tau, n)} \neq \lim_{n \rightarrow\infty} \mathbb{E} X_\tau = 2. $$ So, you will always win, but you have to play forever.

Answer 2

1. First, suppose for simplicity that you bet the same amount $a$ every time. We can treat the roulette game as a binomial distribution with probability $p$ of success (red). For a single round, your winnings change by $+a$ if you win and $-a$ if you lose. For multiple rounds, your winnings are equal to $a$ times (number of wins - number of losses). If you have $k$ wins out of $n$ total rounds, your winnings are equivalently $+a(2k-n)$.

   Let's assume you start with a large amount of money so we can neglect the possibility of going bankrupt. In the long run after $n$ rounds, the distribution of number of wins (a binomial distribution) becomes asymptotically close to a normal distribution (this is the de Moivre–Laplace theorem). The normal distribution has mean $np$ and standard deviation $\sqrt{n\cdot p \cdot (1-p)}$. Your winnings $w$, which are an arithmetic function of the number of wins $k$, are therefore also asymptotically normally distributed:

   $$k \simeq N\left(np, \sqrt{np(1-p)}\right)$$ $$w = a(2k-n) \simeq N\left(na(2p-1), 2a\sqrt{np(1-p)}\right)$$

   To get the win rate, we divide the winnings distribution by $n$: $$w/n \simeq N\left(a(2p-1), 2a\sqrt{p(1-p)/n}\right)$$

   Note that this normal distribution has a constant mean of $a(2p-1)$ and a standard deviation that narrows to 0 as $n\rightarrow \infty$, so that the long term winnings (assuming you can't go bankrupt) are equal to $a(2p-1)$ almost certainly. This is the same as $a(p) + (-a)(1-p)$, the expected winnings for a single wager, Q.E.D.

2. If you can bet a different amount each time, then at any point in the game so far let $a(n)$ be the average amount you bet over all $n$ rounds. By the above reasoning and by linearity of expectation, your average winnings converge onto the expectation for a single game in which you bet the average bet amount $a(n)$.

   (To see this intuitively, imagine listing the rounds you've played so far and grouping them based on the amount you bet during that round. Then within each individual group, you bet a constant amount, so by the previous result, the winnings in that group converge onto the expected winnings for a single round. The winnings overall are the sum over the winnings for each group, weighted by the fraction of the rounds in each group.)

   Of course, depending on your strategy, you can make $a(n)$ converge or fail to converge to a particular value, in which case your expected earnings won't converge to the expected value of a single game. For example, if your strategy is "bet the minimum amount for 1 round, then the maximum amount for two rounds, then the minimum amount for three rounds, then the maximum for four rounds,...", $a(n)$ will never converge.

Answer 3

Here is an answer that considers general betting as you describe, in the situation where your bet is limited by your total money $X_n$ at round $n$. Let $a,b$ be the min and max bet where $0<a<b\leq\infty$. We show that, regardless of your betting strategy, $X_n\rightarrow X_{\infty} \in [0, a)$ with prob 1. In particular, you do not necessarily loose all your money, but you eventually come to a point where your total money is less than the minimum bet $a$ (so you can no longer play).

Setup

There are fixed minimum and maximum bets $a,b$ with $0<a<b\leq\infty$. You start with a given constant $X_0$ dollars with $a\leq X_0<\infty$. Let $B_n$ be the amount you bet on step $n \in \{1, 2, 3, ...\}$ and let $X_n$ be the money you have at the end of step $n$. Assume the rule:

• If $X_{n-1}\geq a$ then you keep playing and $B_n \in [a,\min[b,X_{n-1}]]$.
• If $X_{n-1} < a$ then you stop playing and $B_n=0$. Also $B_{m}=0$ for all $m>n$.

Given you bet $B_n\geq 0$ dollars, you receive back $B_n G_n$ dollars, where $G_n$ is a random gain variable (assume $G_n\geq 0$). Then: $$ X_{n} = X_{n-1} + B_n(G_n-1) \quad \forall n \in \{1, 2, 3, ...\}\quad (Eq. 1)$$

As an example, in your green, red, black situation we have: $$ G_n = \left\{ \begin{array}{ll} 2 &\mbox{ if play $n$ is red} \\ 0 & \mbox{ otherwise} \end{array} \right.$$

Assumptions

The history $\mathcal{H}_n$ is defined for each $n \in \{0, 1, 2, ...\}$ by $$ \mathcal{H}_n = (X_0, X_1, ..., X_n, B_1, ..., B_n, G_1, ..., G_n)$$ Assume that $\{G_n\}_{n=1}^{\infty}$ are i.i.d. with $G_n\geq 0$ for all $n$ and $$E[G_n]\leq 1, \quad 0<Var(G_n)<\infty$$ Assume that $B_n$ depends only on the history $\mathcal{H}_{n-1}$ and not on $G_n$. In particular: $$ E[B_n(G_n-1)|\mathcal{H}_{n-1}] = E[B_n|\mathcal{H}_{n-1}]E[G_n-1]$$

Analysis

Since we cannot bet more than we have, (Eq. 1) implies $X_n\geq 0$ (surely) for all $n \in \{0, 1,2, ...\}$. We get from (Eq. 1) for all $n \in \{1, 2, 3, ...\}$: \begin{align} E[X_n | \mathcal{H}_{n-1}] &= X_{n-1} + E[B_n(G_n-1)|\mathcal{H}_{n-1}]\\ &= X_{n-1} + \underbrace{E[B_n|\mathcal{H}_{n-1}]}_{\geq 0}\underbrace{E[G_n-1]}_{\leq 0}\\ &\leq X_{n-1} \end{align} Thus, $X_n$ is a nonnegative supermartingale and we know by the standard martingale convergence theorem: $$ X_n\rightarrow X_{\infty} \quad \mbox{with prob 1}$$ where $X_{\infty}$ is a random variable. On the other hand, since there is a minimum bet $a>0$ whenever we have at least $a$ money, whenever $X_{n-1}\geq a$ we have: $$E[B_n^2(G_n-1)^2|\mathcal{H}_{n-1}]\geq a^2E[(G_1-1)^2]>0$$ So it is impossible to have $X_{\infty}>a$ (since then we would keep changing by non-diminishing and non-negligible steps, without convergence to any value $X_{\infty}$). So we know that $X_{\infty} <a$. Thus, we eventually stop playing with prob 1 and $X_{\infty} \in [0,a)$.

Answer 4

The second part of the conjecture is also true. There's a version of the strong law of large numbers that applies to sequences of rv's that aren't i.i.d. If you have a sequence $ \{ X_i \} $ where each term satisfies $ E[X_i]=0 $ and each pair satisfies $ E[X_i X_j] = 0$, then if the variances satisfy

$$ \sum \frac{\sigma^2_i (\log i)^2 }{i^2} < \infty,$$

you can conclude that the sample mean $ \bar{X}_n \rightarrow 0 $ almost surely. This is stated in Rao's Linear Statistical Inference and Its Applications in the section on limit theorems. For the roulette problem, the condition on the variances is automatic because of the maximum bet limit. Clearly, the expectations and covariances of the amounts gained at each trial are not zero, though.

There's a dirty trick that can fix this. Since we are assuming unlimited capital for this part of the problem, we can keep the chips we win in a separate pile from the chips we use to place bets (a chip is the minimum bet, or unit). Then simply decree that the chips in the two piles have different "values". Of course, when we go to calculate actual gain and loss, we revert back to their real values, which are equal.

The success probability for each trial is $p$, and the odds they pay you for winning are $\beta:1$. For this example $\beta = 1$ and $p=18/38$. We pretend that the chips we use to place bets are worth $p\beta$ each, and that the chips we receive on winning are worth $q = 1-p$ each. That way, the expected "value" for each trial is $E[X_i] = E[B_i](p(\beta q) - q(p \beta)) = 0$. $B_i$ is the amount bet on the i-th trial, which can only depend on previous outcomes and betting decisions. Similarly, $E[X_iX_j] = 0$ also. So we can conclude that $S_n /n = \frac{1}{n}\sum_{i=1}^n X_i \rightarrow 0$ a.s.

That applies to the pretend "values". To convert back to real values, we work with the number $W_n$ of chips won and the number $L_n$ of chips lost after $n$ trials. The separate piles allow us to track these. After $n$ trials, our "winnings" pile has increased by $W_n$ chips, and the pile from which we place bets has decreased by $L_n$ chips. Note that $S_n = qW_n - p\beta L_n$. The actual number of chips gained (what we're really interested in) is $G_n = W_n - L_n$. The total number of chips wagered after $n$ trials is $T_n = W_n/\beta + L_n$. Starting with $S_n/n \rightarrow 0$, and after some algebra, you can get

$$ G_n/T_n \rightarrow p\beta - q $$ almost surely.

Another way to show it is found in this article: On the Failure of Gambling Systems under Unfavorable Odds