Exercise: Martingale Strategy — Doubling-Up Is Not Free Money

Prerequisites: Martingales (Discrete Time)

Problem

A gambler follows the classical "martingale" betting strategy: start with a bet of 1 unit on a fair coin. After each loss, double the next bet; after a win, reset to 1 unit. Suppose the gambler has bankroll $B$ and plays until they either win at least one round or lose $N$ consecutive times (exhausting their bankroll).

Let $X_n$ be the gambler's wealth after $n$ rounds, with $\mathcal{F}_n = \sigma(\text{rounds } 1, \ldots, n)$, and $X_0 = B$.

1. After the first win, the gambler's net profit for that sequence is $+1$ unit (they recover all previous losses and gain 1). Explain why in one or two sentences.

2. Compute the total loss if the gambler loses $N$ times in a row. Show that this equals $2^N - 1$.

3. Deduce the maximum value of $N$ for a bankroll $B$: the gambler can sustain at most $N = \lfloor \log_2(B + 1)\rfloor$ consecutive losses.

4. Let $\tau$ be the stopping time "first win, or $N$ consecutive losses." Compute $\mathbb{E}[X_\tau - X_0]$ exactly. Show that despite the intuitive "I always win eventually" feel of the doubling strategy, the expected P&L is zero — consistent with the optional stopping theorem.
5. Conceptual. Real casinos have a table limit that effectively caps $N$. Why is the doubling strategy especially dangerous (i.e. why is the actual variance of $X_\tau$ huge even with $\mathbb{E}[X_\tau] = X_0$)?

Hint

For part 4, the probability of $N$ consecutive losses is $2^{-N}$. The strategy gives $+1$ with probability $1 - 2^{-N}$ and $-(2^N - 1)$ with probability $2^{-N}$.