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

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

Part 1

After $k$ consecutive losses, the cumulative loss is $1 + 2 + \cdots + 2^{k-1} = 2^k - 1$ . On the next round the gambler bets $2^k$ . If they win, the net is $2^k - (2^k - 1) = +1$ . Hence any win resets the net profit for the sequence to $+1$ , regardless of how many losses preceded.

Part 2

$N$ losses accumulate total bet $1 + 2 + \cdots + 2^{N-1} = 2^N - 1$ . This is also the maximum drawdown the strategy can sustain before ruin.

Part 3

The gambler can bet $2^k$ on round $k+1$ only if they still have at least $2^k$ unused — after $k$ losses they have $B - (2^k - 1)$ left. The constraint for the $(N-1)$ th loss to be survivable is $B - (2^{N-1} - 1) \ge 2^{N-1}$ , i.e. $B \ge 2^N - 1$ . Hence $N = \lfloor \log_2(B + 1)\rfloor$ .

Part 4

Let $W$ = "win within $N$ rounds" and $L$ = "lose $N$ rounds in a row." Then $\mathbb{P}(L) = 2^{-N}$ and $\mathbb{P}(W) = 1 - 2^{-N}$ .

\mathbb{E}[X_\tau - X_0] = (+1)\cdot(1 - 2^{-N}) + (-(2^N - 1))\cdot 2^{-N} = 1 - 2^{-N} - 1 + 2^{-N} = 0.

Expected P&L is zero. This matches the optional stopping theorem applied to the martingale "gambler's wealth under a predictable fair strategy": any stopping rule that is bounded (and $\tau \le N$ here) preserves $\mathbb{E}[X_\tau] = X_0$ .

Part 5

The variance of the strategy's P&L is

\operatorname{Var}(X_\tau) = 1^2\cdot(1 - 2^{-N}) + (2^N - 1)^2\cdot 2^{-N} - 0^2 = (1 - 2^{-N}) + (2^N - 1)^2/2^N.

For even modest $N$ this is enormous. At $N = 10$ (bankroll $B = 2^{10} - 1 = 1023$ units), the variance is $(2^{10} - 1)^2/2^{10} \approx 1021$ — a standard deviation of $\approx 32$ units for an expected-zero strategy. The strategy converts a small likely gain ( $+1$ with probability $1023/1024 \approx 99.9\%$ ) into a catastrophic rare loss ( $-1023$ with probability $\approx 0.1\%$ ).

This is the textbook example of compound risk: you can manufacture high expected short-run win rates by taking on long-tail loss exposure, but you cannot beat the unconditional mean. At a casino with a non-fair game (house edge), the expected P&L becomes strictly negative and the variance structure is unchanged — doubling-up is dominated in both dimensions. Wall Street's analogue: "picking up nickels in front of a steamroller" strategies like short-gamma vol-selling, which earn premium most of the time and blow up spectacularly on crisis days.

Takeaways

You cannot beat a fair game with predictable bet sizing. The martingale property is preserved under martingale transforms — the gambler's wealth remains a martingale under any predictable strategy.
A high win probability does not mean a profitable strategy. The doubling-up strategy wins 99.9% of the time and still has zero expected P&L, because the 0.1% losing event is catastrophic.
The variance of a zero-mean strategy can be arbitrarily large. Sharpe ratios hide nothing if you only look at mean P&L — you must inspect the variance, and better, the tail.
Real-casino edge makes things worse. With a house edge $\epsilon$ , the strategy's expected P&L is $-\epsilon\cdot(2^N - 1)\cdot \mathbb{P}(\text{some bet is placed})$ , a negative number that scales exponentially with bankroll. Doubling-up not only fails to generate free money; it loses money proportional to how aggressive the bankroll is.