The martingale that "can't lose"

Question

A trader proposes a "sure thing" on a fair, even-money coin. Bet $\$ 1 $. If you lose, bet$ $2 $; if you lose again, bet$ $4 $, then$ $8 $, doubling until your first win, then stop. On the winning flip you recover everything wagered plus$ $1$.

Show the profit really is $\$ 1$ whenever you win, then explain why this is not free money. What is the expected profit with a finite bankroll?

Accepted Answer

The bookkeeping checks out. Suppose your first win is on flip k. You wagered 1 + 2 + 4 + \dots + 2^{k-1} = 2^k - 1 across the losses, and the winning flip pays 2^k (even money on a 2^{k-1} bet returns your stake plus an equal amount, netting the geometric total). Net profit = 2^k - (2^k - 1) = \1. So conditional on ever winning, you pocket exactly \1. That is the seductive part. The catch: an unbounded bankroll is doing all the work. To guarantee the win you must be able to keep doubling forever. With a real, finite bankroll you can afford at most N doublings (a bankroll of 2^N - 1). If you l…

The martingale that "can't lose"

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