You start with $10 and bet $1 per round on a game you win with probability 1/3 and lose with probability 2/3. You stop at $0 or at $20. What is the probability you reach $20?

You start with $10 and bet $1 per round on a game you win with probability 1/3 and lose with probability 2/3. You stop at $0 or at $20. What is the probability you reach $20?

Approach: Use the ratio r = q/p and check that r raised to the current wealth is a martingale, then apply optional stopping at the two absorbing barriers.

1/1025. With p = 1/3 and q = 2/3 the ratio is r = q/p = 2, and 2^{X_t} is a martingale because (1/3)2^{X+1} + (2/3)2^{X-1} = 2^X. Optional stopping at the absorbing barriers gives 2^10 = P * 2^20 + (1 - P) * 1, so the gambler's ruin formula returns P = (2^10 - 1)/(2^20 - 1) = 1023/1048575. Since 2^20 - 1 factors as (2^10 - 1)(2^10 + 1), this collapses to 1/1025 = 0.000976. Doubling a stake at 2 to 1 against is close to hopeless, while the same target on a fair coin would be reached with probability 1/2.

Follow-up: What is the expected number of rounds before you stop?

Key concepts: gambler's ruin, martingale, optional stopping, absorbing barriers.