You bet $1 per round on a fair coin and stop when you are either $10 ahead or $5 behind. What is the expected number of rounds you play?

You bet $1 per round on a fair coin and stop when you are either $10 ahead or $5 behind. What is the expected number of rounds you play?

Approach: The running total is a martingale, so the two absorption probabilities follow from optional stopping. Then use the compensated process S_t^2 - t to convert the squared exit values into the expected duration.

50. The optional stopping theorem applied to the martingale S_t gives 0 = 10 P(up) - 5(1 - P(up)), so P(up) = 1/3 and P(down) = 2/3. Since S_t^2 - t is also a martingale starting at 0, E[S_T^2] = E[T]. The exit values are +10 and -5, so the expected duration is (1/3)(100) + (2/3)(25) = 100/3 + 50/3 = 50. For absorbing barriers at +a and -b the answer is a*b, so widening one barrier lengthens the game linearly in the other. With a biased coin the same identity fails and Wald's first identity gives the duration instead.

Follow-up: How does the expected duration change if the coin wins with probability 0.49?

Key concepts: optional stopping theorem, martingale, absorbing barriers, expected duration.