Solution
The Favoured Gambler
Show the problem again
A gambler starts with $3. Each round she bets $1 and wins it with probability 2/3 or loses it with probability 1/3, independently. She stops when she reaches $0 or $6. What is the probability that she reaches $6? Give a reduced fraction.
Worked solution
The answer is 8/9. Let r = q/p = (1/3)/(2/3) = 1/2. The process r^X (X = current fortune) is a martingale, because p·r^(x+1) + q·r^(x−1) = r^x(pr + q/r) = r^x(q + p) = r^x. Optional stopping gives r^3 = P·r^6 + (1 − P)·1, so P = (1 − r^3)/(1 − r^6) = (7/8)/(63/64) = 8/9.
Source: Gambler's ruin with unequal odds, classical since Huygens and de Moivre. Statement written for AxiomIQ.