Gambler's ruin on a fair coin

Question

You start with \30 and bet \1 repeatedly on a fair coin: each bet wins with probability p = 0.5 and loses with probability 0.5, independently. You stop at \100 (win) or \0 (broke). What is the probability you reach \100? Derive the general fair-game formula for start i and target N.

Accepted Answer

Let hi = P(	ext{hit } N 	ext{ before } 0 \mid 	ext{start } i), with h0 = 0 and hN = 1. Conditioning on the first bet, hi = 	frac12 h{i+1} + 	frac12 h{i-1} \quad\Longrightarrow\quad h{i+1} - hi = hi - h{i-1}. The consecutive differences are all equal, so hi is linear in i: hi = A + Bi. The boundary conditions h0 = 0 and hN = 1 give A = 0 and B = 1/N, hence \boxed{\,hi = \frac{i}{N}.\,} Numbers. With start 30 and target 100, h{30} = \frac{30}{100} = 0.30. So a 30\% chance of tripling to 100 before going broke. The result is beautifully simple: with no edge, your probability of reaching the…

Gambler's ruin on a fair coin

Your answer

More Probability questions