One-third game, climbing from 2 to 5

Question

Each bet wins 1 point with probability p = 1/3 and loses 1 point with probability q = 2/3. You start at 2 points and stop at 5 points (win) or 0 points (broke). What is the probability you reach 5 before 0?

Accepted Answer

Use the biased-ruin formula with odds ratio r = q/p: hi = \frac{1 - r^i}{1 - r^N}. Here r = (2/3)/(1/3) = 2, i = 2, N = 5: h2 = \frac{1 - 2^2}{1 - 2^5} = \frac{1 - 4}{1 - 32} = \frac{-3}{-31} = \frac{3}{31} \approx 0.097. Less than a 10% chance. Being closer to broke than to the target, combined with losing twice as often as winning, makes success rare. Takeaway. The ratio r = q/p = 2 drives the answer \frac{1 - r^i}{1 - r^N} = 3/31; each step you are behind hurts far more than in a fair game.

One-third game, climbing from 2 to 5

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