Strong edge, from 1 point to 3 winning three in four

Question

Each bet wins 1 point with probability p = 3/4 and loses 1 point with probability q = 1/4. You start at 1 point and stop at 3 points (win) or 0 points (broke). What is the probability you reach 3 before 0?

Accepted Answer

The biased-ruin formula with odds ratio r = q/p gives hi = \frac{1 - r^i}{1 - r^N}. Here r = (1/4)/(3/4) = 1/3, i = 1, N = 3: h1 = \frac{1 - 1/3}{1 - (1/3)^3} = \frac{2/3}{1 - 1/27} = \frac{2/3}{26/27} = \frac{2}{3} \cdot \frac{27}{26} = \frac{9}{13} \approx 0.692. You start just one point above the floor, yet the strong edge already makes you a nearly 70% favorite. Takeaway. A large edge sends r = q/p to 1/3, and even a single point of cushion yields \frac{1 - r^i}{1 - r^N} = 9/13; the ratio does the heavy lifting.

Strong edge, from 1 point to 3 winning three in four

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