Steep odds, winning one in four from 2 up to 3

Question

Each bet wins 1 point with probability p = 1/4 and loses 1 point with probability q = 3/4. You start at 2 points and stop at 3 points (win) or 0 points (broke). What is the probability you reach 3 before 0?

Accepted Answer

Apply the biased-ruin formula with odds ratio r = q/p: hi = \frac{1 - r^i}{1 - r^N}. Here r = (3/4)/(1/4) = 3, i = 2, N = 3: h2 = \frac{1 - 3^2}{1 - 3^3} = \frac{1 - 9}{1 - 27} = \frac{-8}{-26} = \frac{8}{26} = \frac{4}{13} \approx 0.308. Even though you start only one step from the goal, the harsh 3:1 loss ratio holds your chance near 31%. Takeaway. A bigger disadvantage means a bigger ratio r = 3, and the win chance \frac{1 - r^i}{1 - r^N} = 4/13 stays low even one step from home.

Steep odds, winning one in four from 2 up to 3

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