Gambler's ruin against the house edge

Question

You start with \50 and bet \1 repeatedly; each bet wins with probability p = 0.49 and loses with probability q = 0.51, independently. You stop at \100 or at \0. What is the probability you reach \100? Derive the general formula for start i, target N, win probability p.

Accepted Answer

Let hi = P(	ext{hit } N 	ext{ before } 0 \mid 	ext{start } i), with h0 = 0, hN = 1, and hi = p\,h{i+1} + q\,h{i-1}. Trying hi = x^i gives px^2 - x + q = 0, with roots x = 1 and x = q/p (check: p + q = 1 makes both work). For p 
e q the general solution is hi = A + B(q/p)^i; the boundary conditions give hi = \frac{1 - (q/p)^i}{1 - (q/p)^N}. (The same formula drops out of optional stopping applied to the martingale Mt = (q/p)^{St}.) Numbers. With p = 0.49: r = q/p = \frac{51}{49} \approx 1.0408, and h{50} = \frac{1 - r^{50}}{1 - r^{100}} = \frac{1}{1 + r^{50}}, \qquad r^{50} = e^{50\ln(51/4…

Gambler's ruin against the house edge

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