Steep odds, reaching three from one chip

Question

A player begins with 1 chip. Each round she wins a chip with probability 	frac14 and loses a chip with probability 	frac34, playing until she has 3 chips or 0 chips. What is the probability she reaches 3 chips before going broke?

Accepted Answer

With p0 = 0, p3 = 1, first-step analysis gives pk = 	frac14 p{k+1} + 	frac34 p{k-1}. The ratio is r = q/p = (3/4)/(1/4) = 3, so pk = \frac{1 - 3^k}{1 - 3^N} = \frac{3^k - 1}{3^N - 1}, \qquad N = 3. At k = 1: p1 = \frac{3 - 1}{3^3 - 1} = \frac{2}{26} = \frac{1}{13} \approx 0.077. Takeaway. The bigger the edge against you, the larger r, and the ruin formula \frac{r^k - 1}{r^N - 1} collapses toward zero: only 1/13 here.

Steep odds, reaching three from one chip

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