Uphill betting, reaching the goal against the odds

Question

A player holds 2 chips. Each round she bets one chip on an unfair game: she wins a chip with probability 	frac13 and loses a chip with probability 	frac23. She plays until she reaches 4 chips (goal) or 0 chips (broke). What is the probability she reaches 4 chips before going broke?

Accepted Answer

Let pk be the probability of reaching 4 before 0 from k chips, with p0 = 0, p4 = 1. One round moves her up (prob 	frac13) or down (prob 	frac23): pk = 	frac13 p{k+1} + 	frac23 p{k-1}. Try pk = r^k for the varying part. Plugging in, r = 	frac13 r^2 + 	frac23, i.e. r^2 - 3r + 2 = 0, roots r = 1 and r = 2. So pk = A + B\cdot 2^k. The ratio here is r = q/p = (2/3)/(1/3) = 2, and matching the boundary values gives pk = \frac{1 - 2^k}{1 - 2^N} = \frac{2^k - 1}{2^N - 1}, \qquad N = 4. Therefore p2 = \frac{2^2 - 1}{2^4 - 1} = \frac{3}{15} = \frac15 = 0.2. Takeaway. With unfair steps the ruin pr…

Uphill betting, reaching the goal against the odds

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