Fair swaps, from 8 crystals up to 20

Question

A collector starts with 8 crystals. Each fair swap gains 1 crystal or loses 1 crystal, each with probability $1/2$ , independently. The collector stops upon reaching 20 crystals or hitting 0.

What is the probability of reaching 20 crystals before running out?

Accepted Answer

Let hi be the probability of reaching N = 20 before 0, starting from i, with h0 = 0 and h{20} = 1. One fair swap gives hi = 	frac12 h{i+1} + 	frac12 h{i-1}, so the gaps h{i+1} - hi are all equal and hi is a straight line, hi = i/N. Therefore h8 = \frac{8}{20} = \frac25. The same answer falls out of fairness: expected final crystals equal the start, 8, and the finish is either 20 or 0, so 20p = 8 gives p = 2/5. Takeaway. With no edge the answer is your fraction of the way to the goal: 8/20 = 2/5.

Fair swaps, from 8 crystals up to 20

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