Chips at a fair table, reaching the cash-out goal

Question

A player sits down with 4 chips. Each round she makes a fair even-money bet of one chip: with probability 	frac12 she gains a chip, and with probability 	frac12 she loses one. She keeps playing until she either reaches 10 chips and cashes out, or drops to 0 chips and is broke. What is the probability she reaches 10 chips before going broke?

Accepted Answer

Let pk be the probability of reaching 10 before 0 when she currently holds k chips. The two ends are certain: p0 = 0 (already broke) and p{10} = 1 (already cashed out). First-step analysis. From k chips (with 0 < k < 10), one round sends her to k+1 or k-1, each with probability 	frac12: pk = 	frac12 p{k-1} + 	frac12 p{k+1}. Rearranging, p{k+1} - pk = pk - p{k-1}: the gaps between consecutive values are all equal. So pk is a straight line in k. With p0 = 0 and p{10} = 1, that line is pk = \frac{k}{10}. Starting from k = 4: p4 = \frac{4}{10} = \frac{2}{5} = 0.4. Takeaway. In a fair (symmetri…

Chips at a fair table, reaching the cash-out goal

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