A favorable game, odds of reaching the target

Question

A gambler starts with 1 token. Each round he wins a token with probability 	frac23 and loses one with probability 	frac13. He plays until he reaches 3 tokens (cash out) or 0 tokens (broke). What is the probability he reaches 3 tokens before going broke?

Accepted Answer

Let pk be the probability of reaching 3 before 0 from k tokens, with p0 = 0, p3 = 1. First-step analysis gives pk = 	frac23 p{k+1} + 	frac13 p{k-1}. The ratio is r = q/p = (1/3)/(2/3) = 	frac12, and the solution is pk = \frac{1 - r^k}{1 - r^N} = \frac{1 - (1/2)^k}{1 - (1/2)^3}, \qquad N = 3. At k = 1: p1 = \frac{1 - 1/2}{1 - 1/8} = \frac{1/2}{7/8} = \frac{4}{7} \approx 0.571. Takeaway. When the walk drifts toward the goal (r < 1), even starting one step up gives a solid 4/7 chance. The formula \frac{1 - r^k}{1 - r^N} handles favorable and unfavorable games alike.

A favorable game, odds of reaching the target

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