Long shot from one chip against a house edge

Question

A player starts with 1 chip and bets one chip per round on a game that wins with probability 	frac13 and loses with probability 	frac23. She plays until reaching 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 = 	frac13 p{k+1} + 	frac23 p{k-1}. The ratio is r = q/p = 2, so pk = \frac{1 - 2^k}{1 - 2^N} = \frac{2^k - 1}{2^N - 1}, \qquad N = 3. At k = 1: p1 = \frac{2 - 1}{2^3 - 1} = \frac{1}{7} \approx 0.143. Takeaway. Against a house edge (r 1), starting far from the goal is punishing: a single chip converts to the 3-chip goal only 1/7 of the time.

Long shot from one chip against a house edge

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