Biased game length, 1 up to 3 with a two-thirds edge

Question

Each bet wins 1 point with probability p = 2/3 and loses 1 point with probability q = 1/3. You start at 1 point and stop at 3 points (win) or 0 points (broke). On average, how many bets until the game ends?

Accepted Answer

Average drift per bet is \mu = p - q = 	frac13, so E[T] = \frac{E[	ext{final}] - 	ext{start}}{p - q}, \qquad E[	ext{final}] = N \cdot hi. With r = q/p = 1/2, the win probability from 1 point toward N = 3 is h1 = \frac{1 - 1/2}{1 - (1/2)^3} = \frac{1/2}{7/8} = \frac47. So E[	ext{final}] = 3 	imes 	frac47 = 	frac{12}{7}, and E[T] = \frac{12/7 - 1}{1/3} = \frac{5/7}{1/3} = \frac{15}{7} \approx 2.143. Takeaway. Even starting one point up, the same recipe applies: E[T] = \frac{N hi - i}{p - q} = 15/7 bets on average.

Biased game length, 1 up to 3 with a two-thirds edge

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