Expected rounds of an uphill game, 2 toward 5

Question

Each bet wins 1 point with probability p = 1/3 and loses 1 point with probability q = 2/3. You start at 2 points and stop at 5 points (win) or 0 points (broke). On average, how many bets until the game ends?

Accepted Answer

The drift per bet is \mu = p - q = -	frac13, and E[T] = \frac{E[	ext{final}] - 	ext{start}}{p - q}, \qquad E[	ext{final}] = N \cdot hi. With r = q/p = 2, the win probability from 2 toward N = 5 is h2 = \frac{1 - 2^2}{1 - 2^5} = \frac{-3}{-31} = \frac{3}{31}. So E[	ext{final}] = 5 	imes 	frac{3}{31} = 	frac{15}{31}, and E[T] = \frac{15/31 - 2}{-1/3} = \frac{-47/31}{-1/3} = \frac{47}{31} \cdot 3 = \frac{141}{31} \approx 4.548. Takeaway. The drift recipe \frac{N hi - i}{p - q} handles any biased game; here it gives 141/31 \approx 4.55 bets.

Expected rounds of an uphill game, 2 toward 5

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