How long does a losing game last, 3 toward 6

Question

Each bet wins 1 point with probability p = 1/3 and loses 1 point with probability q = 2/3. You start at 3 points and stop at 6 points (win) or 0 points (broke). On average, how many bets until the game ends?

Accepted Answer

The drift is now negative: \mu = p - q = -	frac13 per bet. The recipe still holds, 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 3 toward N = 6 is h3 = \frac{1 - 2^3}{1 - 2^6} = \frac{-7}{-63} = \frac19. So E[	ext{final}] = 6 	imes 	frac19 = 	frac23, and both the net displacement and the drift are negative, so their ratio is positive: E[T] = \frac{2/3 - 3}{-1/3} = \frac{-7/3}{-1/3} = 7. You almost always go broke, and it happens fairly fast: about 7 bets on average. Takeaway. With negative drift the sa…

How long does a losing game last, 3 toward 6

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