Expected length of a biased game

Question

Each bet wins 1 point with probability p = 2/3 and loses 1 point with probability q = 1/3. You start at 2 points and stop at 4 points (win) or 0 points (broke). On average, how many bets until the game ends?

Accepted Answer

Two facts combine. First, each bet moves you by an average of \mu = p - q = 	frac13 point. Over T bets your average net move is \mu \cdot E[T], and that must equal how far you end up on average minus where you started: E[T] = \frac{E[	ext{final}] - 	ext{start}}{p - q}. Second, the final position is N = 4 if you win and 0 if you go broke, so E[	ext{final}] = N \cdot hi where hi is the win probability. Using the biased-ruin formula with r = q/p = 1/2, h2 = \frac{1 - (1/2)^2}{1 - (1/2)^4} = \frac{3/4}{15/16} = \frac45. So E[	ext{final}] = 4 	imes 	frac45 = 	frac{16}{5} = 3.2, and E[T] = …

Expected length of a biased game

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