Favorable game, 2 up to 4 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 2 points and stop at 4 points (win) or 0 points (broke). What is the probability you reach 4 before 0?

Accepted Answer

The biased-ruin formula uses the odds ratio r = q/p: hi = \frac{1 - r^i}{1 - r^N}. Now r = (1/3)/(2/3) = 1/2, i = 2, N = 4: h2 = \frac{1 - (1/2)^2}{1 - (1/2)^4} = \frac{1 - 1/4}{1 - 1/16} = \frac{3/4}{15/16} = \frac{3}{4} \cdot \frac{16}{15} = \frac{4}{5}. An 80% chance. With r < 1 the powers shrink instead of grow, and the edge pushes the win probability well above the fair-game value of 1/2. Takeaway. Flip the edge and r = q/p drops below 1, lifting the win chance to \frac{1 - r^i}{1 - r^N} = 4/5; a favorable game rewards you far beyond the naive halfway guess.

Favorable game, 2 up to 4 with a two-thirds edge

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