Unfavorable game, 3 up to 6 with a one-third edge

Question

You play a game where 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).

What is the probability you reach 6 points before 0?

Accepted Answer

For a biased game let r = q/p be the loss-to-win odds ratio. Conditioning on one bet, hi = p\,h{i+1} + q\,h{i-1}. The solutions are no longer straight lines; they stack up in powers of r: hi = \frac{1 - r^i}{1 - r^N}, \qquad r = \frac{q}{p}. Here r = (2/3)/(1/3) = 2, with i = 3 and N = 6: h3 = \frac{1 - 2^3}{1 - 2^6} = \frac{1 - 8}{1 - 64} = \frac{-7}{-63} = \frac{7}{63} = \frac19 \approx 0.111. Barely an 11% chance. The unfavorable odds crush the simple "halfway" intuition: even though 3 is halfway to 6, the answer sits far below 1/2 because losses are twice as likely as wins. Takeaway. With…

Unfavorable game, 3 up to 6 with a one-third edge

Your answer

More Probability questions