Thin edge, risk of ruin with a 55 percent bet

Question

Each bet wins 1 unit with probability p = 0.55 or loses 1 unit with probability q = 0.45. You start with 2 units and play forever against a house with unlimited money. What is the probability you eventually go broke?

Accepted Answer

With an edge and no ceiling, the eventual-ruin probability from i units is P(	ext{ruin}) = \left(\frac{q}{p}ight)^i. Here q/p = 0.45/0.55 = 9/11 and i = 2: \left(	frac{9}{11}ight)^2 = \frac{81}{121} \approx 0.669. A thin edge barely bends the odds: with a ratio q/p close to 1 and only 2 units, ruin is still about two chances in three. Takeaway. When the edge is small, q/p sits near 1 and (q/p)^i stays high; here 81/121 \approx 0.67. Small edges demand large bankrolls.

Thin edge, risk of ruin with a 55 percent bet

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