Smaller cushion, 3 units with a 60 percent edge

Question

Each bet wins 1 unit with probability p = 0.6 or loses 1 unit with probability q = 0.4. You start with 3 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 = 0.4/0.6 = 2/3 and i = 3: \left(	frac23ight)^3 = \frac{8}{27} \approx 0.296. Dropping from 5 units to 3 raises the risk from about 13% to about 30%: two fewer units of cushion cost you two factors of 2/3. Takeaway. Ruin risk is (q/p)^i; thinner bankroll means fewer protective factors of q/p, here 8/27.

Smaller cushion, 3 units with a 60 percent edge

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