Bold play beats timid play at an unfair table

Question

You arrive at a roulette table with \100 and want to walk out with \200. Each even-money bet wins with probability p = 0.474 (a bit less than half, the house edge) and loses otherwise. You can bet however you like each spin, as long as you never bet more than you have. Compare two strategies: "timid" (bet \1 at a time until you reach \200 or go broke) versus "bold" (bet everything you can toward the goal). Which gives a better chance of reaching \200?

Accepted Answer

Bold play. To go from \100 to \200, bet the whole \100 on one spin. You reach the goal iff that single bet wins: P(	ext{bold success}) = p = 0.474 \approx 47.4\%. Timid play. This is biased gambler's ruin from i = 100 to N = 200 with q/p = 0.526/0.474 \approx 1.1097. The success probability is \frac{1 - (q/p)^{i}}{1 - (q/p)^{N}} = \frac{1 - r^{100}}{1 - r^{200}}, \qquad r = 1.1097. Now r^{100} = e^{100\ln r} = e^{100(0.10416)} = e^{10.42} \approx 3.35	imes 10^{4} and r^{200}\approx 1.12	imes 10^{9}, so P(	ext{timid success}) \approx \frac{-3.35	imes10^{4}}{-1.12	imes10^{9}} \approx 3	i…

Bold play beats timid play at an unfair table

Your answer

More Probability questions