Chance the casino loses money over 100 plays

Question

A casino game earns the house an expected profit of $2 per play with a standard deviation of $20 per play. Plays are independent, so by the central limit theorem the total profit over many plays is approximately normal.

Over $n = 100$ plays, what is the probability the total profit is negative (below $0)?

Accepted Answer

Here we care about the total, not the average. When you add up n independent plays, the mean and the variance both add. So the total profit is approximately normal with 	ext{mean} = n\mu = 100 	imes 2 = 200, 	ext{SD} = \sigma\sqrt{n} = 20 	imes \sqrt{100} = 20 	imes 10 = 200. Notice the total's spread grows like \sqrt{n} (not n), so the swings do not keep pace with the steadily accumulating edge, which is why the house wins in the long run. Now find the z-score of the \0 threshold: z = \frac{0 - 200}{200} = -1. From the normal table: P(	ext{total} < 0) = P(Z < -1) \approx 0.1587, about …

Chance the casino loses money over 100 plays

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