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Casino Gambling: Random Variables and Probability Distributions

Casino gambling yields over $35 billion in revenue each year in the US. Casino games of pure chance (e.g., craps, roulette, baccarat, and keno) always yield a house advantage. For example in the game of double-zero roulette, the expected casino win percentage is 5.26% on bets made on whether the outcome will be either black or red. (This implies that for every $5 bet on black or red the casino will earn a net of about .25 cents. It can be shown that in 100 roulette plays on black/red the average casino win percentage is normally distributed with mean 5.26% and std. dev. 10%. Let x represent the average casino win percentage after 100 bets on black/red in double-zero roulette.

a. Find P(x>0). (This is the probability that the casino wins money.)

b. Find P(5<x<15).

c. Find P(x<1).

d. If you observed an average casino win percentage of -25% after 100 roulette bets on black/red, what would you conclude?

Solution Preview

a. Find P(x>0). (This is the probability that the casino wins money.)
Note that x is a normal variable, and let "mean" be its mean. Let sd denote its standard deviation.
This means that z = (x - mean)/ sd is a standard normal variable whose mean is 0 and standard
deviation = 1.

Probability of x > 0 is the same as probability of z > -mean/sd i.e. z > -5.26/10 i.e z > -0.526
This ...

Solution Summary

This solution determines the average casino win percentage.