Question

4.28 An American roulette wheel has 18 red numbers, 18 black numbers, and 2 gree Assume all numbers are equally likely a. What is the probability the number is red? Black? Green? b. The player can make simple bets on the color. A dollar bet on red or blac n numbers. k returns a dollar the variance of this bet? profit if red or black comes up. What is the expected value of this bet? What is c Abet on a single number pays 35 to 1 (if the number comes up, the bettor profits 555 or each dollar bet). What is the expected value of this bet? What is the variance of this bet?

Answer 1

Given roulette wheel with 18 Red numbers, 18 Black numbers and 2 Green numbers, i.e. in all 38 numbers on a wheel.

1. Probability that Number appeared on wheel is Red = 18/38 = 0.473684

Probability that Number appeared on wheel is Black = 18/38 =

Probability that Number appeared on wheel is Green = 02/38 =

1. When player bet a dollar on color Red or Black it will return a dollar

Here if we define random variable X : Gain. Then it will take value +1 or -1

+1 if the color of betting is same as outcome color.

-1 if the color of betting is different than outcome color.

Now expected value of this game is given by

E(X) = +1 P[the color of betting is same as outcome color] -1 P[the color of betting is different than outcome color]

= 36/38 – 2/38 = 34/38 =

It means if player plays say for 100 times then he may win about 89 times.

Variance of this gain based game is

V(X) = 36/38+2/38 – (0.894737)² = 0.199446