In the casino game Roulette there is a wheel with 38 spots T

In the casino game Roulette, there is a wheel with 38 spots. The spots are labeled 1-36, 0, and 00. The wheel is spun and one of the spots is randomly chosen, with each spot equally likely to be chosen. One of the offered bets is to bet that an odd number will be chosen. If an odd number is chosen you win $1, and if any other number is chosen then you lose Si. Let X denote the random variable that denotes the money won/lost on one spin of the wheel (from your perspective). What is the expected value of X? Suppose you play the game for 5 spins. Let X, be the random variable that denotes the outcome of the ith roll, and let X denote the outcome over all 5 spins: i.e. X = X-1 + X_2 - X_3 + X-4 + X_5, . Use the linearity of expectation to compute the expected value of X.

Solution

In the casino game Rouletter, there is a wheel with 38 spots. The spots are labelled 1-36,
0, and 00. The wheel is spun and one of the spots is randomly chose, with each spot equally
likely to be chosen. One of the offered bets is to bet that an odd number will be chosen.
If an odd number is chosen you will $1, and if any other number is chosen then you lose $1.

1. Let X denote the random variable that denotes the money won/lost on one spin of the wheel
(from your perspective). What is the expected value of X?
There are a total of 38 spots.
Out of 38 spots, there will be a total of 18 odd spots, and 20 even spots.
Note that 0, and 00 are considered even numbers.

The probability of winning is: 18/38 i.e., 0.474, and the probability of loosing is 20/38 i.e., 0.526.

So, the expected value of X is: E(X)=xipi = (1) * 0.474 + (-1) * 0.526 = -0.052.