Question

Recall that an American Roulette wheel has 38 possible outcomes, 18 black, 18 red and 2 green. The numbers go from 1 to 36, and the green outcomes are numbered 0 and 00. Someone places 10 bets on "1st 12", that is, that person wins if the outcome is any number between 1 and 12, included. 1. (2pts) What is the probability of losing a ’1st 12’ bet on a single spin? 2. (3pts) What is the probability of winning 7 out of the 10 total bets?

Answer 1

American Roulette

1. The probability of losing a ’1st 12’ bet on a single spin is:

= 0.67

2. The probability of winning 7 out of the 10 total bets is:

= 0.23

Explanation:

Possible outcomes in an American Roulette wheel = 38

Make-up of the outcomes = 18 black, 18 red and 2 green

Green outcomes are numbered 0 and 00

Number of wins for this person = 12

Probability of winning = 0.33 (12/36)

Number of possible losses for this person = 24

Probability of losing = 0.67 (24/36)

1. The probability of losing a '1st 12' bet on a single spin is:

= (36 - 12)/36

= 0.67

2. The probability of winning 7 out of the 10 total bets is:

Probability of winning = 12/36 * 7/10

= 0.23