Question

One of the wagers in roulette is to bet that the ball will stop on a number that is a multiple of 3. (Both 0 and 00 are not included.) If the ball stops on such a number, the player wins double the amount bet. If a player bets $3, compute the player's expectation. (Round your answer to two decimal places.)

Answer 1

The players expectation is -0.1579

Step-by-step explanation:

Consider the Roulette numbers to from 1-36, also the two green zeroes 0 and 00,

Thus, there are 38 places on which the ball can land.

On first 36 numbers, 12 of them are divisible by 3. Also there are 26 numbers which are not divisible by 3.

The probability of a wager winning is: 12/38 or 6/19

Now the probability of a wager to loss is: 26/38 or 13/19

So for a 6/19 chance of winning $6 and a 13/19 chance of losing $3.

Therefore, the expected outcome is:

6 × 6/19 + -3 × 13/19 = 1.8947 - 2.0526 = -0.1579

The expected value comes up with a negative sign. Hence, he will lose money.