Question

In a carnival game, a person wagers $2 on the roll of two dice. If the total of the two dice is 2, 3, 4, 5, or 6 then the person gets $4 (the $2 wager and $2 winnings). If the total of the two dice is 8, 9, 10, 11, or 12 then the person gets nothing (loses $2). If the total of the two dice is 7, the person gets $0.75 back (loses $0.25). What is the expected value of playing the game once?

Answer 1

The expected value of playing the game once is -$0.04

Explanation:

Consider the provided information.

If the total of the two dice is 2, 3, 4, 5, or 6 then the person gets $4 (the $2 wager and $2 winnings).

The total number of outcomes are:

(1,1) (1,2) (1,3) (1,4) (1,5) (1,6)

(2,1) (2,2) (2,3) (2,4) (2,5) (2,6)

(3,1) (3,2) (3,3) (3,4) (3,5) (3,6)

(4,1) (4,2) (4,3) (4,4) (4,5) (4,6)

(5,1) (5,2) (5,3) (5,4) (5,5) (5,6)

(6,1) (6,2) (6,3) (6,4) (6,5) (6,6)

The probability of getting sum 2 is
`(1)/(36)`

The probability of getting sum 3 is
`(2)/(36)`

The probability of getting sum 4 is
`(3)/(36)`

The probability of getting sum 5 is
`(4)/(36)`

The probability of getting sum 6 is
`(5)/(36)`

`P(2, 3, 4, 5 or 6) = (1 + 2 + 3 + 4 + 5)/(36) =(15)/(36)`

Similarly, the probability of getting sum 8, 9, 10, 11, or 12 is:

`P(8, 9, 10, 11, or 12) = (1 + 2 + 3 + 4 + 5)/(36) =(15)/(36)`

The probability of P(7) is
`(6)/(36)`

If the total of the two dice is 2, 3, 4, 5, or 6 then the person gets $4.

If the total of the two dice is 8, 9, 10, 11, or 12 then the person gets nothing (loses $2).

If the total of the two dice is 7, the person gets $0.75 back (loses $0.25).

Thus the required expected value is:

`(15)/(36)*2-2*(15)/(36)-0.25*(6)/(36)`

`(30)/(36)-(30)/(36)-(0.25)/(6)`

`-(0.25)/(6)\approx-0.04`

Hence, the expected value of playing the game once is -$0.04