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 thenthe person gets $4 (the $2 wager and $2 winnings). If the total of the two dice is 8, 9, 10, 11, or 12 then theperson gets nothing (loses $2). If the total of the two dice is 7, the person gets $0.75 back (loses $0.25). What isthe expected value of playing the game once?
A) -$0.04
B) -$0.42
C) $2.00
D) $0.00

Question

Quizplus · Accepted Answer

The expected value (EV) can be calculated by multiplying each outcome by its probability and summing these products. There are 36 possible outcomes when rolling two dice (6 sides per die).- Winning $2 (getting $4 back): This happens if the total is 2, 3, 4, 5, or 6. The probabilities are 1/36 (for a total of 2), 2/36 (for a total of 3), 3/36 (for a total of 4), 4/36 (for a total of 5), and 5/36 (for a total of 6), summing to 15/36.- Losing $2: This happens if the total is 8, 9, 10, 11, or 12. The probabilities are 5/36 (for a total of 8), 4/36 (for a total of 9), 3/36 (for a total of 10), 2/36 (for a total of 11), and 1/36 (for a total of 12), summing to 15/36.- Losing $0.25 (getting $0.75 back): This happens if the total is 7, with a probability of 6/36.Calculating EV:- For winning $2: (15/36) * $2 = $0.8333- For losing $2: (15/36) * -$2 = -$0.8333- For losing $0.25: (6/36) * -$0.25 = -$0.0417Summing these gives the EV of playing the game once: $0.8333 - $0.8333 - $0.0417 = -$0.0417, which rounds to -$0.04.

In a Carnival Game, a Person Wagers $2 on the Roll