In the American version of the Game Roulette, a wheel has 18 black slots, 8 red slots and 2 green slots. All slotsare the same size. In a carnival game, a person wagers $2 on the roll of two dice. A person can wager on eitherred or black. Green is reserved for the house. If a player wagers $5 on either red or black and that color comesup, they win $10 otherwise they lose their wager. What is the expected value of playing the game once?
A) -$0.26
B) -$0.50
C) $0.26
D) $0.50

Question

Quizplus · Accepted Answer

The expected value (EV) can be calculated using the formula: EV = (probability of winning * amount won) + (probability of losing * amount lost). The probability of winning if you bet on red is 18/38 (since there are 18 red slots out of a total of 38 slots), and the probability of winning if you bet on black is also 18/38. The probability of losing (either by landing on the opposite color or on green) is 20/38. If you win, you gain $10 (since you get your $5 back and win an additional $5), and if you lose, you lose your $5 wager. Therefore, the EV for both red and black bets is: EV = (18/38 * $10) + (20/38 * -$5) = $0.26 (approximately). However, since you lose money on average, the correct answer is -$0.26.

In the American Version of the Game Roulette, a Wheel