Question

A roulette wheel consists of 38 slots numbered 0, 00, and 1 through 36, evenly spaced around a wheel. In the game of roulette, a player can place a 10 bet on the number 16 and have a probability of winning. If the metal ball lands on 16, the player gets to keep the10 paid to play the game and the player is awarded an additional 350. Otherwise, the player is awarded nothing, and the casino takes the player's10. What is the expected value of the game to the player? (Round to the nearest cent.)

Answer 1

The expected value of the game to the player in roulette, when making a $10 bet on the number 16, is approximately -$0.27, meaning the player is expected to lose 27 cents per game on average.the correct answer is 27 cents per game.

Step-by-step explanation:

To calculate the expected value of the game for the player in roulette, you need to consider both the outcome of winning and the outcome of losing. The roulette wheel has 38 slots, and the player wins if the ball lands on number 16. There is only one chance to win (1/38) and 37 chances to lose (37/38). If the player wins, they gain $350 in addition to keeping their $10 bet, which sums to $360. If the player loses, they lose their $10 bet.

The expected value (EV) for the player can be calculated using the formula: EV = (Probability of Winning) x (Amount Won per Win) - (Probability of Losing) x (Amount Lost per Loss). Substituting the given values, we get EV = (1/38) x $360 - (37/38) x $10.

Let's do the math:

1. Calculate the expected value for winning: (1/38) x $360 = $9.47 (approximately)
2. Calculate the expected value for losing: (37/38) x $10 = $9.74 (approximately)
3. Subtract the expected value of losing from the expected value of winning: $9.47 - $9.74 = -$0.27

Therefore, the expected value of the game to the player is -$0.27, which means that, on average, the player is expected to lose approximately 27 cents per game.