Question

In American roulette, the wheel contains the numbers 1 through 36 , alternating between black and red. There are two green spaces numbered 0 and 00 . A player places a bet of $9.00 on red to play the game. If the ball lands on red, the player gets a $9.00 for winning and receives the money back. If the ball does not land on red, then the player simply loses the $9.00 placed on the bet. Calculate the expected winnings on one game. Round your answer to the nearest cent.

Answer 1

The expected winnings for one game of American roulette when betting on red is a loss of $0.47, as there is a slightly higher chance of losing than winning due to the presence of two green spaces on the wheel.

Step-by-step explanation:

To determine the expected winnings from one game of American roulette when betting on red, we must consider the total number of red numbers and the green numbers 0 and 00 that will result in a loss. There are 18 red numbers and 20 numbers that are not red (18 black numbers + 2 green numbers), making the wheel a total of 38 numbers.

The probability of winning is the number of red spaces divided by the total number of spaces. So, P(win) = 18/38 and P(lose) = 20/38. If you win, you earn $9.00; if you lose, you lose your bet of $9.00.

The expected value (EV) is calculated as follows:

EV = (P(win) × winnings) + (P(lose) × loss)

EV = ((18/38) × $9.00) + ((20/38) × -$9.00)

EV = ($4.74) + (-$4.74) = -0.47

Therefore, the expected winnings on one game are -$0.47, meaning you would expect to lose, on average, 47 cents per game.

Answer 2

sorry I won't tell anyone about this