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 $2.25 on red to play the game. If the ball lands on red, the player gets a $2.25 for winning and receives the money
back. If the ball does not land on red, then the player simply loses the $2.25 placed on the bet. Calculate the expected winnings on one game.
Round your answer to the nearest cent.

Answer 1

The expected winnings on one game of American roulette when betting $2.25 on red is -$0.12, indicating that on average, the player will lose 12 cents per game.

Step-by-step explanation:

In American roulette, there are 18 red numbers, 18 black numbers, and 2 green numbers (0 and 00). When a player bets $2.25 on red, the player wins $2.25 plus the return of their bet if the ball lands on a red number. If the ball lands on black or green, the player loses their $2.25 bet.

To calculate the expected winnings on one game, we must consider the probability of each outcome and the winnings associated with that outcome. The probability of landing on red (P(red)) is 18/38, because there are 18 red spaces out of 38 total spaces. The probability of not landing on red (P(not red)) is 20/38, because there are 20 spaces that are not red.

The expected winnings (E) can be calculated as follows:

• E = (Winnings when landing on red × P(red)) + (Winnings when not landing on red × P(not red))
• E = ($2.25 × 18/38) + (-$2.25 × 20/38)
• E = $1.0658 - $1.1842
• E = -$0.1184

Rounded to the nearest cent, the expected winnings on one game is -$0.12. This means that, on average, a player can expect to lose 12 cents per game when betting on red in American roulette.

Given this negative expected value, the game is not favorable for the player in the long run.