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 $3.00 on red to play the game. If the ball lands on red, the player gets a $3.00 for winning and receives the money back. If the ball does not land on red, then the player simply loses the $3.00 placed on the bet. If the player places the same bet on red 24 times, what is the player's expected winnings? Round your answer to the nearest cent.

Answer 1

In American roulette, there are 18 red spaces out of a total of 38 spaces on the wheel. This means the probability of the ball landing on red is 18/38 or 9/19.

For each bet of $3.00 on red, the player has a 9/19 chance of winning $3.00 and a 10/19 chance of losing $3.00.

Expected winnings per bet = (Probability of winning * Amount won) + (Probability of losing * Amount lost)
Expected winnings per bet = (9/19 * $3.00) + (10/19 * -$3.00)
Expected winnings per bet = ($1.42) - ($1.58)
Expected winnings per bet = -$0.16

Therefore, for 24 bets on red, the player's expected winnings can be calculated as:
Expected winnings = 24 * Expected winnings per bet
Expected winnings = 24 * -$0.16
Expected winnings = -$3.84

Therefore, the player's expected winnings after 24 bets on red is -$3.84. This means that on average, the player is expected to lose $3.84 after 24 bets on red.