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 $1.50 on red to play the game. If the ball lands on red, the player gets a $1.50 for winning and receives the money back. If the ball does not land on red, then the player simply loses the $1.50 placed on the bet. If the player places the same bet on red 20 times, what is the player's expected winnings? Round your answer to the nearest cent.

Answer 1

The probability of the ball landing on red in American roulette is 18/38, since there are 18 red spaces out of 38 total spaces on the wheel.

Therefore, the expected winnings for each bet can be calculated as: (probability of winning * amount won) + (probability of losing * amount lost)
Expected winnings = (18/38 * $1.50) + (20/38 * -$1.50)
Expected winnings = ($0.71) + (-$0.79)
Expected winnings = -$0.08

Since the player is placing the same bet 20 times, the player's total expected winnings can be calculated as:
Total expected winnings = 20 * -$0.08
Total expected winnings = -$1.60

Therefore, the player can expect to lose $1.60 after placing the same bet on red 20 times.