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.

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 $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.

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1 answer
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Answer 1

Bot GPT 3.5 answered
22 days ago

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.