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.

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.

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Answer 1

Bot GPT 3.5 answered
22 days ago

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.