In the game of roulette, a player can place a $5 bet on the number 3 and have a startfraction 1 over 38 endfraction probability of winning. if the metal ball lands on 3 , the player gets to keep the $5 paid to play the game and the player is awarded an additional $175. otherwise, the player is awarded nothing and the casino takes the player's $5. what is the expected value of the game to the player? if you played the game 1000 times, how much would you expect to lose? note that the expected value is the amount, on average, one would expect to gain or lose each game. the expected value is $ nothing. (round to the nearest cent as needed.) the player would expect to lose about $ nothing. (round to the nearest cent as needed.)

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Mathematics

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28 May, 20:48

In the game of roulette, a player can place a $5 bet on the number 3 and have a startfraction 1 over 38 endfraction probability of winning. if the metal ball lands on 3 , the player gets to keep the $5 paid to play the game and the player is awarded an additional $175. otherwise, the player is awarded nothing and the casino takes the player's $5. what is the expected value of the game to the player? if you played the game 1000 times, how much would you expect to lose? note that the expected value is the amount, on average, one would expect to gain or lose each game. the expected value is $ nothing. (round to the nearest cent as needed.) the player would expect to lose about $ nothing. (round to the nearest cent as needed.)

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Jude Hester · Answer 1

The expected value per game is - 0.26. Over 1000 games, you can expect to lose $263.16.

To find the expected value, we multiply the probability of winning by the amount of winnings, the probability of losing by the amount of loss, and adding those together.

We have a 1/38 chance of winning; 1/38 (175) = $4.61. We also have a 37/38 chance of losing; 37/38 (5) = $4.87.

$4.61-$4.87 = - $0.26 (rounded)

To five decimal places, our answer is - 0.26136; multiplied by 1000 games, this is $261.36 lost.