In the game of roulette, a player can place a $44 bet on the number 22 and have a StartFraction 1 Over 38 EndFraction 1 38 probability of winning. If the metal ball lands on 22 , the player gets to keep the $44 paid to play the game and the player is awarded an additional $140140. Otherwise, the player is awarded nothing and the casino takes the player's $44. 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?

Question

Mathematics

Gabrielle Davies

5 August, 13:48

In the game of roulette, a player can place a $44 bet on the number 22 and have a StartFraction 1 Over 38 EndFraction 1 38 probability of winning. If the metal ball lands on 22 , the player gets to keep the $44 paid to play the game and the player is awarded an additional $140140. Otherwise, the player is awarded nothing and the casino takes the player's $44. 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?

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Madisyn Hill · Answer 1

The expected value of the game to the player is $3,645.0256

If you played the game 1000 times, you would expect win $3,645,025.6

Step-by-step explanation:

The expected value of a discrete variable is calculated as:

E (x) = (x1) * P (x1) + (x2) * P (x2) + ... + (xn) P (xn)

Where x1, x2, ..., xn are the possible values of the variable and P (x1), P (x2), P (xn) are their respectives probabilities.

So, for the game a player can win $140140 with a probability of 1/38 or can lose $44 with a probability of 37/38. Then, the expected value is:

E (x) = $140140 (1/38) + (-$44) (37/38) = $3,645.0526

Therefore, if you play the game 1000 times you can expect to win:

1000*E (x) = 1000*$3,645.0526 = $3,645,052.6