Question

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?

Answer 1

Answer:

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