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?
Accepted Solution
A:
Answer:The expected value of the game to the player is $3,645.0256If you played the game 1000 times, you would expect win $3,645,025.6Step-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) = $3,645.0526Therefore, if you play the game 1000 times you can expect to win:1000*E(x) = 1000*$3,645.0526 = $3,645,052.6