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?

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

10 down (bang1)

5 up

5down (2)

2.5 up

2.5 down (3)

1.25 up

1.25 down (4)

now add them up : )

hoped i lol