Question

a game is played using one die. if the die is rolled and shows a 2, the player wins $45. If the die shows any number other than 2, the player wins nothing.

If there is a charge of $9 to play the game what is the games expected value?

Answer 1

Answer:

The game's expected value is of -$1.5.

Step-by-step explanation:

Expected value:

Probability of each outcome multiplied by the outcome.

One out of 6 sides is 2:

1/6 probability of the player earning 45 - 9 = $36.

5/6 probability of the player losing $9. So

$E = 36\frac{1}{6} - 9\frac{5}{6} = \frac{36 - 45}{6} = -\frac{9}{6} = -1.5$

The game's expected value is of -$1.5.