MATH SOLVE

8 months ago

Q:
# A game is played using one die. If the die is rolled and shows 5 5, the player wins $ 10 $10. If the die shows any number other than 5 5, the player wins nothing. If there is a charge of $ 2 $2 to play the game, what is the game's expected value?

Accepted Solution

A:

The expected value is -$0.33.

The probability of winning is 1/6. The probability of losing is 5/6.

If you win, your prize is 10-2 = 8, since you paid $2 to pay the game. If you lose, you lose the $2 you spent.

The expected value would be the probability of winning, 1/6, multiplied by the winnings, 8, added to the probability of losing, 5/6, multiplied by the loss, -2:

1/6*8 + 5/6(-2) = 8/6 - 10/6 = -2/6 = -0.33

The probability of winning is 1/6. The probability of losing is 5/6.

If you win, your prize is 10-2 = 8, since you paid $2 to pay the game. If you lose, you lose the $2 you spent.

The expected value would be the probability of winning, 1/6, multiplied by the winnings, 8, added to the probability of losing, 5/6, multiplied by the loss, -2:

1/6*8 + 5/6(-2) = 8/6 - 10/6 = -2/6 = -0.33