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?

Question

Gabriel Soto

Mathematics

21 November, 17:44

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?

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Barr · Answer 1

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