Question 220739: A game is played using one die. If the dir is rolled and shows 1, the player wins $1. If the die is rolled and shows 2, the player wins $2, if the die is rolled and shows 3 the player wins $3. If the die shows 4, 5, 6 the player wins nothing. There is a charge of $1.25 to play this game. What is the game's expected value?
thank you!
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! A game is played using one die.
If the die is rolled and shows 1, the player wins $1.
If the die is rolled and shows 2, the player wins $2,
if the die is rolled and shows 3 the player wins $3.
If the die shows 4, 5, 6 the player wins nothing.
There is a charge of $1.25 to play this game.
What is the game's expected value?
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Keep in mind that the player pays 1.25 before he gets any winnings.
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Let "x" be the random variable tracking player's winnings>
Values of "x" are: -0.25,+0.75,1.75,-1.25
The matching probabilities are: 1/6,1/6,1/6,3/6
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E(winnings) = (1/6)(-0.25)+(1/6)(0.75)+(1/6)(1.75)+(3/6)(-1.25)
E(winnings) = [-0.25 + 0.75 + 1.75 - 3(1.25)]/6
E(X) = [-1.50]/6 = -25 cents
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Cheers,
Stan H.
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