Question 300280: ``For a fair game, what should be the charge to play if there is a .002 probability to win $1,000, .008 probability to win $100, and .99 probability of not winning anything?
I am very pressed for time I have to hand in my assignment today at 2pm can someone please help me out? i really appreciate it.''.
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! ``For a fair game, what should be the charge to play if there is a .002 probability to win $1,000, .008 probability to win $100, and .99 probability of not winning anything?
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Let "x" be the price to play.
Random # values: 1000-x, 100-x, 0-x
Probabilities::: 0.002, 0.008, 0.99
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Equation:
Note: "Even" means E(x) = 0
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E(x) = 0.002(1000-x) + 0.008(100-x) + 0.990(-x)
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E(x) = [2000-2x + 800-8x - 990x]/1000 = 0
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[2800 - 1000x] = 0
x = $2.80
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You should pay $2.80 to play the game.
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Cheers,
Stan H.
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