SOLUTION: When you place a $5. bet at the casino, there is a 244 in 495 probability you will win $5. and a 251 in 495 probability you will lose $5. What is your expected value? I don't un

Algebra ->  Probability-and-statistics -> SOLUTION: When you place a $5. bet at the casino, there is a 244 in 495 probability you will win $5. and a 251 in 495 probability you will lose $5. What is your expected value? I don't un      Log On


   



Question 231993: When you place a $5. bet at the casino, there is a 244 in 495 probability you will win $5. and a 251 in 495 probability you will lose $5. What is your expected value? I don't understand this.
Found 2 solutions by vleith, stanbon:
Answer by vleith(2983) About Me  (Show Source):
You can put this solution on YOUR website!
Expected value is given as
expectedValue+=+probabilityOfWinning+%2A+payoff+-+probabilityOfLosing+%2A+betAmount
So plugging the numbers given
eV+=+%28244%2F495%29%2A5+-+%28251%2F495%29%2A5
ev+=+-%287%2F495%29%2A+5
ev+=+-0.071
So you can "expect" to lose just over 7 cents on each bet.
The moral of the story ... the house always wins.

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
When you place a $5. bet at the casino, there is a 244 in 495 probability you will win $5. and a 251 in 495 probability you will lose $5. What is your expected value? I don't understand this.
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Let the random variable "x" represent the players gain.
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Random number values are 5 and -5
Corresponding probabilities are (244/495) and 251/495
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E(x) = (244/485)*5 + (251/495)(-5)
E(x) = 2.46 - 2.54
E(x) = -7.5 cents
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You can expect to lose 75 cents each time you place a bet.
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Cheers,
Stan H.