A casino decides to charge $3 to play the following game: (click for details) what is the probability:?
Posted by admin in Casino on 09 28th, 2010 | 2 responses
3 cards are drawn from a deck of 52. If no hearts are drawn the payer gets nothing; if 1 heart is drawn the player wins $1/50; if 2 hearts are drawn the player wins $7.
How high should the owner make a jackpot on a pull of 3 hearts if he wants the “house” (casino) to average 60 cents a play.
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The house will make 60 cents on average if the player wins 2.40 on average.
Let X denote the number of hearts, then:
P(1)=C(13,1)C(39,2)/C(52,3)=0.436
P(2)=C(13,2)C(39,1)/C(52,3)=0.138
P(3)=C(13,3)C(39,0)/C(52,3)=0.0129
Thus, average player winnings equals E(X) and
E(X)=1.5(0.436)+7(0.138)+k(0.0129)
where k is the jackpot winnings.
We set E(X)=2.4, and solve for k to get k=60.456
So, the owner should set the jackpot equal to $60.46
p(0 ♥) = 39C3 / 52C3 = 9139/22,100 = 0.41353
p(1 ♥) = 13 • 39C2 / 52C3 = 9633/22,100 = 0.43588
p(2 ♥) = 13C2 • 39 / 52C3 = 3042/22,100 = 0.13765
p(3 ♥) = 13C3 / 52C3 = 286/22,100 = 0.01294
expected value(to player) = sum of (payoffs x probabilities) – charge
expected value(to house) = charge – sum of pxp
0.60 = 3 – 0 ( 0.41353) – 1.50 ( 0.43588) – 7 ( 0.137665) – Jackpot ( 0.01294)
0.60 = 3 – 0 – 0.65382 – 0.96355 – 0.01294J
0.60 = 1.38263 – 0.01294J
0.01294 J = 0.78263
J = $60.48
doubtless round it off to $60. why give the chumps the extra $0.48?