A casino owner decides to charge $3 to play the following game. 3 cards are drawn from a deck of 52. If no clubs are drawn the player gets nothing; if 1 club is drawn the player wins $1.5 ; if two clubs are drawn the player wins $7. How high should the owner make the jackpot on a pull of 3 clubs if he wants the house to average winning 50 cents a play?

Question

anonymous · Answer

The question can be subdivided. Easiest is the probability of getting three clubs, which is (13/52)(12/51)(11/50) = 0.01294, very small,
so he  can make this have a substantial pay out P and lose on this bet an average of (0.01294) P per play. 

Drawing no clubs is similarly easy to analyze, (39/52)(38/51)(37/50) = 0.4135. 

Getting the other two probabilities needs binomial distribution analysis.

Good night.

dumbcow · Answer

continuing from above post
-probability for 1 club = 13/52 *39/51 *38/50 = 0.14529

-2 clubs = 13/52 *12/51*39/50 = 0.04588

To determine jackpot, you need Expected Value which is sum of probability*payout
Since avg win for casino is $0.50, the expected winnings for the player is $2.50

$$.01294X + .04588(7)+.14529(1.5) +.4135(0) = 2.5$$
solve for x

dumbcow · Answer

oops there are 3 ways to get 1 club and 2 clubs ... multiply those probabilities by 3
sorry