I am trying to determine the probability of 4 of a kind in pinochle. Pinochle deck consists of 48 cards aces, kings, queens, jacks, 10s and 9s (2 of each suit). Example: 4 of a kind is Ace-hearts,Ace-diamonds, Ace-clubs & Ace-spades. I Believe the correct number of possible 4 of a kind are 4C8*8C44 (70 * 177232627) The total possible combinations are 12C48 (69668534468). 0.178076 Is this correct or am I not factoring in something correctly
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Determining the probability of 4 of a kind in pinochle
- Thread starter Jflynn
- Start date
From what you calculated, there must be 12 cards in a hand. Hence total possible hands is \(\displaystyle \displaystyle _{48}C_{12}\) (you wrote your combinations backwards!).I am trying to determine the probability of 4 of a kind in pinochle. Pinochle deck consists of 48 cards aces, kings, queens, jacks, 10s and 9s (2 of each suit). Example: 4 of a kind is Ace-hearts,Ace-diamonds, Ace-clubs & Ace-spades. I Believe the correct number of possible 4 of a kind are 4C8*8C44 (70 * 177232627) The total possible combinations are 12C48 (69668534468). 0.178076 Is this correct or am I not factoring in something correctly
Do you want 4 of a specific kind, or any 4 of a kind? There is a difference of a factor of 6.
Do you want exactly 4 of a kind, or at least 4 of a kind? The aspect you didn't account for is that the 44 "remaining" cards are of two classes: 4 are interchangeable with your 4 selected, while 40 are different. For "at least" 4, I think you have to add up 4,5,6,7,8 of a kind.
I will consider at least 4 of a specific kind. One (of many!) way is for the first 4 cards dealt to be the selected, and none of the next 8:
P(SSSSxxxxxxxx) = (8/48)(7/47)(6/46)(5/45) * (40/44)(39/43). . .(33/37)
...........................= (8!/4!) * (40!/32!) / (48!/36!) = \(\displaystyle \displaystyle _{36}C_{4}\ /\ _{48}C_{8}\)
This single permutation must be multiplied by 12! representing rearrangement of the 12 cards, but then divided by 4! and 8! for the 4 S and 8 x cards. That is, multiply by \(\displaystyle _{12}C_4\):
P(exactly 4) = \(\displaystyle \displaystyle \dfrac{_{36}C_{4}\ _{12}C_{4}}{_{48}C_{8}} = 0.0773\)
P(exactly 5) = \(\displaystyle \displaystyle \dfrac{_{36}C_{3}\ _{12}C_{5}}{_{48}C_{8}} = 0.0150\)
P(exactly 6) = 0.0015,....P(exactly 7) = 0.00008,....P(exactly 8) = 10^-6
P(at least 4 of a specific kind) ~ 0.094