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If I have a deck of 72 cards with ranks of 1-12 and 6 suits of each rank, how do I calculate the combinations of possible "poker style" hands if players are dealt 6 card hands?

Royal Flush: (ranks 8,9,10,11,12,1) out of six suits?

Straight flush: (1,2,3,4,5,6...etc.)

6 of a kind

5 of a kind

"Modified" full house with 3 of one suit and 3 of another

4 of a kind

3 of a kind and a pair

3 pair

3 of a kind

2 pair

1 pair

1 of each of the 6 suits

Not sure if I'm missing any other possible combinations...

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The total number of hands with $6$ cards is $72 \choose 6$. If there are $N$ ways to get a certain type of hand, then the probability of getting that type of hand is $N/ {72 \choose 6}$.

One example, three of a kind and a pair: The three of a kind can be in any of the $12$ ranks, and the pair can be in any $11$ other ranks. There are $6 \choose 3$ possibilities for the suits of the three of a kind, and $6 \choose 2$ possibilities for the suits of the pair. The last card can be of any of the $10$ remaining ranks, and any of the $6$ suits. So the probability of getting a three of a kind + a pair is $$\frac{12 \cdot 11 \cdot {6 \choose 3} \cdot {6 \choose 2} \cdot 10 \cdot 6}{{72 \choose 6}} \approx 0.0152$$

By the way, you might be missing three hands: the flush, the straight and the four of a kind + another pair.

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