Suits in 7-card hands

Quote: Wizard

I've been thinking about this off and on. I offer the following solution for those like me who are strong in math and weak in programming.

Before I go further, let me remind you the formula for choosing k out of n items, with replacement, is combin(n+k-1,k). Think of the deck having k-1 "repeat" cards.

First, let's ask the question how many classes of 5-card hands are there?

1. There are combin(13,5)=1,287 ways all five cards can be in the same suit.
2. There are combin(13,4)*13=9,295 ways four cards can be in the same suit and one singleton in another suit.
3. There are combin(13,3)*combin(13,2)=22,308 ways 3 cards can be in the same suit and 2 in another suit.
4. There are combin(13,3)*combin(13+2-1,2)=26,026 ways 3 cards can be in the same suit and 1 each in two other suits.
5. There are (COMBIN(13,2)*(COMBIN(13,2)+1)/2)*13=40,053 ways 2 cards can be in two separate suits and 1 in a third suit.
6. There are COMBIN(13,2)*COMBIN(13+3-1,3)=35,490 ways 2 cards can be in one suit and 3 singletons in the other 3 suits.

Add that up and you get 133,459.

If you're going to cycle through the whole deck you have to weight all 133,459 by the appropriate number of suits in each class. Here is a table to help you with that.

Suit distribution	Comments	Ranks	Suits	Product
5,0	All	1287	4	5148
4,1	All	9295	12	111540
3,2	All	22308	12	267696
3,1,1	Matching singletons	3718	12	44616
3,1,1	Non-matching singletons	22308	24	535392
2,2,1	Both matching ranks	1014	12	12168
2,2,1	One matching rank	11154	24	267696
2,2,1	No matching ranks	27885	24	669240
2,1,1,1	Matching ranks	1014	4	4056
2,1,1,1	One matching rank	12168	12	146016
2,1,1,1	No matching ranks	22308	24	535392
Total		134459		2598960

I admit this is not elegant like the solutions posted earlier. It will require more code and just one math error will screw everything up. However, it isn't a memory hog. It should use quite little.

I can confirm the 7-card case has 6,009,159 classes of hands. Upon request, I can supply a similar math table for that.

I'm still working on the bit mapping solution, just for my own programming edification.
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