# of possible flop combinations

[MikeRand2000]

A quick question regarding permutation usage:

1) I'm looking to calculate the number of different flops in hold'em. I have started my analysis as C(52,2) x C(50,3) to represent the # of hole cards and the # of resulting flops, given each hole card pattern. The above formula gives 25,989,600 possible flops (i.e. less than 52!/47! if I cared about the exact order, more than C(52,5) if I didn't even care about hole vs. flop).

2) That being said, I realize that there are between 4 and 24 permutations related to suits. For example, the 25,989,600 number includes 40 ways to make a royal flush. Don't I really only care about the 10 ways to make a royal flush, not the 4 permutations? That being said, I can see the rest of the 25,989,600 really being a smaller subset of outcomes (defined mainly by rank) and specified permutations of suit.

What is the correct way to calculate the # of possible flops given suit permutations? I'd expect a number between 2.5MM and 6MM (given the minimum permutation impact of dividing by 4 and the lower limit of C(52,5).

Any direction would be greatly appreciated.

[gaming_mouse]

What is the correct way to calculate the # of possible flops given suit permutations?

I can't figure out what you mean by this. Your royal flush example makes sense, but I don't know how you'd want to break down what counts as different hands in other cases.

For example, how many possible hole-flop combinations of 4 of a kind would you say there were?

Would it be 13*12*10? But that would count 2/images/graemlins/heart.gif2/images/graemlins/club.gif--2/images/graemlins/diamond.gif2/images/graemlins/spade.gifA/images/graemlins/club.gif as different from 2/images/graemlins/diamond.gif2/images/graemlins/spade.gif--2/images/graemlins/heart.gif2/images/graemlins/club.gifA/images/graemlins/club.gif. Is that what you want?

Similar difficulties become even worse when considering 1-pair on no pair hands.

[MikeRand2000]

Your example points out exactly the problem I'm mentioning. I would want to count all of those similar examples as 1 type of flop.

That's why I think the answer is somewhere south of 6MM: at a minimum every type of flop (26MM) has 4 permutations (shift every suit once).

I can't believe the answer isn't plastered all over this forum.

[gaming_mouse]

[ QUOTE ]
Your example points out exactly the problem I'm mentioning. I would want to count all of those similar examples as 1 type of flop.

[/ QUOTE ]

Which similar examples as one type of flop? All possible 4 of a kinds? Or just all possible 4-kinds made up of 2222A where you hold two of the 2's? I still don't know what your definition for distinct types of flops is? I sort of get the feeling that you might not either, as the question should be much easier to answer once you decide on a rigorous definition.

[MikeRand2000]

Rigorous definition = any two (or more) flops between which I would have no preference one way or another because they are exactly identical in terms of ranks and suits.

Example of 12 permutations of one flop (all of which I want to count as 1 flop):
AcJc 9c4c2h
AcJc 9c4c2s
AcJc 9c4c2d

AhJh 9h4h2c
AhJh 9h4h2s
AhJh 9h4h2d

AsJs 9s4s2d
AsJs 9s4s2c
AsJs 9s4s2h

AdJd 9d4d2c
AdJd 9d4d2s
AdJd 9d4d2h

All else being equal (i.e. nobody at the table has a ridiculous hearts fetish or anything like that), I don't care one bit about the difference between these flops. The point at the end of all of this is to come up with the # of different flop types I need to think about, then collapse those further into flop categories (e.g. top pair, bad kicker, etc.).

I'd call that more relevant than all of these ridiculous analysis pieces where people try to figure out the probability of flopping royal flushes.

[gaming_mouse]

Okay, Mike. I see what you want now. I'll give it some thought and get back to you. While this is an interesting problem, I actually don't see much practical value to it. I mean, how will knowing if there are 1 million of 10 million of these help you at the table?

[gaming_mouse]

[ QUOTE ]

AcJc 9c4c2h
AcJc 9c4c2s
AcJc 9c4c2d

AhJh 9h4h2c
AhJh 9h4h2s
AhJh 9h4h2d

AsJs 9s4s2d
AsJs 9s4s2c
AsJs 9s4s2h

AdJd 9d4d2c
AdJd 9d4d2s
AdJd 9d4d2h

[/ QUOTE ]

Okay. So I see no way to do this except by brute force. You actually have to go through the different categories 1 by 1 and count them:

To begin (work in parentheses is the sum of the answers for the subsections):

* No Pair Hands
** 4-1 suit distribution (1287*5)
*** suited hole cards (1287*3)
**** lowest board card not of your suit.

(13 choose 5) = 1287

**** middle board card not of your suit, same.
**** highest board card not of your suit, same.
*** unsuited hole cards (1287*2)
**** your lower rank is offsuit, 1287.
**** your higher rank card is offsuit, 1287.
** 3-2 suit distribution
etc.....


Doing a complete list would be exhausting, but that is one way to tackle the problem (and the only one I see right now, though there may be an easier way).