Theoretically there is a limited number of possible hands that can exist once the board has been laid in hold 'em. Therefore, theoretically there would exist a mathematical equation to determine that number. In attempting to write out this equation for my own personal whim fullfilment, I ran into a bit of a snag.

How do you count the board cards in hold 'em?

I realize that sounds stupid. Let me explain.

There exists the possibility in hold 'em that a varied number of players from 0-N could use any card laid on the board therefore increasing the possible number of hands.

So would you multiply the board cards by N the number of players or should it be more complicated like a factor based on the probability of a card being a part of a highly favorable hand (whole other can of worms)? How would you deal with the fact that an unknown number of players from 0-N are using each card?

It sounds really stupid, but this problem has been bothering me for a very long time.

Thanks.

I know the answer to this

Lets let the math guys at it

Are you asking how many distinct boards there are in holdem? This is pretty easy to calculate.

Or are you asking what the potential number of 5 card hands people could have, given all of those boards? This is also calculatable... This really just simplifies down to how many distinct 5 card hands there are period.

Well I was thinking myself how you said how to handle 0-N players.
I usually treat N as the number 10. And I replace 0 with 1. 0 players are no players. 10 players is a regular game.

Now I memorized this formula.

2 handed is more of a feel game of moving and giving and using simple patterns.

10 handed is a game of patience where you need to be confident you can make a move against so many odds. 6 handed is kind of a subtle mix but the combinations of them can be pretty very different from the first two.

I guess you can calculate by looking at the board and realizing that with that board how many people can change it. I guess all of them. So I guess that makes sense.

How many five card possible hands exist with any given board. Although, now that I am thinking about it again. I realize that the question isn't dependednt on the number of players and that was where I was getting stuck.

The answer is either calculated based solely on the number of possible hands with 5+2N cards in play or equal to N.
Thanks.

Well I guess you would take 5 cards for the board. and there are 2 cards each person. the five cards would be altered by each two once. and there are n people. so is the number whatever N is?!?!?!

Ah I see what you're getting at. Yes, the number won't be dependent on the number of players at all. I suspect you might have to break boards down by various characteristics and I think there would be lots of them. Doing it for a *given* board isn't too bad but I don't know how easily you'd find a general solution.

It might be interesting to try a brute-force approach to this, with computing. Enumerate each holdem board, and each possible hand with that board, and see if patterns emerge. I admit to using computers a crutch in this regard but that's probably where I'd start. I'd do a few hundred boards and see if anything pops up.

it's N?! it's N?!? right?!?!

This answer is dependant if I am referring to the question I assume you are asking. I beleive I am simply giving the process to solve that problem with this.

P(even)=outscomes for event/total # of possible outcomes

Finding out whar goes in the numerator and denominator of this probability requires finding the number of ways to rearrange the outcomes. This leads you to:

A permutation , which is an order of elements in a set, chosen without replacement..

here are 3 ways to choose something out of a set. Once you pick the first one, there are only 2 left. And once the 2nd element is chosen there are only one left. It is a finite s(sample space). Already determined by 3*2*1=6 permutations of that set.

The number of possible ways to rearrange k objects is k*(k-1)*(k-2)*..*(2)*(1)

The n! the person said above reprents a factorial. The exclamation point commands multiplying by k, and then k-1, k-2, and on down 2 times 1 for any k greater than or equal to one

Formula for counting the number of permutations of k items chosen from n items(w/o replacement) is P^nk=n!/(n-k)! ( the n is to the power of the probability and k is to the base of the probability, i dont know how to make the k go down or get smaller on the comp). That is the permutation formula.

You can take on the choosing and rearrange the process using these two simple steps given below:

1)find the total number of ways to rearrange all objects.
- For ex, we have 6 objects. 6!=720
2) divide out the number of ways to rearrange the objects you cant select.
for this examples sake, leave out 2 people (2!=2). So the numebr of ways to select 4 objects out of 6 you can take 720/2=360. 6 choose 4.

Now when selecting an item, ask, does the order matter? if it doesnt you use a combinations. Order matters inpermutation,

. Say you have ten people in a class. You need to choose 3 people to win this gift, and all the gifts are the same. So should order matter? No. How many ways can you choose? 10!/(10-3)!=10*9*8= 720. the 7! in the denonimator canceled out 10*9*8*7/7*9*8 all the way down to 1. Therefore 10*9*8 will prodive you the answer.


If not referring to your question, my apologies.

I think OP is asking, how many possible hands can there be with n players and 5 board cards, where each player can use 0 to 5 of the board cards.

Take one player. He has C(7,5) possible hands = 21. That applies to every player. So, at this level, the number of possible hands for n players is 21^n. This means, that after the river card is dealt, there are 21^n five-card poker hands that are possible among the n players. This, of course, assumes a specific dealing of hands and a specific board.

If you then want to count how many ways of distributing 2 cards to n players and 5 cards to the board, that is another level. The number of ways to do that is also calculable. Assume just two players. The number of possible 2 card hands to the two players is C(52,2) C(50,2)/2!. That leaves 48 cards remaining, 5 of which make up the board. This means that there are C(48.5) board possibilities. So, the number of total hands and boards is C(52,2) C(50,2) C(48,5)/2 and for each of these possibilities, there are 21 x 21 hands that can be made between the two players. The result is about 6 x 10^14, a pretty big number. Note that this number includes equivalent hands in the sense that Ah Td has the same value as Ac Ts if neither make a flush with the board.

Extending this to n players is fairly straightforward.

you guys can't explain it logically and your math goes nowhere. Stop posting.

bye

NH wp sir.