question that i should know the answer to....but dont.

How do you figure out how many combinations of certain hands there are? I have usually just been guessing with estimating pocket pairs are half as likely as 2 unpaired cards.....

pokerstove is your freind. There are 6 ways to have each PP, 12 ways to have unpaired unsuited cards, and 4 ways to have unpaired suited cards.

if you mean the number of all possible hands its 169.

nah, bode nailed it. thanks.

Originally Posted by will641

if you mean the number of all possible hands its 169.

Nope, there are 1326. And it makes a difference.

Its really simple. Kind of like counting outs. Say you have KQ and you want to know how likely the other player is to have AA,KK,AK for some reason.

First we do AA. For someone to be dealt AA they have to be dealt one ace, and then another ace. There are 4 ways to be dealt one ace. There are then three ways to be dealt a remaining ace. So the number of ways of being dealt AA is 4x3 = 12. But AhAd is the same hand as AdAh so we divide by 2 because we've counted each hand twice. In other words there are 6 combinations of AA.

KK works the same way except we have a K already so there are 3 ways of being dealt the first K, and then 2 ways of being dealt the second K. Total combinations of KK are 3x2 /2 = 3.

Same again for AK. There are 4 aces and then 3 Kings. But this time we dont have to divide by 2 because we arent repeating any counts so there are 4x3 = 12 combos.

so in general a PP has 4x3/2 = 6 combos and an unpaired hand has 4x4 = 16 combos.

Originally Posted by Pelion

so in general a PP has 4x3 = 12 combos and an unpaired hand has 4x4 = 16 combos.

OOP's you want to try again?????

good old perms and coms huh...

there are 6 possible combinations of each pocket pair. in terms of pocket pairs, offsuit hands, and suited hands, there are 169 total hands (ie poker tracker style). if you want to know the number of combinations including both rank and suit, i first choose one card from 52 and then choose one card from the 51 remaining and divide by two.

52C1 * 51C1 = 52 * 51 = 2652 and then you divide by 2 giving 1326.

the reason you divide by two is because 8c7h == 7h8c. ie the ordering of the cards can be chosen differently but it does not matter.

*edited to remove stupidity*

Let me see if I can get this right.

4 4
4 4
4 4
4 4
4 4
4 4

So there are only 6 possible combos. If there were 5 suits:

4X 4
4X 4
4X 4
4X 4
4 4
4 4
4 4
4 4
4 4
4 4

There would be 10 combos.

i.e. if there are four suits, number of possible pairs is 3+2+1, and so on.

Originally Posted by Pelion

so in general a PP has 4x3 = 12 combos and an unpaired hand has 4x4 = 16 combos.

Originally Posted by Trainer_jyms

OOP's you want to try again?????

Originally Posted by Pelion

jyms i cant work out what you mean.

Originally Posted by biondino

Let me see if I can get this right.

4 4
4 4
4 4
4 4
4 4
4 4

So there are only 6 possible combos.

yea oops. Jyms is right you have to divide the PPs by 2 to account for 4d4h being the same hand as 4h4d. You dont have to divide by two for the unpaired hands though.

Originally Posted by biondino

i.e. if there are four suits, number of possible pairs is 3+2+1, and so on.

Yup. To generalize it's:
c!
n!*(c-n)!

where c is the number of cards and n is the number of cards chosen

For your example the number of suits is equivalent the c and n = 2 so it simplifies to:
c*(c-1)
2

which is equivalent to your answer:
(c-1) + (c-2) + (c-3) + ... + 2 + 1

Also while using it to set up ranges, it's quite important which cards are on the board and which you are holding.

Let's say the board is Kh4c5s and we are holding As3s.
Now there are only 9 combos left of AK for villain instead of 16 because we hold an A and there is a K on the board, so possible holdings of AK are 3x3=9.