Hand combinations

[pyroponic]

I read this somewhere I think in Sklanky's "Poker, Gambling, and Life." How exactly do you calculate the total combinations your opponent is holding?

For example you hold KK, and you want to know the total combinations your opponent could have of AA, AKo, and AKs.
Or if you have AQ, the total combations your oppoenent could have AQs, QQ, KK, AA, Ako, AKs.

You get the idea, thanks.

[LetYouDown]

What exactly are you asking? Your opponent can hold C(50,2) possible hands. Are you looking only for hands that beat you? That's obviously dependent on what you hold.

[AaronBrown]

For the first problem, there are eight Aces and Kings in the deck, and you hold two of them. The remaining six can be held in 6*5/2 = 15 ways.

Since there are four A's, 4*3/2 = 6 of these hands are AA. There are two outstanding K's, that's 2*1/2 = 1 way to hold KK. Four A's and two K's gives 4*2 = 8 AK hands (you don't divide by two because the cards are different). 2 of the 8 are suited, because there are only two available suits and each one specifies an exact hand. So 6 are unsuited AK.

In the second example, there are 12 cards, and again you hold 2. So there are 10*9/2 = 45 ways for your opponent to have only A's, K's and Q's in his hand. By the logic above there are 6 KK, and 3*2/2 = 3 AA, and also 3 QQ. There are 4*3 = 12 AK and KQ, and 3*3 = 9 AQ. 3 of the AK and KQ are suited, 9 are not. If your AQ is suited then 3 of the other AQ are also suited, otherwise only 2.

[pyroponic]

I'm sorry, i'm trying to figure out some sort of algorithm and
can't seem to derive one from the information you've given me. Is there a formula for a) total combinations, and b) number of ways to have a specific hands (for example the number of ways to have AKo when you hold KK).

I suppose you could take b and divide that by a to calculate an estimated probability one could hold that specific hand.

Thanks

[treeofwisdom7]

[ QUOTE ]
For the first problem, there are eight Aces and Kings in the deck, and you hold two of them. The remaining six can be held in 6*5/2 = 15 ways.

Since there are four A's, 4*3/2 = 6 of these hands are AA. There are two outstanding K's, that's 2*1/2 = 1 way to hold KK. Four A's and two K's gives 4*2 = 8 AK hands (you don't divide by two because the cards are different). 2 of the 8 are suited, because there are only two available suits and each one specifies an exact hand. So 6 are unsuited AK.

In the second example, there are 12 cards, and again you hold 2. So there are 10*9/2 = 45 ways for your opponent to have only A's, K's and Q's in his hand. By the logic above there are 6 KK, and 3*2/2 = 3 AA, and also 3 QQ. There are 4*3 = 12 AK and KQ, and 3*3 = 9 AQ. 3 of the AK and KQ are suited, 9 are not. If your AQ is suited then 3 of the other AQ are also suited, otherwise only 2.

[/ QUOTE ]

are you a math teacher? np


i remember from ken warren who said that for non pairs there are 16 combos and pairs have 6 combos but i never figured out how to get that i can get the pairs.. 4*3?2 but non pairs? 8*7/2 which doesnt come out

[treeofwisdom7]

oh wait i think i know what it is. 8 cards * 1 of the 4 of the other .. ex Aor K. if hte first card is A then you have 4 remain kings so 8*4/2 = 16 thanks ken warren

[AaronBrown]

Sorry, I was too cryptic.

If you want to know the number of hand combinations your opponent can have from N cards, you take N*(N-1)/2. So if you don't hold any Aces, the number of ways to get a hand composed only of Aces, N = 4, is 4*3/2 = 6. If you hold one Ace, it goes down to 3*2/2 = 3, because now N = 3. If you hold a pair of Aces, N = 2, and there is only 2*1/2 = 1 way for him to hold a pair.

You wanted to know how many hands there were composed only of Aces and Kings, given that you hold a pair of Kings. N = 6, since there are four Aces and two Kings available, so 6*5/2 = 15.

To see how many of them are AA (composed only of Aces) we take N = 4, 4*3/2 = 6. For a pair of Kings, N = 2, so we get 2*1/2 ways to hold KK. 15 - 6 - 1 = 8 has to be the number of ways to hold AK. We can also get it by multiplying the number of Aces available (4), by the number of Kings available (2).

To figure suited hands is a bit more complicated.

(1) If there is four of one rank available, then the number of suited hands equals the number of available cards of the other rank. That's the case here. There are four Aces available and two Kings. Since there are four Aces, the number of suited AK hands is the number of available King suits, which is 2.

(2) If there are fewer than four of both suits, then you have to count the overlaps. For example, if you hold AK there are 3*3 = 9 other possible AK combinations. If your AK is suited, then three other suited AK combinations are available. But if your AK is unsuited, then only two suited combinations remain. Either way, the number of unsuited AK's is 9 minus the number of suited AK's.