number of starting hands

I often see books say that there are only 169 starting hands for texas hold em. Can somebody explain the math of that to me? I know it's 13x13, but that doesn't seem quite right to me, because that doesn't take into account the possibility of suiting your hand.

For example...think of 2 through A of two suits, say hearts and spades(inconsequential which suits we choose, because the hand can either be suited(A [img]/images/graemlins/heart.gif[/img], K [img]/images/graemlins/heart.gif[/img]) or unsuited A [img]/images/graemlins/heart.gif[/img], K [img]/images/graemlins/spade.gif[/img], [img]/images/graemlins/diamond.gif[/img], [img]/images/graemlins/club.gif[/img] s are all the same). The A [img]/images/graemlins/heart.gif[/img] could be paired with 2 [img]/images/graemlins/spade.gif[/img] through A [img]/images/graemlins/spade.gif[/img], and also the 12 other [img]/images/graemlins/heart.gif[/img] cards, for a possible 25 hands. Now go to the K [img]/images/graemlins/heart.gif[/img], it could pair with any of the spades and any of the 11 other heart cards(it already paired with the ace) for 24 possible hands....and so on with the Q [img]/images/graemlins/heart.gif[/img], J [img]/images/graemlins/heart.gif[/img], etc. So the equation becomes 25(A)+24(K)+23(Q)+22(J)+.....14(3)+13(2)= 247 The number in the parentheses is the heart card, preceded by the possible combination of cards remaining for that card.

So is it 169 starting hands or 247?

[uuDevil]

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So is it 169 starting hands or 247?

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It's 169 if you don't distinguish hands like AcKs from AdKc or AsKs from AhKh or 2h2c from 2d2c.

One way to do the math is to realize there are C(13,2)=78 possible nonpairs (if you don't care about suits) and 13 possible pairs. But if you want to distinuguish between suited and unsuited nonpairs, then there are 78x2=156 suited or unsuited, unpaired hands. So there are 156+13=169 possible hands.

Using your method, if you ran through all the possible card combinations, you would get a total of 1326, which is the same as C(52,2), the total number of ways to choose 2 cards from a full deck of 52 cards (in this case AhKh is distinct from AdKd, etc.).

UUDevil...thanks for answering, but I still have some reservations about the math behind this. Im not quite sure I understand your explanation either...can you dumb it down for me?

When I look at this, I'm only looking at the possibility of having any two card combination and allowing for the additional strength of a suited starting two cards. That is why you only need to pick from any two suits. You would end up the same whether you use [img]/images/graemlins/spade.gif[/img] [img]/images/graemlins/diamond.gif[/img], [img]/images/graemlins/spade.gif[/img] [img]/images/graemlins/heart.gif[/img], [img]/images/graemlins/spade.gif[/img] [img]/images/graemlins/club.gif[/img], etc...in that any AK could either be suited or unsuited; what the suit combo is makes no difference.

Here's how I look at arriving at the 169 figure. Picture(or do) this....pull out a complete set any two suits (2 thru A). Lay one suit out in column and the other in a pile, A(top) down to 2. Now pull the A out of the pile and put in next to the top card in the column...you will have an AA(unsuited) pair. Now move it down the column, AK(unsuited), move down, AQ(unsuited), etc...for a total of thirteen possible combinations. Do this for all thirteen cards in you pile, and you will have 169 possible combinations. However, we are missing the fact that the A in the pile could also be dealt out with the K(Q,J,t,etc) in the pile, giving you a suited starting hand, which definitely increases the value of the starting hand.

Counting the possibility of suited hands is how I arrived at the 247 figure(it doesn't matter preflop what the suit is since the odds of three to the flush on the board are the same regardless of suit). And, preflop, A [img]/images/graemlins/spade.gif[/img]K [img]/images/graemlins/heart.gif[/img] rates the same strength as A [img]/images/graemlins/diamond.gif[/img]K [img]/images/graemlins/heart.gif[/img], A [img]/images/graemlins/diamond.gif[/img]K [img]/images/graemlins/club.gif[/img]...etc.

Of course, maybe I'm just over-analyzing a simple point...

[Cooker]

You are double counting some offsuit hands in your method. Looking at aces you have 13 offsuit cards (including the K) and 12 suited cards. Now looking at Kings you now incorrectly write 13 offsuit cards plus 11 suited cards = 24. This has included AKo 2 times. Done correctly, there are actually 12 offsuit cards and 11 suited cards for K to prevent the double counting. You should have decreased by 2 for each card instead of 1, so your sum should read

25(A) + 23(K) + 21(Q) + ... + 3(3) + 1(2) = 169

Still, this is a very clumsy way to do this and the previous poster did it in a much nicer way. However, this clearly displays your error.

[Henke]

You are double counting some hands. By your way of calculating it, it should be 2*sum(k,k=1..13)-13 [img]/images/graemlins/smile.gif[/img]

Look at it this way, with the piles and stuff. First time you count: AA, AK, AQ, AJ, AT ... A2 that's thirteen. But then you go on counting: KA, KK, KJ, KT ... K2 for thirteen more combinations. But now you have counted AK twice, which is wrong.

So the hands with an A high will be 13, but those with K high will be 12 and so on, untill you reach the duce, which is only one hand with no card higher than two... That's why you get 13+12+11+10+...+1=91 for all pairs and say offsuit cards. Then just add the suited once (substract the pairs and multiply by two).

A hint though, uuDevils method using combinations is much easier. If you want to analyze poker hands mathematically, look up the subject of combinations (look here for a short introduction).

[Henke]

You may type faster, but I have more posts!
[img]/images/graemlins/grin.gif[/img]

[uuDevil]

The other posters have shown where you are double-counting. If you don't trust the combinations calculation to find the number of unpaired hands, think of it this way [from a post by AaronBrown]:

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If you don't have a pair, one card can be any of 13 and the other can be any of 12. But you have to divide this by 2, because K-9 is the same as 9-K. So 13x12/2 = 78.

You have to multiply the 78 by 2 because the hand can either be suited or unsuited.

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[Orpheus]

The math has been handled quite ably, so let me suggest a visual proof with useful applications:

Draw and label a 13x13 grid. the squares above (and including) the "pocket pair diagonal" include every possible unsuited TXHE hand. The squares below the PP diagonal represent every possible suited hand -- i.e. it replicates (in reverse order) every combination above the diagonal. Suited pairs are, of course, impossible.

You can use this fact to quickly construct compact specialty charts. I keep xeroxed sheets of blank labeled grids in my drawer. They've come in handy. That's all I'll say.

[Cooker]

I may type faster, but that is more than compensated by the number of stupid sounding typos I make. I have been somewhat withdrawn from the forums, since I am trying to spend more time playing, reviewing my own play, and rereading some books. When I am a little happier with my level of play at the games I frequent, I may return to more frequent posting. I found that in the past, posting has helped my game, because explaining a line forces you to understand the concepts behind every play.

Ahhh....I now see the error of my ways. Thanks to all for explaining it in a manner I could understand.

Good luck