I understand the general idea, but I'm struggling with individual card representation. Example: The included opening range has 240 total combos. However, I'm attempting to learn how many of a specific card is in each range. So, in this range, how many Kings are there? I count a total of 54 combos that contain a King. Is that the correct way to think about it?

However, the problem I'm having trouble wrapping my brain around is if I do that same thing to the rest of the cards, there are a lot more combos than 240, which makes sense because there will be some overlap, but then it's a completely different number to have to remember. I'm assuming this is a common question, but I can't really find any good framework to follow. Am I missing something easy, or am I on the right track?

There are three types of hands for the purpose of combos: pocket pairs, off suit non pairs, and suited non pairs. First pairs: consider a particular pair, say AA. There are four possible aces for the “first” card and three for the “second”. If the order in which they were dealt mattered (that is AsAh is somehow different than AhAs) there would be 4x3=12 combos. Since order is irrelevant and for each possible suit combo there are two “orders” possible we must divide this by 2 giving 6 possible combos for pairs, such as AhAc, AhAs, AhAd, AcAs AcAd and AsAd.

Now for non pairs, such as AK for instance, there are 4 possible aces and 4 possible kings, for a total of 16 combos of AK. Obviously 4 of these combos are suited (AhKh AdKd, etc). Therefore, by subtraction there are 16-4 = 12 combos of offsuit AK.

If you are counting combos while you have a hand, you can reduce the possible combos your opponent might have (also applies if you see board cards). For example if you have AhQs, your opponent’s range would only include 3 combos of AA (3 for “first” ace x 2 for “second” divided by 2 as discussed before). His range would also only have 12 combos of AK (3 aces x 4 kings) of which only 3 suited combos are possible (he can’t have AhKh), so therefore his range contains 9 combos of AKo and 3 of AKs, rather than the normal 12 and 4 respectively. When you hear people on here discuss “blockers” this is what they are talking about. Villain is less likely than normal to have AA or AK in his range because of the fact that you have AQ.

https://imgur.com/m88ASlC

For visual thinkers.

Thanks for the responses. I understand there are 16 total combos of AK, 12 being unsuited and 4 being suited. I understand there are 6 combos of each pair. That all makes sense. However, is there an easy way to identify how many Kings or Aces are in each range? If an Ace or a King is top pair on the board, is there a way to quickly tabulate how many combos of an A or K, or do you have to think about every possible hand combo they have in real time?

I was hoping to do work away from the tables where I could roughly calculate how many combos of each card a specific position would generally have.

You just have to add all the combos of king-containing hands to get a total. In your example, he has Kxs with x being 8 or higher, Kyo with y being J or higher, and KK in his range. There are 6 possible suited hands, 3 offsuit hands plus KK in his range that have kings. That would be 6x4 (suited) + 3x12 (offsuit) + 6 (KK) = 24+36+6=66 combos with kings. (Looks like you missed either one offsuit or three suited combos when you did your count)

Since his range has 240 total combos, you can conclude that the probability he has at least one king is 66/240 = 27.5%.

Perfect, that's what I was looking for. Much appreciated!