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How-To: Calculating Hand Combinations for Advanced Players
The title of this thread is an obvious reference to various Twoplustwo book titles and does not imply that this post is for advanced players only.
In How-To: Calculating Hand Combinations, a post I made a few months ago, I outlined how to figure out how many combinations there were of a single hand on a given board (for example, there are 12 combinations of JT on J32 if we hold AA). With a little bit of practice it becomes second nature and is a staple in any type of range analysis.
Instead, what we're looking at here is something like finding the number of combinations of all suited broadways on AhKhKc3c2d when we hold 9c9d. From the previous post I linked to, we could break down each hand like this: AKs(2), AQs(3), AJs(3), ATs(3), KQs(2), KJs(2), KTs(2), QJs(4), QTs(4), JTs(4), for a total of 29. With my findings today, we say 40 - 11 = 29 and have our answer a in a couple of seconds instead. The trick is to figure out where the 40 - 11 comes from, and that's what I'm going to get to here. And surprising to me at least, it doesn't require much (if any) memorization.
Warning: If you have not read this post and completely understand it, then the following information is probably going to seem overwhelming and confusing. As such, I'm going to break this topic up into a number of sections, the first of which is suited connectors. The rest of the sections will come in my next post in this thread.
Introduction to Suited Connectors
My Holy Grail is to be able to quickly manipulate ranges and perform calculations with those ranges mentally. To that end I've spent a reasonable amount of time working on how to simplify these processes for certain typical "parts" of a range. Here I'm going to outline a quick way to do calculations like these with a range of suited connectors. For the examples I'm going to give here, I'll be using {QJs-54s} which is 8 hands and 32 possible combinations before cards are dealt. Later I'm going to make this more general so it can be applied to any continuous range of suited or offsuit connectors with n gaps.
I want to stress that this is not difficult at all to do, despite the examples with lots of numbers and such below. The examples are just here to make sure there are no questions about how or why this works.
The idea is that we're going to start with the number 32 and subtract some amount based on the cards that are exposed. In our example range {QJs-54s}, I'll refer to any Queens or Fours that show up as "border" cards. Similarly, if we were just looking at {T9s-65s}, Tens and Fives would be "border" cards. Cards in between the border cards I refer to as "inside" cards, and cards that aren't inside or border cards are outside cards. So with {QJs-54s}, the border cards are Queens and Fours, the inside cards are Jacks through Fives, and the outside cards are everything else. Outside cards don't affect our calculation.
There are 4 "rules" to do this with that are pretty much common sense once you think about it for a moment. Our count always starts with however many possible combinations there were before any cards were dealt. I'll illustrate these 4 "rules" in examples.
Example 1: {QJs-54s} on Q 9 2
* Rule 1: A border card gives us -1 from our count.
* Rule 2: An inside card gives us -2 from our count.
* Rule 3: An outside card doesn't change our count.
The Queen is a border card, the Nine is an inside card, and the Two is an outside card, for a total change of -3 from our count, so there are 32 - 3 = 29 possible combinations of {QJs-54s} on this board. Now let's verify: QJs(3), JTs(4), T9s(3), 98s(3), 87s(4), 76s(4), 65s(4), 54s(4). And the total number of combinations is 3+4+3+3+4+4+4+4 = 29.
Example 2: {QJs-54s} on T T 4
The board being paired doesn't affect our calculation. The Tens are inside cards and the Four is a border card for a total change of -5 from our count, so there should be 32 - 5 = 27 possible combinations of {QJs-54s} on this board. Now let's verify: QJs(4), JTs(2), T9s(2), 98s(4), 87s(4), 76s(4), 65s(4), 54s(3). And the total number of combinations is 4+2+2+4+4+4+4+3 = 27.
Example 3: {QJs-54s} on T 9 5
* Rule 4: A suited connector on board gives us +1 to our count.
Here all three cards are inside cards, and we have one suited connector shown on board (T 9 ) for a total change to our count of -2-2-2+1 = -5. Then there should be 32 - 5 = 27 possible combinations of {QJs-54s} on this board. Once again we'll verify: QJs(4), JTs(3), T9s(3), 98s(3), 87s(4), 76s(4), 65s(3), 54s(3). The total is 4+3+3+3+4+4+3+3 = 27.
Example 4: {QJs-54s} on Q J T 9 9 when our pocket cards are A J .
There is 1 outside card (Ac), 1 border card (Qs), and 5 inside cards (Jh, Th, 9h, 9c, Jc). In addition, there are 2 suited connectors we can make with the exposed cards (JhTh, Th9h). The 5 inside cards give us -10, 1 border card makes that -11, 2 suited connectors on the board make that -9, and 32 - 9 = 23, so there should be 23 possible suited connectors on this board. For the last time, let's verify: QJs(1), JTs(2), T9s(2), 98s(2), 87s(4), 76s(4), 65s(4), 54s(4). Our total is 1+2+2+2+4+4+4+4 = 23. Ta da!
And that's all for the suited no-gap connectors.
Edit: Typo in example 2.
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