How-To: Calculating Hand Combinations for Advanced Players

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01-01-2009, 06:20 PM
spoonitnow

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 :h: 9 :s: 2 :d:

* 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 :h: T :s: 4:c:

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 :h: 9 :h: 5 :c:

* 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 :h: 9 :h:) 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 :s: J :h: T :h: 9 :h: 9 :c: when our pocket cards are A :c: J :c:.

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.
01-02-2009, 10:18 AM
Monty3038

Wait... In example two, following rule two, the count should be -4, not -5, right?

Oh, no wait... 4 is a BORDER card, not an OUTSIDE card... that is where the problem is... needs a quick edit there spoon...
01-04-2009, 10:15 PM
daven

Re: How-To: Calculating Hand Combinations for Advanced Playe

Quote:

Originally Posted by spoonitnow

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.

i'm interested in this grail you quest for.... the new work you're developing is impressive, well, I'm impressed anyway. You're doing the 99% perspiration that most of us are too lazy to do, genius.

I wonder how many keys there are to poker, range in all its flavours is obviously one. Situation, exploitation, optimisation must also be relevant. And the seemingly ephemeral of how these fit together - Right now I'm psyched to dive into all of this, but need to postpone until I finish work end-April (excuses... i could easily fit this in now, but my priorities are different)

I hope solving these riddles generates the $$ it deserves, thanks for sharing these diamonds
01-04-2009, 11:47 PM
spoonitnow

This is my 3000th post. Yay. I actually started typing up the broadways portion of all of this and then came across what I'm assuming is a special case that's going to require some more investigation that I don't feel like doing right now, so we'll do the pocket pairs instead.

Pocket Pairs

Looking at pocket pairs is pretty easy if you understand how the combinations work (remember 6310 from my original combinations post). Our base count is 6 x # of hands, so if we're looking at {22+} it's 6 x 13 = 78. If a card appears just once, it's a -3. If a card appears twice, it's a -5. If a card appears three or four times, it's a -6. Like I said, pretty easy.

Example 1: We hold AK on K83. How many combinations does {22+} make up?

We start with 78. The Ace is -3 to give us 75. The two Kings are a -5 to give us 70. The Eight is a -3 and the Three is a -3 for a total of 64. To check: AA(3), KK(1), QQ-99(4x6=24), 88(3), 77-44(4x6=24), 33(3), 22(6), and 3+1+24+3+24+3+6 = 64.

Example 2: We hold AA on 986. How many combinations does {JJ-22} make up, and what portion of those are sets?

We start with 10 hands x 6 = 60. The Aces are -5, and the other three cards are -3 each (-9 total), for a total of 46. There are three sets that have three combinations each, so 9/46 of those combinations are pocket pairs, or about 20%.

Looking Ahead to Broadways

What's interesting is that so far the broadways portion of all of this seems to have some really "pretty" things happening, and if you've done much with math, even just arithmetic, you'll sort of have an idea of what I mean. Here are a couple of examples:

Note: I'll use the notation {B} to refer to the range of broadways, which is any two cards Ten and up, excluding pocket pairs.

When we consider {B} on a flop of A72r, we can simply consider a hit to be any Ace, and a miss to be everything else. As it turns out, there are a total of 144 broadway combinations on this board (and any board like Bxxr where B is a broadway card and each x is a non-broadway card), but 48 hit, and 96 miss. This means that exactly 1/3 of {B} hits the pair, and exactly 2/3 of {B} does not.

In a similar case, if we consider {B} on a flop of AA2r, we can again consider a hit to be any Ace, and a miss to be everything else (and like the previous example, this is the same for any paired board of the form BBx). Here we have 128 possible combinations, where 32 hit and 96 miss, and again it comes out to nice even numbers since 32/128 = 0.25 = 25%. So {B} hits this flop exactly 25% of the time, and misses it exactly 75% of the time.

Once we start considering two-broadway flops like QJ2r and also two-tone flops, {B} has more than two shades of strength that aren't as easily defined as "hit" and "miss". It seems that it's easiest to understand how the range hits on a rainbow board and then to understand what changes when you switch the suits of one of the cards to make it two-tone.

Since I'm not ready to go to bed yet, I'll keep typing along.

If we consider {B} on QJ2r and similar flops, then there are a few ranges that "hit" but that are of various strengths. First off, there is a total of 129 combinations that make up {B} on this board. Of those that hit a pair or better, 9 are two-pair, 36 are top pair, and 36 are second pair. The 48 that remain are all gutshots, except in the case that the two broadway cards on the flop are connected (like in QJ2r) where 16 of them are open-ended straight draws, and the remaining 32 are gutshots. It's likely important to note that no hand from {B} ever completely misses an unpaired flop of type BBx.

Looking at two-tone flops of type Bxx, there are two possible changes from their rainbow counterparts depending on if the broadway card on the flop shares a suit with another card on the flop or not. On A :s: 7 :s: 2 :h:, 1/4 of the missed broadways become flush draws. Now the distribution goes from {48 hit, 96 miss} to {48 hit TP, 6 flush draws, 90 miss}, and improves. A flop of A :h: 7 :s: 2 :s: isn't much different, except now four of the top pair combinations are also flush draws. The distribution goes from {48 hit, 96 miss} to {4 TP + FD, 44 TP, 6 FD, 90 miss}.

I'll continue with this later.
01-05-2009, 02:40 AM
OhBollocks

5 spades
01-06-2009, 12:01 PM
Erpel

Quote:

Originally Posted by spoonitnow

Example 2: We hold AA on 986. How many combinations does {JJ-22} make up, and what portion of those are sets?

We start with 10 hands x 6 = 60. The Aces are -5, and the other three cards are -3 each (-9 total), for a total of 46. There are three sets that have three combinations each, so 9/46 of those combinations are pocket pairs, or about 20%.

Fantastic post, thanks - btw, I think you'll find AA is not in a {JJ-22} range so the -5 here doesn't apply.
01-06-2009, 07:26 PM
RML604

NM, Erpel beat me to it.
01-06-2009, 11:29 PM
spoonitnow

Quote:

Originally Posted by Erpel

Quote:

Originally Posted by spoonitnow

Example 2: We hold AA on 986. How many combinations does {JJ-22} make up, and what portion of those are sets?

We start with 10 hands x 6 = 60. The Aces are -5, and the other three cards are -3 each (-9 total), for a total of 46. There are three sets that have three combinations each, so 9/46 of those combinations are pocket pairs, or about 20%.

Fantastic post, thanks - btw, I think you'll find AA is not in a {JJ-22} range so the -5 here doesn't apply.

Yeah I'm retarded sry I'll fix it later.
01-07-2009, 11:24 AM
settecba

Quote:

Originally Posted by spoonitnow

Quote:

Originally Posted by Erpel

Quote:

Originally Posted by spoonitnow

Example 2: We hold AA on 986. How many combinations does {JJ-22} make up, and what portion of those are sets?

We start with 10 hands x 6 = 60. The Aces are -5, and the other three cards are -3 each (-9 total), for a total of 46. There are three sets that have three combinations each, so 9/46 of those combinations are pocket pairs, or about 20%.

Fantastic post, thanks - btw, I think you'll find AA is not in a {JJ-22} range so the -5 here doesn't apply.

Yeah I'm retarded sry I'll fix it later.

Great post.
Something else to fix:

Last examples, there are 24 combos of FD(1/4*96), not 6. Those are 6 hands, 4 combos each.
01-07-2009, 02:22 PM
Erpel

Quote:

Originally Posted by settecba

Quote:

Originally Posted by spoonitnow

Quote:

Originally Posted by Erpel

Quote:

Originally Posted by spoonitnow

Example 2: We hold AA on 986. How many combinations does {JJ-22} make up, and what portion of those are sets?

We start with 10 hands x 6 = 60. The Aces are -5, and the other three cards are -3 each (-9 total), for a total of 46. There are three sets that have three combinations each, so 9/46 of those combinations are pocket pairs, or about 20%.

Fantastic post, thanks - btw, I think you'll find AA is not in a {JJ-22} range so the -5 here doesn't apply.

Yeah I'm retarded sry I'll fix it later.

Great post.
Something else to fix:

Last examples, there are 24 combos of FD(1/4*96), not 6. Those are 6 hands, 4 combos each.

Let me know if I get too nit picky, but I rather suspect you'll find that suited clubs, diamonds and hearts do not constitute flush draws - only spades do.
01-07-2009, 02:33 PM
settecba

thats right erpel, now im retarded...