Application of hand combinations.

[d0zer]

This is a concept I've never understood on a solid mathematical level.

Say you've got 33 on a 37A rainbow board, and you're confident that the nitty-ass villain you're up against is holding either AK or AA.

Now (assuming your read is 100% accurate), if it was equally likely that villain is holding AK as he is to be holding AA, then felting this would be basically 0EV, since you're a huge dog vs half his range and a huge favorite vs the other half.

But there are more combinations of AK than there are AA, so he's more likely to be holding AK then he is AA, so felting this becomes +EV.

...am I on the right track here? How many more combinations of AK are there than paired hands?

The implications of this then are that you have to weigh the likelihood of unpaired cards in villain's range higher than you do paired cards, right? but how much more?

[Ghaleon]

There are 16 ways to make AK and 6 ways to make AA.

So I would say, weigh the unpaired hands 2.67 times more

[PapalRage]

but with the A on the flop, there are only 3 combos of AA left and 12 combos of AK left. so if your read is accurate, he will have AK 4 outta 5 times.

[d0zer]

but with the A on the flop, there are only 3 combos of AA left and 12 combos of AK left. so if your read is accurate, he will have AK 4 outta 5 times.

Oh interestings...that throws a bit of a wrench in things.

So on a AKx board there's 9 AK combos left, and 3+3 = 6 combos of AA and KK.

And on a Kxx board there's 12 AK, 3 KK & 6 AA.

This is great info for when your 'inner nit' starts assuming 'set'.

[Evansgambit]

I don't like to think in terms of percentage being dealt preflop as this is at times not realisitc.

While you can mathematically be dealt 6 different and unique AA. You find that in the actual game, there are only two AA hands at best, using all the aces . For example, when a person has the ace of hearts, another person can not have that ace of hearts.

You have 33 on a flop of 37A, and assuming you think the villian has either AA or AX (X=King, Queen, Jack). There is only one hand your afraid of and that's AA - the only combination of AA using what ever remaining aces in the deck you like.

But with a flop of 37A, he could have AX in three different ways, using any of the remaing three aces.

So, assuming you know your player. You just have to weigh it up and call. You lose to one hand AA, but beat the remaining three hands AX.

Although he's more likely to be dealt AK than AA, but according to your question, given the cirumstances that it is equally likely for the villian to hold AA as AK, equally likely as in 50% AK and 50% AA. Then it becomes a coin flip on your part.

For me to many people play AK stronly in this spot, that its hard to decide. Factoring in, of course, how s/he likes to play the nuts.

[Ragnar4]

But you're felting here everytime, no matter what your read is, and if you're coinflipped for his read versus AA, and AK... you're STILL felting.

[d0zer]

But with a flop of 37A, he could have AX in three different ways, using any of the remaing three aces.

Ax is dealt in a lot more than three different ways. It's 3 times however many remaining cards the 'X' has. So on a 37A board there's 12 ways to have AJ, and 9 ways to have A7, so there are (12 * 10) + (9*2) = 138 different combos of Ax, and only 3+3 = 6 combos of AA or 77, so if you put Ax in villain's range, this is an easy all-in with a set (as it usually is with a set).

[Evansgambit]

well about the AX. What your saying from a probability point of view is correct.

But think about it, how many aces you got remaining in the deck? Three. So in reality, he can only have one of these three aces to have AX.

I think for me, there's alot of confusion between probability of being dealt, and ACTUAL physical hands available.

Again, i see all these numbers, especially with the number of hands that dominate AJ for example. Maths will tell you 12 ways for this and 9 ways for that. But in reality, you only got 3 remaining aces, and the pairs KK, KK, QQ, QQ, and the one JJ.

This is just the way i'm thinking. if someone can explain it to me, then i welcome it.

[d0zer]

What we're doing is putting an opponent on a range of hands, then seeing how we compare against that range in a loosely quantifiable way.

Yes there are only 3 aces out there, but if you've got say AQ, and you're pretty sure he'd only play AA & AK like this it's an easy fold. What about if he'd only play AA/AK/AQ/AJ/AT like that. THEN is it an easy fold? That's when I'd do the hand combination math & see how many hands I beat vs how many beat me, and compare that with the price I'm getting to showdown.

To complicate things, sometime (maybe even often), you have to assign probabilities to certain hand groups.

When you're limiting your opponent's postflop ranges, you have to consider the possibility they're holding one or both of the remaining 'hot' card on the board, but also consider all the other hands they may be playing like this. Sometimes there really are no other hands (or a ridiculously low probability of them), like in a 3-bet pot with a nit who always gives up without hitting who leads on a A33 board.

As you get better and better, you get more accurate with the ranges you assign villains.

[bigslikk]

I'd deem the "hand combinations" idea as one factor of many when creating a read.

e.g.
"Hmmm... He reraised preflop, and he's pretty solid, so AK and AA both make sense given the action of the hand. I'd normally lean towards villain having AA in this spot given the reraise but since that hand has a lower probability of being dealt out I'll decide it's 50/50 either way."

I would then proceed to felt it; Villain flips over his hand.

77.

[Stacks]

So on a 37A board there's 12 ways to have AJ, and 9 ways to have A7, so there are (12 * 10) + (9*2) = 138 different combos of Ax, and only 3+3 = 6 combos of AA or 77

Hey dozer... Can you explain this little bit of math to me please? I just learnt a little about hand combos today and I realize how an unpaired hand (say AK) has 16 combinations and a pair has 6. Given the board and it having one of the cards we feel the villian would play this way (A) we now have 12 combos for AK (4 aces * 3 kings = 12) and 3 combos for AA. However, I fail to see where you are getting the 10 and 2 used in your math (I do understand how you got 12 for AJ and 9 for A7 though)?

On a second glance my only guess here on your 10 and 2 would be that on that board there are 10 Ax hands you could have and no be making 2 pair and 2 Ax hands that you could have and make 2 pair. Is that correct?

[Evansgambit]

Ok, i think i got something here. maybe, maybe not.

This is only a random probability thing, just the odds, nothing to do with preflop play and player analsysis. We will use probability and combinations from maths class.

A deck has 52 cards - agree. We have seen already 5 cards, that leaves 47 cards remaining. These 5 cards are: 333A7.

Now would the probability that villian has AA be dependent on what comes on the flop, turn and river? The answer I believe is yes, since the probability of the flop is dependent on the hole cards. This in turn, whilst not changing the actual outcome, changes the random preflop odds. If the flop comes with two aces, the chance of him having AA is slimmer, and if the flop is AAA, the chance of him having AA is zero. So a relationship exists. (I could be wrong of course)

Now to determine just the random chance of him holding AA. Its simply a matter of the combinations of AA available from the 47 cards, divided by the total number of two card combinations with the remaining 47 cards.

AA as we all say, can be dealt in 3 ways. Eg. As Ad Ac can form As Ad, As Ac and Ad Ac. (Not to be confused with reality, where you can ONLY have two AAs maximum =P). This is a combination (nCr), so the order in which the aces are dealt is irrelevant, thus As Ad is the same as Ad As.

Next the number of total possible 2 card combinations is simply 47 options for the first card, and 46 options for the second. Thus, 47 * 46 = 2162.

Probability = required outcome/total outcomes = 3/2162 = 0.138% which gives us the odds that this villian is holding AA.

Whereas the chance of AK which is the same as AX (other than A3 or A7) is: 12/2162 = 0.555%. Thus AK is 4 times more likely to be present than AA.
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Now suppose, we put this villian on an Ace for sure as is what's started by the original poster, since villian would need at least an Ace to remain competitive. It is then a matter of working out the odds for the second card.

Because we assume the villian has an ace, we can say that we've seen 6 cards from the deck. These are 333 A7 A. Leaving 46 cards remaining. Notice, as we've put villian on an ace, the chance of him having AA preflop suddenly increases.

It is now, 2 remaining aces/46 cards in the deck = 4.348% the villian has AA. Leaving 95.652% of the time by random chance, villian does not have AA, but AX.

Now based on odds alone, and not considering player tendencies and capabilities - I gotta call and pay AA off.

[d0zer]

On a second glance my only guess here on your 10 and 2 would be that on that board there are 10 Ax hands you could have and no be making 2 pair and 2 Ax hands that you could have and make 2 pair. Is that correct?

Yeah that's what I was going for. I'm still new to this stuff myself though, so ...

[Evansgambit]

Actually i think i'm wrong in the AX having 12 possibilities. too much thinking lol. will review some other time. cheers

[d0zer]

Actually i think i'm wrong in the AX having 12 possibilities. too much thinking lol. will review some other time. cheers

Well you're right in that any single Ax has 12 possibilities, but all the Ax's have a whole lot more. Which makes sense because if you're up against some maniac who's got Ax, 2pr, sets in his range on a A73 board, and you've got AQ, intuitively you should know that you're ahead most of the time.

When you add up all the dozens of hand combinations of the Ax's you beat, you see what you knew intuitively expressed mathematically.

[RML604]

Now would the probability that villian has AA be dependent on what comes on the flop, turn and river? The answer I believe is yes, since the probability of the flop is dependent on the hole cards. This in turn, whilst not changing the actual outcome, changes the random preflop odds. If the flop comes with two aces, the chance of him having AA is slimmer, and if the flop is AAA, the chance of him having AA is zero. So a relationship exists. (I could be wrong of course)

I'm not positive about this, but I think your thinking is backwards. If the flop comes AAA, you're right that there's no chance of him having AA. But since the hole cards are dealt before the flop, isn't it correct to think of it in terms of him having AA means there is a 0% chance of the flop coming AAA?

You can use the flop to better deduce what cards he may (or more likely may not) be holding, but I don't think the flop necessarily changes the odds of what he's holding. After all, how can action A change the odds of action B when A comes after B?

So what does this mean? I don't know. But I think what it means is that if an A is on board, that doesn't necessarily mean that a player has slimmer odds of holding AA. You can say he only has 3 ways to make AA now instead of 6, so he's 50% less likely to hold AA, but he did have 6 ways to have AA when the cards were dealt.

Just because an A is showing doesn't mean he couldn't have had that card in his hand. It just means that he doesn't now.

I'm not sure if I'm right about this, especially when you consider the case of the board coming AAA, because this does mean that the probability of villain holding AA is 0%.

Anyone agree with anything I'm saying, or am I way off on this?

[BankItDrew]

i've never replied because i have no idea wtf the problem is

[d0zer]

i've never replied because i have no idea wtf the problem is

I don't have a problem, do you have a problem? Cause if you wanna problem, we can have a problem.

[sweetlemon69]

FFIIGGHHTT!!!!!!!

[Keilah]

Now would the probability that villian has AA be dependent on what comes on the flop, turn and river? The answer I believe is yes, since the probability of the flop is dependent on the hole cards. This in turn, whilst not changing the actual outcome, changes the random preflop odds. If the flop comes with two aces, the chance of him having AA is slimmer, and if the flop is AAA, the chance of him having AA is zero. So a relationship exists. (I could be wrong of course)

I'm not positive about this, but I think your thinking is backwards. If the flop comes AAA, you're right that there's no chance of him having AA. But since the hole cards are dealt before the flop, isn't it correct to think of it in terms of him having AA means there is a 0% chance of the flop coming AAA?

You can use the flop to better deduce what cards he may (or more likely may not) be holding, but I don't think the flop necessarily changes the odds of what he's holding. After all, how can action A change the odds of action B when A comes after B?

So what does this mean? I don't know. But I think what it means is that if an A is on board, that doesn't necessarily mean that a player has slimmer odds of holding AA. You can say he only has 3 ways to make AA now instead of 6, so he's 50% less likely to hold AA, but he did have 6 ways to have AA when the cards were dealt.

Just because an A is showing doesn't mean he couldn't have had that card in his hand. It just means that he doesn't now.

I'm not sure if I'm right about this, especially when you consider the case of the board coming AAA, because this does mean that the probability of villain holding AA is 0%.

Anyone agree with anything I'm saying, or am I way off on this?

Now that we've seen the ace we have more information. When we decided preflop that he had 6 ways to make AA that was because we didn't know that at least 1 A was still in the deck. He only had 3 ways to make AA, we just didn't know it.

Now that we know it, we go back and revise to say that he only could have had 3 combos of AA.