Combinatorics and You: Relativity in Hand Reading |
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Posted Thu Jan 26, 2006 2:34 pm GMT by snoogins47
This post is spurred on by a post I made in General last night.
I posed a question that I had seen on another forum/blog:
Let's say UTG makes a moderately tight raise from {AA-99,AK,AQ,AJ,KQs}.
On which of the following flops is he most likely to have top pair or better? On which is he least likely to have top pair or better?
Axx
Kxx
Qxx
Jxx
8xx
The correct answer is Kxx. We'll get to figuring that out in a second.
We all know that to get an accurate idea of our hand strength, we have to narrow our opponent's possible hands to a "range" or hands that he may have, and compare our chances of winning vs. that range. We instinctively know that we have to weight hands that are more likely, and downplay hands that would be an "oddball:" if a certain hand is unlikely, why should our performance against it play heavily into our analysis?
One factor that can often be overlooked, and that is intensely important in getting an accurate idea of how you stack up, is the sheer distribution of pocket-card possibilities. Luckily, in a Hold'em context, this calculation is a breeze! Let's take a look here:
Facto #1: Pairs are unlikely. Any specific unpaired hand is ~2.67 times more likely than any specific paired hand.
Facto #2: Suited hands are even more rare: any specific generic (as in, ignoring suits) hand, is 4 times more likely than its generic counterpart: (AKs, vs AK)
Facto #3: Within a specific realm of hand, the unsuited variety is 3 times more likely than the suited one (AKo vs. AKs)
Facto #4: A dead card, which I'm defining here as a card that cannot possibly be in your opponent's hand (you hold it, it's on the board, you snuck it out of the deck when he wasn't looking) has a more pronounced effect on pairs, than on offsuited hands.
For instance: If you hold one Ace, the chances of your opponent holding Ace-King drop by 25%. The chances of your opponent holding AA, however, is cut in half.
Facto #5: TWO dead cards, do more. If you hold AA, your opponent has only 1/6th the chance of holding AA that he normally does. His chance of holding any specific hand including an Ace, such as AK, are cut by a "mere" 50% from the original probability.
Facto #6: Three dead cards obviously makes pairs impossible, and cut the "specific offsuit hand" probability to 25% the original. This doesn't come up quite as often, but it's useful for like, flopping sets.
This math seems very engaging, and without a real look into how it's done, probably seems quite high level. In fact, I'm doing it/checking it in my head as I'm typing this out. It's not hard.
Back when I used to be a workin' stiff, I would spend hours doodling, figuring, and everything. I worked in a small motel, that on weeknights, would sometimes only have 4 or 5 guests in the whole thing. I'd have 8 hours to kill. It leads to crazy things. During this time, I had played some poker and read a bit, but I was fairly new to the theory aspect. At first, I had to figure these out by hand, but as I worked with them more and more, they stuck with me.
Basically, it's a simple combinatorial problem, but don't be scared: that word is probably harder to use than the math is.
Say that our opponent, we've decided, has to have either AA, or AK. How does the distribution look?
It's quite simple. For AA, we need to only concern ourselves with the Aces. There are four aces in a deck of cards.
As Ac Ad Ah
How many possible ways can these cards make one pair? We know he can't have two of the same card, so we have to look at how many combinations of "AA" there are.
As Ac
As Ad
As Ah
Ac Ad
Ac Ah
Ad Ah
That's them. 6 of them. There are six ways our opponent can make AA.
What about AK?
With four Aces and four Kings, it's easy to do this one. Each A, can be matched up with one of the four Kings, to give him AK.
4x4 = 16.
16 ways our opponent can make AK.
These ALWAYS remain constant. Therefore, with that range of hands, our opponent is 2.67x more likely to have AK than AA. This is a very significant difference. For instance:
THE BAD WAY: We have QQ. We think he has AK or AA, and reason that half the time we're a 4 to 1 underdog, and half the time we're even money. (.5 x 50%) + (.5 x 20%)This puts us at about 35% to win. THIS IS WRONG
The RIGHT way: Accounting for the increased likelihood of AK, we reason that 16 times out of 22(22 is the total of possible holdings here) we are 50/50, and that 6 of those 22, we only win 20% of the time. This puts us at nearly 42% to win! That's a pretty big deal. (16/22 is roughly 73%, so 6/22 is roughly 27%)
So now that we can see the basics here, and have an idea how to work with them, here's the numbers of possible combinations to remember. I'm going to call them "frequency factors"
| Code: | UNPAIRED HANDS:
Ignoring Suit, with Zero relevant cards dead: 16
Ignoring Suit, with One relevant card dead: 12
Ignoring Suit, with Two relevant cards dead: 8
Ignoring Suit, with Three relevant cards dead: 4
ANY combination that is suited, with zero dead: 4
ANY combination that is suited, with one suit dead: 3
ANY combination that is suited, with two suits dead: 2
ANY combination that is suited, with three suits dead: 1
(This is tricky to explain. What this means, is that if one of the diamonds is dead, it is impossible to have that hand in that suit: if the Ad is dead, he cannot hold AKd. This doesn't change if the Kd is ALSO dead.)
Specific suited hand (for instance, AK of diamonds); 1
Specific suited hand with one card dead: Impossible
PAIRED HANDS:
Pair with Zero cards dead: 6
Pair with One card dead: 3
Pair with Two cards dead: 1
Pair with Three cards dead: N/A
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What to do with these? You can quickly use them to get an estimate of how likely certain hands are at the table, and you can use it to get specific amounts while away. So, let's practice using an example.
You make an awesome super-badass power re-raise with AK. A reasonable opponent pushes all in, over the top. You reckon that he will only do so with AA, KK, or AK. What is the distribution?
We have one of the Aces and one King accounted for, so, that leave 3 ways he can make AA, 3 ways he can make KK, and 12 he can make AK.
Hence, AK is twice as likely as our opponent holding a pair. He's a 2:1 underdog to have a pair, even given this tight range of hands.
Now, how do we figure out the original question? Well, it's simple really. Take all his hands, assign the correct frequency factor. Mentally, or physically. Now look at each board. Add up the frequency factors of each hand that fits the criteria: top pair or better. The board with the LOWEST sum, is the board on which he is least likely to fit the criteria.
Since this post is almost entirely off the cuff, I'm afraid I haven't been able to devise a 'system' to do the math in a quick and dirty fashion at the table, though I've spent a few minutes thinking about it. I will get back to you if I can derive some sort of formula that gives a reasonably correct answer. One way to apply it in hand analysis:
Take your range of hands.
Add up the frequency factors: this sum will be how many combinations of hands fall into this range.
For each hand, divide that specific hand's frequency factor, with the sum total of combinations. You'll get a decimal. That's the percentage of all the possible combinations in the range, that = that specific hand.
More simply, if your range were 100% correct and there was no necessary weighting, that is the precise percentage of the time that he will hold the hand in question.
I will probably add to this/edit it in the future, and hopefully come up with a good simple way to facilitate these calculations at the table. I hope somebody finds this interesting, and possibly *gasp* useful.
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Posted Thu Jan 26, 2006 2:55 pm GMT by Soup_dog
Great post! This is the kind of stuff I love. Thanks a bunch!
Posted Fri Jan 27, 2006 6:22 am GMT by tame_deuces
This is incredibly useful stuff and one of the things which took some time to sink into my head when I started playing hold'em.
Another principle which one can superimpose on top of this is Bayes Theorem. http://en.wikipedia.org/wiki/Bayes_theorem
It's not a difficult principle, basically it just means you update a belief after observing evidence. All poker players do it, but many often forget it. It means if you figure a player for a typical hand range preflop, then you can narrow down his handrange by observing later actions AND observing the fallen cards and the cards in your own hand. So basically you do what Snoogins has showed in his post + you subtract unlikely hands that villain is to have based on later actions, the common mistake is to start including hands that you earlier deemed unlikely into the equation again.
For example against a tight player this is a much worse flop for your AA: T J 5, than this: J J 5. But many people forget to start with villain's preflop range and then narrow it down based on what cards fall on the flop and villain's actions, instead they just conjure up a new horrible hand range for the villain there and then on the flop. Imagine this hand:
You raise preflop, villain re-raises from the blinds, You call
(knowing villain is tight, you figure his re-raising hand range from the blinds to be about JJ+,AKo+, with an odd weird hand now and then. You call for deception).
Flop comes J J 5.
villain bets, you raise, villain pushes
Now, some pokerplayers forget their preflop ranging on villain and they might actually lay this down. Or they might call because they 'have a hunch'.
Ofcourse, few human poker players are so rigid that we could be so certain of being ahead as we are here, but this is things worth having in mind.
I don't have Snoog's way with numbers, so I'll stay away from them.
Posted Sat Jan 28, 2006 10:25 pm GMT by snoogins47
A favorite "teaser" of mine is a simple conditional probability question, that I believe I first heard read from Ed Miller, but I'm realy not very sure of that part.
It's not that complex, but it's a nice way to kinda get your head around the concepts without having a mathematical background, which is exactly the kind of thing I always need.
You're playing Russian Roulette with your favorite gun-toting maniac. He pops two bullets into the six-chambered revolver, spins away, aims, and click. He's fine. He hands the gun to you.
Assuming you don't enjoy self-inflicted gunshot wounds to the head, or, at least assuming you are competitive and want to beat him, do you just aim and pull, or do you want to spin the chamber first, THEN fire?
Out of 6 chambers, 2 have bullets, so you're a 2:1 favorite to survive if you spin. You hit a bullet 33.33333threethreethree% of the time. Not too bad, really. "I bet my life that you miss your flush"
But what other information do we have? We know he just pulled the trigger, and did not die: more specifically, we can use the knowledge we have of the PRIOR state of the gun, to give us a better idea of the current state.
Let's do a stupid little diagram.
| Code: |
O O O O X X
1 2 3 4 5 6
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Since we just SAW that he pulled the trigger, the gun advanced, and there was no bullet, that narrows down our possibilities: there are four empty chambers, so there are only four specific states the gun could have been in prior to now. Chamber 1, 2, 3, or 4. Since his firing advanced the gun, the gun must right now be ready to fire Chamber 2, 3, 4, or 5. So, only Chamber 5 is lethal, and we are a 3:1 favorite (75%) to survive if we pul the trigger without spinning.
The Bayesian logic in poker thing tends to be something along the lines of this: say our opponent will only cap the betting with AA/KK. You hold AQ. The combination of these two bits of information shows us that our opponent is a 2:1 favorite to hold KK.
Now the flop comes A 7 2 rainbow. Using this new information, we deduce that our opponent is actually 6 times more likely to hold KK than AA, and we are a very solid favorite against what he might hold.
We can't stop there though. If our opponent puts in fourteen or fifteen bets, I assure you the chance of us having the best hand is almost zero, regardless of our past information.
Posted Sun Jan 29, 2006 1:30 am GMT by Phil14312
I've read this thread 4 times and I am just now beginning to arrive at these concepts. 4 more and maybe I will understand.
Posted Mon Jan 30, 2006 12:38 am GMT by supafrey
oh snoo - never enough emphasis put on the meta game. put down the calc textbook you dork.
Posted Mon Jan 30, 2006 12:39 am GMT by snoogins47
| supafrey wrote: | | oh snoo - never enough emphasis put on the meta game. put down the calc textbook you dork. |
Aren't you the one who claimed that "meta game" was a term I made up?
Posted Mon Jan 30, 2006 12:46 am GMT by supafrey
no, I just claimed that you using poker lingo like that was you overcompensating for your tiny manhood. like how men can never explain home repairs without overcomplicating it to women with big words.
THAT'S how i'm hypocritical, you douche.
we both know i'm right, too. scientism gambling MORE PLEASE.
Posted Mon Jan 30, 2006 11:47 am GMT by suitedaces84
| snoogins47 wrote: | A favorite "teaser" of mine is a simple conditional probability question, that I believe I first heard read from Ed Miller, but I'm realy not very sure of that part.
*snip*
Since we just SAW that he pulled the trigger, the gun advanced, and there was no bullet, that narrows down our possibilities: there are four empty chambers, so there are only four specific states the gun could have been in prior to now. Chamber 1, 2, 3, or 4. Since his firing advanced the gun, the gun must right now be ready to fire Chamber 2, 3, 4, or 5. So, only Chamber 5 is lethal, and we are a 3:1 favorite (75%) to survive if we pul the trigger without spinning. |
Now let's say we play the same game against the same maniac. Except this time he puts one bullet in, spins the barrel then puts another bullet in. Again, he puts the gun to his head pulls the trigger and is fine. He hands the gun to you, should you spin the barrel, fire or does it make a difference what you do?
This one is a little tougher.
Posted Mon Jan 30, 2006 4:20 pm GMT by snoogins47
I kinda want to wait to see if anybody else figures it out before I write the answer, but I don't think anybody is as dorky as I hope they'd be.
Posted Mon Jan 30, 2006 4:24 pm GMT by supafrey
spin it.
didn't do any math, but i'm going to post the "common sense" answer i'm figuring. (Which is usually wrong and part of the joke)
His turn means that the bullet was not in the first chamber that he fired. This eliminates one bullet position, leaving you with a 2/5 chance of blowing your brains apart. (you can't shoot from the same chamber twice)
A random spin gives you the original 2/6 (1/3) chance instead. +EV
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