View Full Version : Betting Any Two Hole Cards

04-25-2005, 01:20 PM
For whatever their reasons might be, there are players who are willing to enter a NL HE hand with any two hole cards. Let's call them Villain, and of course we are Hero.

Hero has a set of acceptable opening hands, perhaps based on his position at the table, or otherwise. Villain has no such restriction on his play. In fact, once Villain ascertains Hero is one who thrives on overcards (the usual case, when one chooses opening hands tables), Villain can exploit Hero quite often. However, the reverse also applies. I certainly don't want to get into their respective strategies, since that has nothing to do with my question.
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What is the distribution of best hands, considering only hole cards and the flop?
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I'm asking this question only for the Villain's hands, since in his case these are 5 truly random cards. I suspect the answer to this question is available in published form, since stating the question is so easy! The answer would be of the form:

7 high card: 0.0197%
8 high card: 0.0273%
A high card: 0.654%
2/2: 1.259%
3/3: 1.627%
A/A: 7.935%
3/3 and 2/2: x%
4/4 and a lesser pair: y%
etc etc
Royal flush: z%

This is just an example; numbers written down are meaningless, except to support the example.

For my purposes, I'm not particular as to whether the results were obtained theoretically, or using Monte Carlo. Simply speaking, what are the results of having random hole cards, in terms of best hands thru the flop?

Every set of tables of opening hands has a different answer to this question. Finding best hands thru the flop is then a function of choice of opening hand sets. It seems reasonable to me that a person making a decision about opening hands tables would be interested in this answer, before finalizing the choice. When the hand begins, the person either enters the hand or he doesn't; best hands thru the flop include his acceptable hole cards plus 3 other random cards. This problem is solvable only using Monte Carlo, I'd guess. Presumably the answer offers hands superior to those Villain obtains, but then again Villain is in many hands Hero isn't in. If Hero is so tight that he plays only 20% of hands, then Villain will be in with Hero only 20% of the time, and Villain will be in and Hero out, 80% of the time. If all other players at the table have opening hands tables just like those of Hero, then Villain will pick up the blinds 80% of the time, uncontested.

Life isn't that simple, and neither is poker. These things are situational, too. If everyone has folded and only the button remains, you are going to strongly consider going into the hand, and who cares about the opening hands tables at that point?

Thanks for your comments!


04-25-2005, 01:25 PM
this seems like a very interesting question, but I think it is slightly ill-posed -- you share 3 cards with the Villian (i.e. the flop), which will change the distribution of what he can and can't have. For example, the probability that Villian has 7-high on a AKQ board is 0.

So what's the difference between your hand and his? Starting cards /images/graemlins/smile.gif

04-25-2005, 03:02 PM
Thanks for your inputs!

At the actionable point, which is when the decision to go into the hand or not is made, Villain doesn't have to make the decision; he just goes in. Hero, on the other hand, has 94 unsuited, and chooses on the basis of his opening hand tables to not enter the hand. The Villain, being in the hand with his random cards, then is the beneficiary of this mighty fine flop, whether he's deserving or not.

In the Monte Carlo treatment of the hand, of course you are correct! At the very least a flop of AKQ offers best hand of Ace. The Monte Carlo is executed on the basis of just one of the two, Villain or Hero. If the focus of this calculation were Villain, cards are shuffled and then all 5 cards, for the combined Villain hole cards and the flop, are selected at random. Then the best hand is figured out on the basis of only those 5 cards, and statistics are incremented. This process is repeated 1 million times, and then the distribution of best hands becomes the output result (the answer).

On the other hand to handle the Hero Monte Carlo calculation best, it would be wise to do it on the basis of the exact opening hand tables Hero uses. Each hand Hero might have is done in proper proportion. Pairs can be made in 6 ways, suited connectors in 4 ways, and unsuited nonpairs come up 12 ways.

Let's use an example. A/A is included in Hero's opening hands list. Then his hole cards are set to be A/A exactly 6 times, and the Monte Carlo computation then proceeds for the remaining 3 random cards of the flop, just like it was done with Villain. K/K is then set 6 times, Q/Q 6 times etc. In this way, when one has proceeded thru Hero's entire opening hand tables exactly once, the result is precise regarding sampling from Hero's opening hands, and the Monte Carlo limits to the 3 random cards. Then the procedure is repeated again and again, until in excess of 1 million hands have been included. Takes more programming time, but it's a better Monte Carlo computation.

You're correct that Hero's cards are selected on one basis, and Villain's on another. Hero is tightly constrained, and Villain knows it (at least after a few dozen hands). Villain has all sorts of options available to him that Hero doesn't. Will the best hands distribution for Hero be better than that of Villain? Presumably so, since the result is biased towards better cards. Will this advantage Hero has counteract 100% advantages Villain has? That's where the discussion would go, after we had best hands distributions for both Hero and Villain, but without these it's premature.

Does that stop persons from venturing opinions? Not at all! Opinions are a dime a dozen, and most would pronounce authoratively that the tight Hero play will overcome loose Villain play. That's where this is going to, but first we need those distributions!