Challanging Ed Miller's Criticism of Lee Jones

[razor]

In a game where only 3 players see the flop there are players capable of folding weak/mediocre Aces.

[John Biggs]

I think you've put in a lot of effort to calculate on some very flawed assumptions. The results are therefore not just questionable but useless.

[Mike Haven]

the results are not questionable, and no correct results are ever useless

it may be that some players won't call with, say, A6o - that would affect real-life results to a minor extent

name the scenario and my programme should be able to give you the figures within moments

i am giving facts not recommendations - what you decide to do armed with the facts is up to you

[jedi]

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Well, this HAS already been taken into account. You're playing against a player that will take his losing hand too far.

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Just time for a quick note.
Too far can simply mean a desire to see the flop with Ace/Little...or even to see the turn, or thru the river...to automatically assume that the Mystery Hand--be it Ace/Little (with a likely win) or 5,6 off--will always go thru the river is what the calculation is based on. That's simply not realistic.


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Why is this not realistic? This is Small Stakes HoldEm, written for loose low limit games. This is NOT Hold Em for Advanced Players. To not take advantage of people's propensity to play sheriff and call down your bets is a serious leak at this level. If you have a read on a player as being tight post flop and won't call without top pair or better, THEN you might have a point, but the typical player at these levels WILL pay you off.

[Ed Miller]

i have stated that in my calculations i assumed any player receiving Ax will see the flop - therefore, with this criterion, if only one player saw the flop and the flop was AAA there is a 38.3% chance that he has Ax - this seems odd at first glance, but you have to remember that the other eight players were dealt two cards pre-flop that were not Ax

This is not correct Mike. Again, Bayes' Theorem bites you in the ass. [img]/images/graemlins/tongue.gif[/img]

38.3% is merely the PRIOR probability that someone was dealt an ace. But the fact that only one player has called gives you a TON more information about the distribution of hands.

If that's not clear, then think about it this way. Assume that you know all nine of your opponents will play any ace, as well as a host of other hands like two big cards, a lot of suited hands, and some connected hands.

The flop comes AAA.

How likely is it that you are against an ace if all nine people called preflop?

How about if only one person called preflop?

What if zero people called preflop (yes, I know, there would be no flop in that case... assume you were rabbit-hunting)?

These numbers are NOT all 38.3%. [img]/images/graemlins/smile.gif[/img]

Here's a general rule for everyone. As soon as you get ANY more information about a situation, the probability CHANGES, and the prior probability is no longer correct. The number of people calling preflop is very significant information.

[John Biggs]

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This is not correct Mike. Again, Bayes' Theorem bites you in the ass.

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My man!!! This is what I intuitively thought. If Mike doesn't have time (though it looks like he's pretty dedicated) I'll do the math myself.

[John Biggs]

Since a three-way pot (you with KK versus two opponents) was stipulated in the example, that's what seems most relevant here. So what's the true probability that if you hold KK, see the flop with two other players who will play any Ace vs. a raise, and the flop comes Axx, that you are up against one or more Aces in their hands?

First, note that in fact the stipulation that they will play any Ace vs. a raise is totally irrelevant! This is because we have no information about what hands they won't play vs. a raise, therefore we are working essentially with random hands.

With only two opponents, our cases are as follows: no Aces, one Ace in the hands of each opponent, one opponent with AA, or one opponent with AA and the other opponent with the case Ace.

One approach is to add these cases together to get a total of ways. Is there a simpler approach? Yes; it's to regard our opponents' four cards as a four-card block and calculate how many ways there are to deal this block such that it contains at least one Ace. The easiest way to do this, in turn, is to calculate how many ways you can deal no Aces to this block, then subtract that number from all ways to deal the block.

We have seen five cards, therefore 47 cards remain in the deck from which we will deal a four-card block. If we take out the three remaining Aces, that gives us a 44 card deck to deal a block with no aces:

(44,4) = 135,751

Total ways to deal the four-card block:

(47,4) = 178,365

Subtract to get the ways to deal a block with one or more Aces:

42,614

Now we divide by total ways, then multiply by 100 to get our percent chance we're up against at least one Ace:

24%

Therefore Kings are a pretty darn good hand in a three-way pot versus what are essentially two random hands by the terms of Mike's argument. In a tougher game, we'd need to rethink our assumptions.

[Mike Haven]

We have seen five cards, therefore 47 cards remain in the deck from which we will deal a four-card block.

no - this is one area where i say you are wrong

we have "seen" five cards PLUS the seven folders' cards, nineteen in all - we know that the latter fourteen cards do not include any that are Aces - (it doesn't matter what they are; only that they don't include any Aces)

[razor]

This is silly...

[John Biggs]

Look, you've got to get a grip here. Go back and look at how I did the calculations. You'll see that the 47-card deck I'm talking about contains the three Aces in question.

If you're interested in probability, I would recommend that you begin doing some reading on the subject. Right now your understanding is primitive in the extreme - misunderstanding would be more like it.

Any good textbook on either statistics or probability will teach you how to do combinations, which are a very useful tool; in addition, a Web search (or even a search of these forums) will turn up many hits on how to calculate poker hands and situations specifically. There is also the "Hold'em Odds Book," available from Amazon, in which the author teaches you combinations and leads you through them exhaustively. The key is postulating the correct scenarios before doing the math, which you have utterly failed to do.

I gather from a previous post that you use a software program to do your calculations. You'll do much better if you ditch the program, learn probability, and do your calculations by hand (or with the help of a spreadsheet) for a while.