the "What the [censored] am I trying to ask here?" post.

[MyPTBB100isPOOP]

Really may be way off base here, but oh well.

Lest assume that on a board of AcKcQdJd9h villian could only have a set/air and we have 22.
On river villian bets 100 into 100 pot, so 33% of his range needs to be a bluff for us to breakeven.
If we plug the sets into pokerstove, we see that they make up 2.3% of all hands.
I assume this means we need to find an additional 1.15% hands he would bluff with to call here.
So we decide he is betting: 76s,65s,54s,43s and now 33% of his hands are bluffs, we can call.
Am i right so far?


This is where i become greatly confused: Which thought is more accurate?

Because 1 of each of the set cards is out on the board, he can only actually have 3 combinations of each pair. Giving him 15 possible hands that arent bluffs. Meaning we need to find 7.5 different hands that he would bet with as a bluff.

OR

Because he can only actually be holding any one combination at a time, he can only have 5 possible hands that arent bluffs. Meaning we need to find 2.5 different hands that he woudl bet with as a bluff.

I was going to expand on this and ask another question, which was basically the same thoughts but in relation to the suited cards, and off suited cards that he would be bluffing with, but i feel that I may be way out in left field on this one, and want to see if i can get some response B4 posting any further. Is it crazy that I am even thinking about this process? If not am i going about it the wrong way?

[jay_shark]

Correct .

If 15 hands are value bets , then villain should bluff with about 7.5 hands . Obviously villain would choose 7 or 8 hands since he cannot bluff with 7.5 hands . Also , since he is giving you 2:1 odds , you are indifferent to calling him with your 2's . He gains by bluffing but keep in mind that playing optimal game theory strategy is not always in your best interest .

If you would fold 1 out of 5 times even though he's giving you 2:1 odds , then he should bluff with more hands and vice versa .

[holdem2000]

[ QUOTE ]
Because 1 of each of the set cards is out on the board, he can only actually have 3 combinations of each pair. Giving him 15 possible hands that arent bluffs. Meaning we need to find 7.5 different hands that he would bet with as a bluff.

[/ QUOTE ]

This one is most correct... the first method you mentioned is off because given the current board the starting hands which led to sets are no longer 2.3% of all hands, in the same way that on a flop of AAA the starting hand AA is not the opponent's hand 1/221st of the time. The last method you mention is wrong because all hands are not equally likely. There are only 3 ways to have AA on your board but 12 ways to have 73o.

The last caveat to the approach you outline is referred to as Bayesian analysis, and I recommend reading Mathematics of Poker for an extensive treatment (I think in part I). The intuitive idea is as follows:

In your example, it is true that there are 15 combinations of hands that make sets, and in some sense we do want to find 7.5 hand with which he bluffs in order for a call to be correct. Say that we found a set of 7.5 hands that he bluffs with... what if he only bluffs with each of these hands 1/4 of the time he has them? Then instead of needing 7.5 we would need 30 hands that he bluffs with (because he only bluffs 30 * 1/4 = 7.5 times when he gets those 30 hands). But his bluffing frequency can be confounded in other ways: say that we decide that the only hand he bluffs with is 73o, and that he bluffs every time in this spot when he has 73o. Since there are 12 ways he can have 73o this at first may seem like a call. But we have more information that we've been ignoring so far: the play from the previous streets on the hand (or really any other information we have at this point, including timing tells, physical tells, etc.). What if he had 3-bet you preflop, and you were 100% certain that he wouldn't have done this with 73o? Then in fact there zero combinations of 73o that he could be holding rather than 12. Or, if we think that half the time that he was dealt 73o he would have played it exactly as he actually did play the hand before the river, then there are effectively 6 possible combinations of 73o he could be holding (of the 12 possible original combinations, he would only play 6 as we have seen him play so far). Again, we may at this point think we're done, there are 15 set combinations and only 6 effective bluff combinations: we're not getting 2:1... but we have one more thing to consider: we have to reduce from normal combinations to 'effective' combinations not just of his bluff hands but also of his sets. Say first that we think all of his set hands would be played as he did preflop, but that the flop and turn had been checked by both players, and we think he would only play a set this way 80% of the time. Then the effective number of set combinations he can hold at this point is 15 * .8 = 12. So we believe that effectively he has 12 possible combinations of cards which make a set for him, and effectively 6 combinations of bluffs, so as in your original assessment we're ahead exactly often enough to call if we wish getting 2:1 odds.


Even something as simple as evaluating how often your opponent is bluffing can be pretty complicated. In a real game, you'd want some sort of heuristic to reduce your calculations, if you could directly apply this reasoning at all. More important probably is making the general concept intuitive after enough play and hand analysis.