How to determine the probability of your opponents hand?

[wuarhg]

Say I have A/images/graemlins/spade.gifQ/images/graemlins/spade.gif in MP, folded to me and I raise. A Tight smart player 3-bets on the button (16/7) and I call.

Flop comes A/images/graemlins/diamond.gif K/images/graemlins/diamond.gif 2/images/graemlins/club.gif

Now what goes through my head is what are his possible holdings? He could have anything from AA-99(Maybe even down to 77-66 but that's irrelevant) or AK. I beat QQ-99 at this point.

I bet, He raises...
I've seen people do calculations on the probability that he has AA,KK or AK. How exactly is this done? I've tried reading in TOP but I need clarificiation. Does anyone care to explain? /images/graemlins/smirk.gif

[J.R.]

Well, there are 6 ways to make a pocket pair (AA or KK) and 16 ways to make other card combos, included suited combos (AK). Just discount these numbers given the card you know he doesn't have.

There are 2 aces and one K accounted for, which you know your opponenet doesn't have. So there is one way to make AA, 3 ways to make KK, and 6 ways to make AK.



BTW, "(Maybe even down to 77-66 but that's irrelevant)"- not likely at all if his pfr is around 7% (basically AA-TT (30 hands), AK(s)-AJ(s)(offsuit and suited combos)(48 hands), ATs (4 hands), KQ(s) (16 hands). This is 98 hand combos, which is a bit more than 7% (about 93 hands). Then consider most people don't 3-bet wth half as many hands as they raise with, and you have a much narrower hand range. Sure, people do anomolous things, but TAGs don't 3-bet other TAGs with middle pairs very often.

[marand]

There are 6 ways for each pocket pair. When 1 of the cards is out, there are only 3 ways left. If 2 cards are out of that rank he can only have that pair 1 way.

You know that 2 aces are out. (You have one and one is on the board.)
1 king is out and 1 queen is out.

Possible holdings of your opponent:
Hand, Combinations
AA 1
KK 3
QQ 3
JJ 6
TT 6
(He will probably not raise with a smaller pair? Depends on how tight he thinks you are?)

Since 2 Aces are out and 1 king is out he can have AK for a total of 2*3 = 6 combinations. (2 aces left and 3 kings left)

So hands that beat you:
AA 1, KK 3, AK 6 == total of 10 hands

Hands that you beat:
QQ 3, JJ 6, TT 6 == 15 combinations

So from this calculation it seems like you are likely ahead after the flop.

But, the question is if he would raise you on the flop if he has a hand that you can beat... will he raise you on the flop with a hand like QQ?

[wuarhg]

Great replies, this clears it up a bit.

Say I call the flop raise and the turn brings T/images/graemlins/heart.gif
I check and he bets. There is now about 5.75BB in the pot. I just picked up a gutshot. I have additional 4 outs to the nuts and I might still have the best hand. He might also have made a set with TT.

So the likelyhood that i'm behind:
AA 1
KK 3
AK 6
TT 3
Total of 13 ways.

And likelyhood I am ahead decreases somewhat.
QQ 3 ways
JJ 6 ways
Total of 9 ways.

so now i'm 1.4 to 1 to be ahead, correct? This means i'm ahead around 30% of the time(1.4 in 2 that i'm behind, I can't seem to calculate this correctly)? and i'm getting 5.75 to 1 pot odds. This is not sufficient based on pot odds alone but how do you guys count this when there's a 30% chance (if this is correct) that you're ahead? Do you add it to the pot odds something like 5.75 * 1.3 (if there's a likelyhood that you're ahead 30% of the time)?