This is an awfully tough situation. Thanks for posting it. Here is why I would have trouble throwing this hand away: I count nine hands we are currently behind: AA, KK, JJ, 99, AK, AJ, A9, KJ, and QT. The nines and the QT seem less likely given the play of the hand, but I will includ them to be conservative. Given that you hold AQ, I count 49 combinations of cards that would give an opponent one of these hands. This is out of 46*45 = 2,070 possible hands they could have been dealt (considering that we know 6 cards they don't have). 49/2070 = 2.37%, so if we ignored the play of the hand, there is a 2.37% chance that any individual opponent was dealt a hand that is beating you, or collectively an 9.47% chance that one of your opponents was dealt one of these hands.

Obviously, all of the betting makes it more likely that someone has one, but assuming you will call one bet on the river if you call two on the turn, you are getting 5 to 1 to call. For simplicity let's assume all of the outs cancel each other (i.e. a third flush card could sink you, but a Q is an out against AJ or A9, a T is an out against most of the other hands, etc.). Then the question is, is there a 1 in 6 chance that we're ahead?

Even if we assume that the action indicates that it is eight times more likely than a random distribution of hands would dictate that someone holds a hand that is beating us, that leaves us with a 75.7% chance we are behind, so we should still call. Given that the short stack may be tilting a bit and the other guy is pretty aggressive, I think I would end up calling, but I wouldn't like it.

Edit: I think maybe my math is all wrong. Please ignore. [img]/images/graemlins/blush.gif[/img]