[jbrennen]

I think you played it perfectly.

Assume he had two unpaired wheel cards in the hole at the start, and that he wouldn't bet 4th the way he did with a 4 in the hole. So his hole cards consist of two of: A, 2, 3, 5. If you enumerate all of the possibilities, there are 20 different ways he could have an Ace under out of 52 different hands. So figure 5/13 chance that he paired his Ace.

So 5/13 of the time, you've got a live draw to win.

Assume that you'll only call a river bet if you get a Ten or lower on the river which doesn't pair you. That leaves 18 cards out of 37 unseen where you'll put in 2 bets. Let's just say that 1/2 of the time you'll call a river bet.

So your expected outlay on 6th and 7th is 1.5 big bets. If you catch your card on the river, and if the Ace paired your opponent, you're probably about a 60:40 favorite (that's just rough, but probably close).

So your expected return from the pot is equal to 15.60 big bets (ending pot size), times 1/2 (chance that you catch your river card), times 5/13 (chance that your draw was live), times 6/10 (chance that your made live draw holds up to win). That's 1.8 big bets.

So I'm pretty sure that the call on 6th is correct, with an EV of about +0.3 big bets.



If you expand Villain's hole card range to two cards 6 or under, calling 6th becomes -EV, but if you expand it again to two cards 7 or under, calling becomes +EV again (because he might have paired 5th street). I see no reason to assign Villain the exact very specific range of hole cards that makes our call -EV. Nothing else in his play indicates that he would be willing to play (A-6)-J but not (A-7)-J.