Cryptologic
Pot Limit Omaha Ring game
Blinds: $0.15/$0.25
9 players
Converter

Stack sizes:
UTG: $25.25
UTG+1: $17.80
MP1: $23.25
MP2: $38.25
MP3: $6.30
CO: $105.65
Hero: $21.05
SB: $28.05
BB: $12.60

Pre-flop: (9 players) Hero is Button with 9 Q T 8
UTG folds, UTG+1 calls, 2 folds, MP3 calls, CO calls, Hero calls, SB folds, BB checks.

Flop: J 7 A ($1.4, 5 players)
BB checks, UTG+1 bets $1, MP3 calls, CO calls, Hero calls, BB folds.

Turn: 2 ($5.4, 4 players)
UTG+1 bets $4, 2 folds, Hero calls.



Results:
Final pot: $38.5

I knew if i hit he wasn't going to lay down his set.

So on turn i counted his 12$ left as in the pot as well.

Now,

Is this a ridiculous logic to justify this call or is it sensible?

Looks fine. 8, 9, T or K = 13 outs. 11 outs to the nuts.

You're actually getting correct pot odds to call without any implied odds.

Am i?

On turn 13 outs if he doesn't have hearts. If he have hearts i have 9 outs.

13 outs = %26 on turn right?

I am getting a little more than 2 to 1.

This still confuses me a lot, Can you show the math in a clear way please?

Sorry, you have 10 outs to the nuts, not 11.

10/44 = 22.7% <-- You will hit the nuts this % of the time.
13/44 = 29.5% <-- You will hit a straight this % of the time.


To call the turn you are putting $4 into $5.4 + $4 = $9.4. To calculate what pot odds you need you use Pot size/Bet size.

(9.4 + 4)/4 = 3.35.

1/3.35 = 0.2985 = 29.9%. So, if you assume any straight will win, regardless of flushes, you are getting pretty much exact pot odds to call the turn.

If you need to hit the nuts, you will need some implied odds, so working backwards from the above calculation:

22.7% = 0.227 = 1/4.4

4.4 = (9.4 + 4 + X)/4 where X is the extra amount you need off him on the river

17.6 = 9.4 + 4 + X

4.2 = X


You need to make $4.20 more on the river if you lose every time you don't hit the nuts.

It's all clear now.

Thanks a lot mate.