Total Omaha newbie here:

Situation me: K K Q J and opponent has A 9 6 4 and hits an Ace.

FLOP: 3 A T

I believe I have 8 outs for absolute nuts straight (2 kings, 3 queens, 3 jacks). From trying to apply my Holdem knowledge I would say the percentage I have for winning given all in on flop is 32% requiring pot odds of 2.2 to 1.

There is also a chance of runner runner 9 and 8 for straight to King - but that would be insignificant to my chances?

When I input these details into:
https://www.cardplayer.com/poker-too..****ulator/omaha

It calculates my winning percentage as 37.8% requiring pot odds of 1.6 to one for a call.

MY QUESTION: where does the extra value come from?

Some rough calculations on my part suggests that the fact that the Omaha emulator knows all my opponents cards cannot account for the difference.

PS: I tried including a screenshot from the Omaha emulator to make my question clearer and failed - your help tells me there is an "Manage Attachments" icon at the bottom of posts - I could not find it.

With all cards known, there are 41 possible turns and 40 possible rivers.

The nut straight has (8/41) + (33/41)*(8/40) = 35.61%.
The backdoor straight, since a 9 is dead, has (4/41)*(3/40) + (3/41)*(4/40) = 1.46%.
You can backdoor a boat with running tens or threes, (3/41)*(2/40), times two for each = 0.73%.

Add that all up and you get your 37.8%.

Hi Rei: thank you very much for that - thats pretty clear.