PL Omaha hand probability

I am in a tournament with KQJ6 double suited on BB and get free play with 1 limper. Flop comes A-10-7 with two spades giving me nut flush draw and a straight draw as well. I end up getting all-in on flop and make my flush to win. My opponent had A-10-7-4 for three pair and no spades in his hand. I am wondering- by how much is he ahead, if at all?

[Guy McSucker]

Great! an opportunity for me to use the hand evaluation software I just got hold of...

According to this software, which I may be using incorrectly,
of the 820 possible turn/river combinations, you'll win 411 and lose 409!

Seems like it's quite close... /forums/images/icons/wink.gif

Guy.

OK, my calculation is hand made so their might be some misstakes.
You have 4 ways to win, a flush, the nut straight,
a backdoor low straight (8-9, 9-8) and a backdoor set of sixes.

Two assumptions made, you don´t have the six of spades
(one more straight card for you) and your opponent do not have a backdoor flushdraw.

With this setup you win 862 times out of 1640 or 52,6%.


Please, feel free to correct me if I´m wrong.

Purple Haze

download calculator from:
http://koti.mbnet.fi/jraevaar/pokercalculator/

[Ignatius]

Your results are correct for the absolute worst-case of minimum str8-flush possibilities and his opponent having a completly live backdoor flush.
<pre><font class="small">code:</font><hr>
$ pp KsQsJc6c AhTd7d4h -b AsTh7s
Omaha, 2-handed, pot 2, cost 1, board As-Th-7s, full deck, 820 boards
Ks-Qs-Jc-6c: 50.12% 1:1 (EV +0.00) 411 wins 0 splits 409 losses
Ah-Td-7d-4h: 49.88% 1:1 (EV -0.00) 409 wins 0 splits 411 losses
</pre><hr>
In the best case he could be as much as a 5:4 favorite
<pre><font class="small">code:</font><hr>
Omaha, 2-handed, pot 2, cost 1, board As-Ts-7h, full deck, 820 boards
Kh-Qs-Js-6h: 55.12% 5:4 (EV +0.10) 452 wins 0 splits 368 losses
Ac-Td-7d-4c: 44.88% 4:5 (EV -0.10) 368 wins 0 splits 452 losses
</pre><hr>
cu

Ignatius