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  #1  
Old 01-23-2007, 07:01 PM
Abramovic Abramovic is offline
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Default Omaha Problem

What is the most amount of outs you can have on the turn while being behind in Omaha?

For example 25 in Holdem, how many in Omaha?
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  #2  
Old 01-23-2007, 07:28 PM
luckyjimm luckyjimm is offline
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Default Re: Omaha Problem

They did this thread a while back....
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  #3  
Old 01-23-2007, 07:29 PM
Abramovic Abramovic is offline
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Default Re: Omaha Problem

Any idea how I can find it?
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  #4  
Old 01-24-2007, 07:08 PM
Silent A Silent A is offline
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Default Re: Omaha Problem

First of all, multi-way, it's possible to have zero outs on the turn while holding the nuts.

best I could come up with quickly for heads-up:

Result

http://twodimes.net/h/?z=2335060
pokenum -o 2c 2s 3s 3c - kh qh 9d 8d -- jh th 6d 5d
Omaha Hi: 40 enumerated boards containing 6d 5d Jh Th
cards win %win lose %lose tie %tie EV
3s 2s 3c 2c 10 25.00 30 75.00 0 0.00 0.250
9d 8d Kh Qh 30 75.00 10 25.00 0 0.00 0.750

30 outs

I'm sure there's something better out there
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  #5  
Old 01-24-2007, 08:30 PM
iggymcfly iggymcfly is offline
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Default Re: Omaha Problem

Just changing the 9 to a 7 on that board will get you two more outs since the nine's already an out from the high open-ender. If you do that in this scenario, the drawing hand has 32 outs and is 80% to win.

I think that's as many outs as you can possibly have on the turn, but I could be wrong. Maybe something where you have two flush draws, an overpair, and a wrap against a low two pair so you can get board pair outs?
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