#1
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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? |
#2
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Re: Omaha Problem
They did this thread a while back....
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#3
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Re: Omaha Problem
Any idea how I can find it?
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#4
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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 |
#5
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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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