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Shaun
09-29-2002, 04:48 AM
hand A: KhQh9hQd
hand B: AcJs10c5s
hand C: Kc5hKsJc

Flop: AhThQs

Who has the best of it when all three go all-in? By How much?

Gabe
09-29-2002, 05:41 AM
A 50%
B 15%
C 35%

irchans
09-29-2002, 11:29 AM
Here is what two dimes says.

twodimes.net results (http://www.twodimes.net/poker/?g=o&amp;b=Ah+Th+Qs&amp;d=&amp;h=Kh+Qh+9h+Qd+%0D%0AAc+Js+Tc+5s %0D%0AKc+5h+Ks+Jc%0D%0A%0D%0A%0D%0A)

09-29-2002, 08:55 PM
Who has the best of it when all three go all-in? By How much? Hand A by a lot.

For a \$1 unit basis per showdown:(based on only 100,000 trials)
hand A: KhQh9hQd : wins \$0.4896 per hand per dollar bet
hand B: AcJs10c5s : loses \$0.5462 per hand per dollar bet
hand C: Kc5hKsJc : wins \$0.0566 per hand per dollar bet

Flop: AhThQs

09-29-2002, 09:31 PM
Dear irchans:

What does EV mean? Please define it. Hand B pays the freight. Hand A win about 8.65 times as much money as hand C. Neglecting ties, I can see that you have to win 1/3 of the time on balance (that is have an EV of .333) to break even for this problem.

[Note: this is a cached result]
Running: pokenum -o kh qh 9h qd - ac js tc 5s - kc 5h ks jc -- ah th qs :
Omaha Hi: 666 enumerated boards containing Qs Ah Th
cards

For NP = 3 players showdown:
Hand EV Winnings per hand per dollar bet
A 0.495 = (EV - 1/NP) * NP = EV/NP - 1 = \$0.486
B 0.151 = (EV - 1/NP) * NP = EV/NP - 1 = -\$0.546
C 0.353 = (EV - 1/NP) * NP = EV/NP - 1 = \$0.059