Omaha 8 PL hand

[Shaun]

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]

A 50%
B 15%
C 35%

[irchans]

Here is what two dimes says.

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

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

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

these are good answers....

[irchans]

Dear Carl,

TwoDimes is defining EV to be (win%)+ 1/2*(tie%).

I assume the EV means "expected value". One way that (win%)+ 1/2*(tie%) could be expected value is the following game. Three players walk up to a table with $1.00 in the center. They are given the hands 'kh qh 9h qd', 'ac js tc 5s', and 'kc 5h ks jc'. The cards 'ah th qs' are placed in the middle. Now the players agree to give the $1.00 to the best omaha hand after dealing 2 more cards. If there is a tie, the will split the $1.00 equally between the ties.

Does that answer your question?