PLO probability someone has aces when hero has 1 ace (8 max)
Join Date: Nov 2021
Posts: 1
Hi Guys, what is the probability someone else in the table has aces (8 max) when we are playing PLO and I (hero) have one ace?
Is there an software you use for such calculations?
Join Date: Aug 2011
Posts: 5,081
Method 1: [# of villains] * P(arbitrary villain holds >1 aces) = 7*[45 + 3*C(45,2)]/C(48,4) ≈ 10.85%
(Multiplying by the # of villains works because there can't be two villains simultaneously holding two aces.)
Method 2: [# of valid binary arrangements of the 3 A's] / [total arrangements] = [7*4 + 7*6*C(4,2)*4 + 7*C(4,2)*(48-28)] / C(48,3) = same
Join Date: Jan 2009
Posts: 4,987
I got the same result as heehaw. What I found to be interesting was that the probability one or more opponents get at least one ace is a whopping 93.4%. Hopefully, heehaw or someone else can confirm or correct this result
I don’t know if there is free software that deals with PLO deals, but I would think there are some that solve hypergeometric probability problems, often applicable to this type of question. Certainly, you can use a program like Excel which has various probability functions or advanced calculators that have factorials, which are used to solve combinations.
Join Date: Aug 2011
Posts: 5,081
Re: at least one ace, I get the same result: 1 - C(20,3)/C(48,3) = .9341
Join Date: May 2012
Posts: 4,707
I was gonna give an answer. but more difficult than I thought... hold-em is easy calculation... and someone holding only one ace is easy I think too... the problem with Omaha is you have to keep track of someone else having just one ace. then that changes the odds of others having 2 aces..... oops, I might be totally confused in this.
I'm gonna try to give answer in time, but others will know right away.
