Playing tourney Omaha hi/lo split last night, I hit a Royal Flush on the flop, we were 9 handed at the time and I'm wondering if anyone out there knows the odds on that happening
In Omaha you must play exactly two cards from you hole cards
First you need the two royal cards in your hole cards. You don't want a 3rd as it blocks the royal.
You could technically have two royal draws in your 4 cards
First look at the case of a single 2 royal in you hand and other two card are not blockers or a draw to another nut flush.
combin(52,4) = 27,0725 possible starting hole cards
combin(5,2) = 10 ways to get 2 of the nut flush
combin(4,1) = 4 suite
combin(47,2) = 1081 other card combination that do not block your royal
Royal hole card draw chance = 8% (higher than I thought it would be)
Now need exactly one board to make the royal on the flop
combin(48,3) = 17,296
Flop a royal = 0.006%
Hole x flop odd = 0.001% = 1 / 108,290 royal flop in Omaha
The chance of two royal draws is 0.44% and then you double the flop odds for a total of 1 / 1,951,025
Flop a royal in hole'em = 1 / 649,740
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Same as other answer. I thought it might be wrong so looked at it from another perspective and came to the same number. – paparazzo Apr 19 '17 at 18:22
The number of different poker hands is '52 over 5' which is 2598960. Four of them are royal flushes, so the chance of a poker hand to be a royal flush is 649740. Because you have four holes, you can make six different hands, so the odds of this happening are about equal to 6 in 649740, so 1 in 108290.
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I was confused over how to handle that as there is now 7 cards out with the 3 flop plus 4 hole cards. But that kind of makes sense. – paparazzo Apr 17 '17 at 0:47