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
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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
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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
