Probability of a RF in Omaha

[BigBluffer]

This was asked a little while ago in B&M and the OP was chastised for posting in the wrong forum. [img]/images/graemlins/smile.gif[/img]

Playing Omaha, given that you are dealt 2 to a royal, what are the odds of completing the royal?

Also, before the deal, what are the odd of making a royal in Omaha? Clearly, these odds will be much higher than in the first situation.

[BruceZ]

[ QUOTE ]
Playing Omaha, given that you are dealt 2 to a royal, what are the odds of completing the royal?

[/ QUOTE ]

C(45,2)/C(48,5) = 1,728.6-to-1.

That is C(45,2) ways to choose the 2 board cards that are not part of the royal, divided by C(48,5) total possible 5 card boards. This assumes that neither of your other 2 hole cards are part of this royal, nor both part of another royal.


[ QUOTE ]
Also, before the deal, what are the odd of making a royal in Omaha? Clearly, these odds will be much higher than in the first situation.

[/ QUOTE ]

4*C(47,4)*C(5,2)*C(4,2)/C(52,4)/C(48,5) = 10,828-to-1.

That is, 4 different 5 card royals, times C(47,4) ways to choose the other 4 cards, times C(5,2) ways to choose which 2 royal cards are hole cards, times C(4,2) ways to choose which 2 non-royal cards are hole cards, divided by C(52,4)*C(48,5) total possible ways to get 4 hole cards and 5 board cards.

[BigBluffer]

Wow! Awesome! Thank you.

It might take me the rest of the day to remember my college stats enough to thoroughly digest your reply. [img]/images/graemlins/smile.gif[/img]