Chances of a Royal in Omaha Vs. Hold 'Em

I'd like to sort of re-open a discussion about permutation and odds in relation to the chances of hitting a Royal Flush by the end of a hand in Hold 'Em versus Omaha

How do you calculate this knowing that in Omaha you need to use two hole cards from your hand and three from the middle when you also need to invalidate all of your hands where you have more than 2 hole cards to a royal flush.

That last part is crucial. I can't find calculations online that take the cancelling of those hands into consideration.

My theory is that when you include that in the calculation, the chances of hitting a royal by the end in Omaha may be worse than in Hold em because you can't play the board and you can't hit it when you have more than two cards to a royal in your hand.

For Omaha, I get a probability of 0.000092345 =~ 1 in 10829.

For Texas Hold 'Em I get 0.000032321 =~ 1 in 30940

So, it's easier to get the royal in Omaha.

Anyone agree/disagree?

I calculated the probability of Royals in Omaha and got the same number as Wizard.

Royal combinations in Omaha
= 4*C(5,2)*C(47,2)*C(3,3)*C(45,2)
= 42807600

Total combinations in Omaha
= C(52,4)*C(48,5)
= 463563500400

Probability of royals in Omaha
= 42807600/463563500400
= 0.00009234463
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I did consider a player hand that has two royal cards of one suit and another two royal cards of a second suit, such as AsKsAhKh
but because there cannot be two royal flushes in a total of 9 cards (4 player +5 board), I decided that the trivial calculation that I did above was correct.

To be complete, this is also in agreement with Wizard:

Probability of Royals in Texas Hold'em
= 4*c(47,2)/c(52,7)
= 0.00003232062

Quote: gordonm888

To be complete, this is also in agreement with Wizard:
link to original post

Thank you Gordon for the confirmation on both.

Thanks! I guess I got thrown off by google results that show up first for some reason. Cheers!