2 Royal flushes in 1000 hands

[PokerPimp]

Hi i having been searching for the odds of hitting a royal flush in texas holdem. All i can find is 650k-1 but it seems thats just for 5 cards stud. What i am wondering is what are the odds for 7 cards. I dont expect someone to figure out the odds of hitting 2 within 1000 hands of eachother but i would just like to know the odds of hitting ONE lol i got 100x the bb bonus so that was 50$ bonus both times which wasnt to shabby. If you could help me out i was just curious thanks guys.

cheers

[Silent A]

This isn't an exact way to calculate this, but it'll be in the right ball park:

On the river, there will be 28 different ways you can make a 5 card hand in hold'em.

SO the odds of getting a Royal Flush in any given hand is approximately:

1 in 649,740/28 = 1 in 23,205 (assuming you always see the river)

The probability that you won't get any royal flush's in a given 1000 hands is:

(23204/23205)^1000 = 0.957820307

Which means the odds of getting at least one royal is about 1 in 23.7.

The probability of getting exactly 1 royal in any 1000 hand span is:

1000*(1/23205)*(23204/23205)^999 = 0.041278241

So the probability of getting 2 or more royal flushes in a 1000 hand span is:

1 - 0.957820307 - 0.041278241 = 0.000901452, or 1 in 1109

So you'd expect to play in the ballpark of 1 million hands of Hold'Em before you's ever get two royals in a 1000 hand stretch.

[BogusPomp]

The probability of seeing a royal flush after 7 cards is

(4 * 47 * 46 )/ (52 choose 7) = ~0.00006464124 or about 1 in 15 470

To find the chance of this happening exactly twice in 1000 hands, use EXCEL BINOMDIST (2,1000,0.00006464,false) = 0.001956684

[BruceZ]

[ QUOTE ]
The probability of seeing a royal flush after 7 cards is

(4 * 47 * 46 )/ (52 choose 7) = ~0.00006464124 or about 1 in 15 470

To find the chance of this happening exactly twice in 1000 hands, use EXCEL BINOMDIST (2,1000,0.00006464,false) = 0.001956684

[/ QUOTE ]

That needs to be P = (4*47*46/2)/C(52,7) = 4*C(47,2)/C(52,7) =~ 0.00003232062
or 1 in 30,940. Then the probability of exactly 2 in 1000 is C(1000,2)*P^2*(1-P)^998 or BINOMDIST(2,1000,P,FALSE) =~ 0.000505226 or 1 in 1,979.

If you must use both hole cards, then the probability of a royal is only P = 4*C(5,2)*C(47,2)/C(52,2)/C(47,5) =~ 0.00002125857 or 1 in 47,040. The probability of exactly 2 in 1000 is C(1000,2)*P^2*(1-P)^998 or BINOMDIST(2,1000,P,FALSE) =~ 0.0002209986 or 1 in 4,525.

[lgas]

I don't know but I got two royal flushes last night on poker stars within 1600 hands. That is all.