Royal vs Royal

[TheBlackKing]

My Wife and I decided to play 17 card chinese poker (3 card hand, 4 card Badugi hand, 5 card 2-7 hand, 5 card poker hand). Loser has to take out the trash.
The very first hand we play I check my cards and see a royal flush. So I start trashtalking her that I would win the bottom 100% sure see just sits and looks at me smiling.
Long story short we both have a royal (but mine was in spades) she had to take the trash out.
My question is: if you make 2 piles of 17 cards from a deck of 52 How big are the odds of royal vs royal ?
my poker sense says around one in 100.000.

Regards BK

[Tom1975]

Ok, I've taken two stabs at this and it's turning out to be much harder than I first thought. That, or there's a simple way to solve this that I'm missing. I get the odds of one player being dealt a royal as 4*C(47,12)/C(52,17)=0.00952381 but I don't know where to go from here.

I also calculated the number of ways to select 34 cards which contain A-10 of 2 suits, and then counting the number of ways to divide them into seperate piles where each contains a royal. The problem I see is that it's possible for there to be 3 or 4 royals out there and I'm not sure how to take this into account.

[Siegmund]

I would start by imagining a deck with three "suits", the five top spades, the five top hearts, and the 42 other cards. The chance that Player 1 gets a spade royal and Player 2 gets a heart royal:

Player 1 gets 5, 0, and 12 of these; player 2 gets 0, 5, and 12; remaining in the pack are 18 plain cards.

1*1*42C12 / (52C17) * 1*1*30C12 / (35C17)

~ (1/1984) * (1/52.46) ~ 1/104084.

Now, Player 1 can actually have any of four suits, and player 2 any of the remaining three suits: so multiply by 12: approximately 1 in 8674.

The above double-counts hands where someone has two royals at the same time, but there are very few such hands - you could take the time to calculate it, but the final answer will be in the neighborhood of 1 in 8700 still.

[bigpooch]

I get 12*[C(42,12)/C(52,17)]*[C(30,12)/C(35,17)] =
0.000115258821.

For two royal flushes for the first player and one for the
second:

C(4,2)*2*[C(37,7)/C(52,17)]*[C(30,12)/C(35,17)]=
0.0000001073098.

The second player could have two royal flushes and the first
only one, so for someone to have two royal flushes, the
probability is roughly double the above number (forgetting
about both players having two royal flushes).

Then, the approximate probability for two players to have a
royal flush is about 0.000115044201 or about 8691.3 to 1
against, so it's not as long as one might think.

[TheBlackKing]

Thank you guys very much.
Even though 1 : 9000 isnt really rare we still laugh about it.
Regards,

BK

[infinity235]

You used suits as a tiebreaker??