odds of two people getting dealt AA?

[all_in15111]

This situation came up in a game i was playing yesterday, and i was just wondering if anybody knew the odds of that happening?

[gaming_mouse]

(10 choose 2)*(6/(52 choose 2))*(1/(50 choose 2))

About 1 in 6016

[all_in15111]

thanks alot

[Siegmund]

It's one of those rare things that really does happen every so often.

I hadn't ever seen it until this past week - when it happened in two consecutive sessions of B&M play! The second time I was involved - and still made a nice tidy profit on the hand. The final pot was split, of course, but we had a 5-way capped preflop, thanks to some limpers who got taken for a nasty ride by me raising from the small blind and the other guy limp-reraising from sixth or seventh seat.

[billyjex]

Hey gaming mouse, here's one for ya:

I saw AA vs. AA today in a single table qualifier to the million guaranteed this weekend (which I won /images/graemlins/grin.gif)

Anyways, what are the odds of AA vs. AA being dealt, and one of the A's making a four flush to take the pot?

Sucked for one guy.

[gaming_mouse]

[ QUOTE ]
Hey gaming mouse, here's one for ya:

I saw AA vs. AA today in a single table qualifier to the million guaranteed this weekend (which I won /images/graemlins/grin.gif)

Anyways, what are the odds of AA vs. AA being dealt, and one of the A's making a four flush to take the pot?

Sucked for one guy.

[/ QUOTE ]

That's awesome. I hope you rubbed that all over his face.

Anyway, let's do 4 or 5 flush, since either has the same effect.

4*((12 choose 4)*36 + (12 choose 5))/(48 choose 5) = 0.0434782609

Now multiply that by the original answer:

About 1 in 138368.

Nice work.

[WantedMan]

So gamingmouse...you are my mathmatical god...wow...just wow
i got a long way to go
/images/graemlins/confused.gif

[gamble4pro]

If you are dealt AA, the odds of someone else being dealt AA are n/C(50,2) = n/1225, where n is the number of your opponents.
The odds of two people being dealt AA (from outside view, i.e. not taking into account any hole cards) are C(n,2)*C(4,2)/(C(50,2)*C(48,2))

[gamble4pro]

Sorry, the last formula is C(n,2)*C(4,2)/(C(52,2)*C(50,2))