Not sure if this is the right forum to ask this, but how do I calculate the odds of AA vs KK on a flop of AAK being dealt in a heads-up hand?

I saw this ridiculous hand in a ridiculous movie over the weekend, and I want to quantify just how unlikely this is.

There is more than one answer to this kind of question, because it's not really fully specified.

For example, one answer is "how likely is this to happen on the next card dealt"

Another might be "What's the chance this would happen at least once if you played X hands"

Also the question is how many people at the table - does it have to happen to you and person B or just any 2 people at the table?

Another variation might be, does it have to be AA and KK? Isn't AA and QQ about the same thing, or 99 and 66 etc?

Something close to this happened at the WSOP a few years ago, Sammy Farha had AT and another guy had TT on a AAT board. First hand of the WSOP, they got it all in.

I remember a couple years ago seeing a WSOP broadcast where 77 and 76 got it in on a 766 flop. Dirty. I have no idea what the odds of this are, no matter how you ask it.

So in the movie, it was a heads up hand, and it was the first and only hand that the players played. And it was obviously a cooler hand for dramatic purposes. So I guess my question is what are the odds of this specific situation happening in one hand?

Here's my attempt at the math, could somebody see if it's correct reasoning?

Player A being dealt AA:
4/52 * 3/51 = 1/221

Player B being dealt KK:
4/50 * 3/49 = 6/1225

Flop being AAK:
(1/48 * 1/47 * 2/46) * 6 = 1/8648

Is my flop calculation correct? I'm treating each Ace individually whereas there are two remaining Kings, then multiplying by 6 since it doesn't matter what order the flop comes out.

If those odds are correct, I'd multiply them together for:

1/221 * 6/1225 * 1/8648 = 1/390,204,967

Does that look right?

Odds of getting AA vs KK dealt HU = 2!*(4/52)*(3/51)*(4/50)*(3/49)

Odds of flopping AAK after = 3!*(1/48)*(1/47)*(2/46)

2*(4/52)*(3/51)*(4/50)*(3/49)*6*(1/48)*(1/47)*(2/46) = 3/585307450

So about 1 in 200 million.

Juk

Why the 2! for the AA vs KK HU?

I just assumed either player could get dealt the AA/KK hand.

Juk

Hmm k. Thanks!

Actually you were correct! I counted the A in both spots then permuted it again

Another way to get the value is: of the 48C3 possible flops only 2 can contain AAK (ie: each using both available As and either K) and 48C3/2 = 17296/2 = 8648

Juk

Nice. Thanks!