I have a hard time wrapping my brain around Demi's method. I like counting methods better, so I'm going to throw out a second way to solve the problem (although it's more complicated than I initially thought).

Terminology:
x! = x(x-1)(x-2)...(2)(1). (a.k.a. factorial)
5! = 5 * 4 * 3 * 2 * 1 = 120

x!! = x(x-2)...(3)(1). (a.k.a double factorial, will always be odd for us)
5! = 5 * 3 * 1 = 15

(x, y) = x! / y!(x-y)! (a.k.a combination or unordered permutation)

(4, 2) = 4!/(2! * (4-2)!) = 24/(2*2) = 6
in words, choose 2 out of 4 (e.g. , , , )
, , , , ,
for 6 pairs (note: = )

How many ways to deal AA, AA, KK?
How many ways to deal 10 people cards?
(52, 20) is the total number of ways to pick 20 cards out of a full deck
(20-1)!! is the number of ways to split these 20 cards into 10 hands

Multiply for the total number of hands:
(52, 20) * (20-1)!! = 8.2492 * 10^22
(this is the denominator for our final probability)

Now for the numerator:
There are 3 ways to get 2 pairs of As, (, ), (, ), (, )

There are (4, 2) = 6 ways to get 1 pair of Ks

We have used 6 cards, now we need 7 more hands:
(46, 14) pick 14 cards from the remaining 46 cards
(14-1)!! number of ways to split these cards into hands

(46, 14) * (14 - 1)!! = 3.2416 * 10^16

For the numerator we combine these three values for the total number of deals with 2 pairs of As, and 1 pair of Ks:
3 * 6 * 3.2416 * 10 ^ 16 = 5.8349 * 10^17

(we could remove the hands that contain AA, AA, KK, KK, but that is ~0.1% so I'm going to ignore it)

Total Probability:
AA, AA, KK deals divided by Total number of deals
5.8349 * 10 ^ 17 / 8.2492 * 10^22 = 7.0733 * 10 ^ -6 or
0.000 007 073

Probabilities to odds (see Demi's post):
0.999 992 927 to 0.000 007 073
141,381 to 1

Demi: this is off by a factor of 2 from yours. Either I made a mistake, or your calculations don't account for swapping the A pair hands. If that's the reason this is an example of why I like counting, I'm less likely to miss special cases.