Here's my question: You're sitting on KK. What's the probability that someone at a table of n (remaining) players has AA? I want to leave n up to 8 flexible because, depending on position, reads, etc., there are various players you can exclude even at full table (obviously the folds).

Here's my thinking: If there's only one player, it's easy, just 12/(50*49) = 0.5% = 1/204.2

Now, for 8 players, I don't see an easy way to do this by the old "one minus" method. But it seems to me that it should look something like this:

The probability of the first player having AA is just the result we just saw. Similarly for all 8. So, you could just multiply by 8. But, if you do that, you've counted some situations twice, namely those where 2 players at the table have AA. So, I'll now just call p the probability that a given player has AA on random hands for everyone except me (I have KK).

Now, there are (counting here ordered hands), there are 4! different ways for 2 players to get AA. So, the probability of that happening for 2 GIVEN players is 4!/(50*49*48*47).

And, among n players, there are n!/[(n-2)!*2] permutations of 2 players.

So, for n players, I come up with the following formula:

n*p - {(n!/[(n-2)!*2])*[4!/(50*49*48*47)]}

This yields the following percentages:

8 players: 3.9% = 1/25.6
7 players: 3.4% = 1/29.2
6 players: 2.9% = 1/34.1
5 players: 2.4% = 1/40.9
4 players: 2.0% = 1/51.1
3 players: 1.5% = 1/68.1
2 players: 1.0% = 1/102.1
1 player: 0.5% = 1/204.2

I feel sure this is roughly right, but I'm not completely confident that my method is 100% clean here...

I'm sure you'll find what you are looking for here :
http://www.math.sfu.ca/~alspach/comp35/
it seems a little bit more complicate

I think your numbers are off slightly but close enough to get a general view.