With AK, what are the chances someone has KK or AA

I stumbled across this and did some calculations, which I will explain. Question 2 appears to be a bit harder - I haven't finished that one yet.

I will now calculate the probability that exactly one player holds KK (or AA), and noone else hold the other top pocket pair.

Assuming there are n players on the table and you are holding AK. Then there's 50 cards left.

There are (n-1) players who could hold KK. There's 6 different combinations for KK in this situation since the hero holds one already. There's 48! other combinations of cards the other players could have.

This means there are (n-1) * 6 * 48! combinations in which one player holds KK. But from 48! we need to subtract the combinations in which AA appears.

It becomes (n-1) * 6 * (48! - (n-2) * 6 * 46!) - all possible combinations that have KK but not AA. Dividing by 50! gives the probability we are looking for.

And we get: Link The probability for n players that exactly one holds KK and noone AA (and the Hero AK). So, for 10 players (including the hero), the answer is appr. 2.15%. Link.

So the probability that at one other player holds KK or AA and noone else has the other top pocket pair, is twice as much: Appr. 4.3%

If the original question was asking for the probability that at least one player holds KK or AA, then simply add the probability that both KK and AA have been dealt to the other players (see below), leading to appr. 4.36%.

The probability that one other players hold KK and another AA is this Link

Edits: Some corrections.