7-Stud odd

[ChipLeader]

I read that when there are two of a card showing, for example sake we'll say theres 2 aces showing as the doorcard, there is a better chance a player has a pair of aces than if only one is showing.
On this logic I was wondering if there is 3 aces showing, how often will one of these players have a pair of aces?

[Precision1C]

# of aces chance of at least one pair of aces
1 11.53%
2 15.51%
3 12.24%

The way I derived these numbers is by using combinatoric probability of there being no aces amongst the down card and substracting that from one to determine the chances of there being at least one ace among the unknown cards. For example in the case with 3 up aces the chance for no aces amongst the 6 down cards is 1 - (C(48,6)/C(49,6)). Where C(N,K) = N!/(K!*(N-K)!))