[SwissPoker]

I was wondering how Mike Caro arrives at 97.3 to 1 to a full house in 5draw (no joker) when drawing 3 to 2 aces?

I tried to figure it like that:

Since I keep two aces 50 cards are left. Thus there are C(50,3) = 19'600 combinations of 3 cards out of 50.

First, I calculated the 3-of-a-kind combinations ...
12 ranks * C(4,3) = 48

Second, I calculated the one pair + aces combinations ...

12 ranks * C(4,2) * 2 aces = 144

Added 144+48 = 192

That is, 192 combinations of a total of 19'600 combinations make my full house.

19'600/192 = 102.08

= 101.08-to-1 or about 1%

Am I on the right track or what am I doing wrong?
Thanks for any help.