Probability of making a full house

[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.

[TomCollins]

You aren't accounting for discards. Remember, yiour discards aren't shuffled back in, so you can be sure you won't draw them.

So there are 9 ranks with 4 cards left, and 3 ranks with only 3 cards left.

[SwissPoker]

Thank you very much for your help.

I corrected the calculation ...

Total of combinations: C(47,3) = 16'215

9 ranks with 4 cards left:
9 * C(4,3) = 9 * 4 = 36
9 * C(4,2) * 2[aces] = 9 * 6 * 2 = 108

3 ranks with 3 cards left:
3 * C(3,3) = 3 * 1 = 3
3 * C(3,2) = 3 * 3 * 2[aces] = 18

36+108+3+18 = 165 combinations of a total of 16'215 combinations make the full house.

16'215/165 = 98.3

= 97.3-to-1 or about 1.02%