Use the inclusion/exclusion technique: add the individual probabilities of seeing ace, plus of seeing deuce, etc. before seeing a face card. There are of course ten probabilities to add, all of them the same: 4/16 = 1/4; since there are ten of them the total is 10/4 = 5/2.
Then subtract out all the pairwise probabilities of seeing an ace or a deuce, an ace or a trey, etc. before the first face card. Each pairwise probability is 8/20 = 2/5, and there are C(10,2) = 45 of them so the total is 90/5 = 18 to be subtracted out.
Continue to triples, quadruples, ... , any of all ten, alternately adding and subtracting the totals.
let individual how many sum
1 1/4 10 5/2
2 2/5 45 18
3 12/24 120 60
4 16/28 210 120
5 20/32 252 315/2
6 24/36 210 140
7 28/40 120 84
8 32/44 45 360/11
9 36/48 10 15/2
10 40/52 1 10/13
The probability of getting at least one of each numeric denomination before any face card is therefore
5/2  18 + 60  120 + 315/2  140 + 84  360/11 + 15/2  10/13 = 1/286
~= 0.00349650349653097
