Probalility of a set when holding a pair.

[VivaLaViking]

Greetings,

I am verifying the integrity of the two major online gaming sites and I will share the software I have written and the results of my findings.

I want to verify my calculations first. My question is if you are dealt a pair of down cards what percentage of flops will result in a set without quads or a full house?

My calculation uses combinatorial math written as (x C y) stated as; choose y cards from x choices.

To continue:
To choose the set only tho cards remain, (2 C 1).
Any card must not result in quads, (48 C 1).
Any card must not result in quads or a full house, (44 C 1).
Total number of flops while holding a pair is (50 C 3).

(2 C 1)(48 C 1)(44 C 1) = 4,224
(50 C 3) = 19,600

4,224 / 19,600 = 21.55%

Can anyone verify this result?

TY