Question:
A deck of 52 cards, 2 cards are dealt. We assume that the card we have
in our hand are not a pair.
For example we get: '3 of hearts' and '2 of clubs'.

What is the probability that if you draw 3 cards at random from the remaining 50, you'll get at least another 2 and at least another 3? In other words: having two 'not pair' cards, what is the probability of making at least two pairs (or full house) if you draw 3 cards from a deck of the remaining 50 cards?

What I did so far:
2 Cards, unpaired, are now in our hands. 50 cards remaining.
Currently 6 cards can make one pair. So when the first card hits the
table, the chance of 'not making' a pair is:
(50-6)/50

When the second card hits the table (that becomes the 4th visible card
in total), 49 cards are remaining, but only 3 of these can make a
pair. So when the second card is placed on the table, the chance of
not making a double pair is:
(49-3)/49

Now what confuses me is: I could multiply (50-6)/50 *
(49-3)/49 = 2024/2450
Then 1 - (2024/2450) would give me the chance on having (at least a
double pair): 426/2450... this would mean that once out of 5.75
times would result in a double pair... this looks pretty high and this
is only after 2 cards have been dealt... one is remaining, so I'm
wondering: What did I overlook?

What do I do with the third card? How do I include this in this
calculation?
Keep in mind: we need *at least* (in this example) a 3 and a 2 on the
table. If on the table would appear: 3-3-2, this should be included in
the probability.