I checked the related questions How many ways to get at least one pair in a seven card hand? and Dealing a 5 card hand with exactly 1 pair, but didn't see an answer to my precise question.
It's easy to count the number of 7-card hands from a 52-card pack with at least one pair or triplet or 4-of-a-kind (quadruplet?): $C(52,7)-4^7C(13,7)$, with the latter term being the number of hands that have no repeated ranks. The fraction of hands with at least one such multiplet is $508027\,/\,643195$ or about $0.790$.
I want the number of hands that have at least one pair that is not also a triplet and not also a quadruplet. Empirically, the fraction of such hands is about $0.738$.
My unsuccessful attempt at a theoretical calculation goes as follows:
There are $52$ choices for the first card. To complete the mandatory pair, take one of the three remaining cards of the same rank and different suits (three choices), and eliminate the remaining two of the same rank, leaving $48$ cards. Draw the remaining five cards in $48!\,/\,43!$ different ways (multiplets in the remaining five are fine). Reckon as equivalent all $7!$ permutations of the total hand, for a final count of
Unfortunately, this is $44519904\,/\,7$, not only non-integer, but yielding a fraction $0.0475$ of $C(52,7)$, much too low.
What is wrong with my counting argument?
What is the right counting argument?