There are many approaches to answer such a question. Of course, they should all arrive at the same answer. I will ignore flushes in what follows. Incorporating the possibility of one or both players having a flush can be done, but I will leave that for others.

Break it up into cases of how many pairs each person has.


Case 1: Each player has the same 7 distinct ranks [1,1,1,1,1,1,1]

Prob = C(13,7)*{[C(4,2)*C(2,1)]^7}/2 / [C(52,7)*C(45,7)/2]
= 30,743,691,264 / 3,035,546,247,333,600

which is approx once in every 98,737 deals.


Case 2: Each player has the same 6 distinct ranks (with same pair) [2,1,1,1,1,1]

Prob = C(13,6)*C(6,1)*[C(4,4)*C(4,2)]*{[C(4,2)*C(2,1)]^5}/2 / [C(52,7)*C(45,7)/2]
= 7,685,922,816 / 3,035,546,247,333,600

which is approx once in every 394,949 deals.


Case 3: Each player has the same 5 distinct ranks (with same two pairs) [2,2,1,1,1]

Prob = C(13,5)*C(5,2)*{[C(4,4)*C(4,2)]^2}*{[C(4,2)*C(2,1)]^3}/2 / [C(52,7)*C(45,7)/2]
= 400,308,480 / 3,035,546,247,333,600

which is approx once in every 7,583,018 deals.


Case 4: Each player has the same 4 distinct ranks (with same three pairs) [2,2,2,1]

Prob = C(13,4)*C(4,3)*{[C(4,4)*C(4,2)]^3}*[C(4,2)*C(2,1)]/2 / [C(52,7)*C(45,7)/2]
= 3,706,560 / 3,035,546,247,333,600

which is approx once in every 818,965,900 deals.


GRAND TOTAL:

Prob = 38,833,629,120 / 3,035,546,247,333,600

which is approx once in every 78,168 deals.