
Larry C. answered 08/17/19
Computer Science and Mathematics professional
For this type of question, it's easier to figure the chance of no pairs in 5 cards. Then, the chance of at least one pair would be 1 minus that probability.
So, the 2nd card would have a 48/51 chance of not pairing the first card, the 3rd card 44/50 of not pairing either of the first two, 4th card 40/49 and 5th 36/48. This works out to around 50.1% so the chance of at least 1 pair is around 49.9%.
The chance of no pair in 2 hands is around 25.1 so indeed the chance of at least 1 pair between the two hands is around 74.9%