Odds of Two People having one DOB and another two people having a different DOB in a Group

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So the odds of in a room of 14, two people sharing a DOB, and another two people sharing another DOB is 4.97%.

However, I cannot figure out how to get to 4.97%. I calculated it being 1.9% by doing the following:

•the number of ways to choose the two duplicated birthdays: (365/2) 
•the number of ways to assign the earlier birthday to two people, and the later birthday to two other people: (14/2)(12/2)         
 •the number of ways to assign 10 out of the remaining 363 days to the rest of the people: 363!/353!

Dividing this product by the number of ways to distribute birthdays without restrictions yields the final probability of 1.9%.

Where is the 4.97% coming from?