[LetYouDown]

Glanced briefly, there's at least one problem. Your denominators of C(52,5) and C(52,2) are incorrect. You have knowledge of at least 2 cards...so those are impossible on the board.

There are C(50,2) possible hands for him to have. So the odds of him having an overpair to your 10-10 is:

[4 * C(4,2)]/C(50,2) or 24/1225 or about 1.96%.

Assuming he has K-K, the odds that you'll hit one 10, without a K on board should be:

2 * C(44,4)/C(48,5)

271502/1712304 = ~15.86%

Excluded hitting both 10's, as K's are essentially irrelevant at that point. If the board comes 10-10-K-K-X, it's worth the story.