I like your thinking here a lot. I agree their is 12 ways to make the nut straight. Isn't there 9 ways to make a set on the flop ? And 15 ways to make a set including the turn and river?

I guess you could count 7-x of diamonds. And I think there are 9 ways to have that hand on the river.

Let's assume that he always completes the small blind and all hands are equally likely. (probably not true)

Let's also assume he doesn't call with two pair on the river. The only two pair he might call with is one with aces. So let's just say he only calls when he has aces and sevens. (because i don't think he would raise on the flop with mid or low pair and I don't think he would have bet the turn with k-7)

Going with your idea:
He calls (or raises) and I lose -
12 ways with the nut straight
9 ways with the flush (7-x diamonds)
Total - 21 ways

He calls and loses -
12 ways - set (Let's say he doesn't bet kk on the turn - passive)
9 ways - aces and sevens
Total - 21 ways

He could reraise on a bluff, so you need better than 50-50 to raise but since he is passive I can discount that factor somewhat. Looks pretty close.