math question regarding headsup situation

[Wynton]

I posed this question in the probability forum, but no replies yet. I pose it here because I think the situaton arises frequently at six-max tables.

Say you're headsup on the flop (without a pocket pair), and fail to flop a pair. And say the flop itself does not contain a pair. What would the odds be that our opponent neither has a pocket pair nor has flopped a pair himself?

And what would those odds be if you still fail to make a pair on the turn?

Ultimately, I'm trying to get a mathematical basis for dealing with headsup situations when it's impossible to make assumptions about our opponents' holdings, e.g., the opponent is loose and open-completes from sb or defends his bb from a open-raise from the button (or sb) with virtually anything.

[witeknite]

No taking into account that he will have a pair on a KJ9 flop more than on a 752 flop I came up with no pair 62% of the time on the flop.

He doesn't have a PP 94% and 2 hole cards don't make a pair on the flop ~66% of the time. So 62%. On the turn it drops to 54%.

WiteKnite

[Wynton]

Did you consider the condition that hero also does not have a pocket pair and has not made a pair on the flop, when making these calculations? Does that condition make much of a difference at all, if any?