[PairTheBoard]

That looks right to me. I suppose to be precise you would make p the probabilty that no one has a hand good enough for pot odds to call, and f the average win frequency for those hands that do have pot odds to call. But similiar to your treatment of the Blinds it may not be worth the complication when X is relatively large.

You can then calculate p using the same math technique as when you calcuated the Ranking for Median Best Hand.

Ideally it would be nice to have a spreadsheet with win percentages for all 2 Hand matchups. I suspect K used something like that when he did his calculations. You could then automate a lot of solutions. I myself would not want to individually enter every hand matchup in a 2-dimes type program to get the figures though. However I think you could get a good enough aproximation with a simplification by way of Hand Matchup Types. There are only a few Hand Matchup Types and all matchups of the same type have fairly close win frequencies. Sklansky had a thread going here recently which basically asked what your average win freqency would be if you could see your opponent's hand and only played when you had an edge against it. I was able to make a pretty good aproximation to the correct answer by just considering Hand Matchup Types.

For non pocket pairs you are up against:
1. 2 Undercards
2. 1 Undercard and 1 Middle card
3. 2 Middle cards
4. 1 Overcard and 1 Undercard
5. 1 Overcard and 1 Middle card
6. 2 Overcards
7. An underpair
8. A Middle Pair
9. An Over pair

With Pocket pairs you are up against the Hand Types:

1. Under cards
2. 1 undercard and 1 overcard
3. overcards
4. An underpair
5. An overpair

I think I got them all. You can pick examples of each Matchup type. I would use differing suits with minor straight chances. Then use those win freqencies for all hands of the same type. Figure all the greater and lesser flush and straight potential hands cancel each other out - aproximately.

I think such a settup would allow you to pump out a lot of good aproximate solutions without a whole lot more work.

PairTheBoard