[gaming_mouse]

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22% seems like a low # though? That means 39% of the time one person will pair up. Thus the person with the high card has a 22+39 = 61% chance of winning?

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No. You can say that 78% of the time at least one of them will pair up. It does not follow that 39% of the time exactly one will pair up. You have to do a different calculation to solve that.

But it seems the question you are interested in is: How often does the player with the high card win, when that player has not improved to a pair?

So imagine yourself on the river with your A-high hand, say. You now want to know the chance your oppo has paired.

There are 15 unseen cards which your oppo could hold to make a pair with the board, out of a total of 45 unseen cards. We also assumed that your oppo's cards are different ranks than yours, which means that there are really only 39 unseen cards from which his hand can be composed.

Thus your oppo could hold any one of (39 choose 2) = 741 possible hands. However, we also assumed that your oppo does not hold a pocket pair. There are 5*3 + 6*4 = 39 pocket pairs possible. So there are really 702 possible hands for your oppo.

How many don't contain a card that pairs the board? (24 choose 2) = 276, not counting possible pocket pairs. There are 24 possible pocket pairs, giving our oppo 252 possible hands that don't pair that board and that agree with our assumptions. Thus the chance that your high card wins is:

252 / 702 = 0.358, about 36%

Note that this does not take into account the information you glean by your oppo's betting patterns. For example, if he has checked raised the turn, you should probably give him credit for a pair.

gm