Can someone confirm the following for hold’ em:

Probability of exactly one pair by the river. Pair must include one or both hole cards (e.g. dealt a pair). Board pair+, straights, flushes allowed but don’t count as a hit. Sets, trips, quads involving hole cards not allowed.

Prob= 0.4094

I thought I understood the question, but now after re-reading it again, my first reading was way wrong.

Anyway, I wrote a simple program to tally the number of 1-player NLHE deals that have exactly one pair using at least one card from the player's hole cards (any higher poker ranking is excluded).

My program reported that there are 6,454,980 such deals. So this, in terms of a ratio, is the numerator.

I am not sure if it is easy to tally (or calculate) the denominator given the "exclusions" you mentioned in OP. I don't have time to do anything more on this question at the moment.

And possibly I misinterpreted (or miscoded) what you are looking for. So, feel free to provide further clarification and/or critique my result above. I will check back later for what it is worth.

The exclusions are only for the numerator or hits. Straights, flushes are allowed for good hits as long as there is exactly one pair using at least 1 hole card.

Good hands: J J |A 3 5 5 9 (board pairs ok)--- J Q | J A 3 5 7 --- J Q |7 8 9 T J (straight ok)

Bad Hands: J J |A 3 5 7 J (set)--- J Q | A 3 5 5 9 (no pair w/hole card) --- J Q | J Q 2 4 6 (2 pair)

I used a relatively simple equation

Okay, completely disregard my previous post. I completely misunderstood what you were asking. My bad, though you could have been a little clearer originally.

Anyway, with a better understanding of what you are asking (and the code already written), I ran a simulation of 10,000,000 deals of 1-player NLHE and kept track of the situation you are seeking.

I got 4,093,182 "successes" out of the 10,000,000 deals -- which appears to confirm your result posted above.

OK - thx.