Here is my current thinking, please let me know if it is correct. The result doesn't seem intuitive and I feel like I'm missing something, hence I am reaching out here.
52 - 4 = 48 cards remaining. (4-1) * 4 = 12 "outs" for at least a pair 1 - (36 * 35 * 34 * 33 * 32) / (48 * 47 * 46 * 45 * 44) = 0.7798 that at least 1 out is drawn on board, since it is equal to 1 minus the probability that none of the outs are drawn (I don't the "1 -" but it felt natural). The chance of straights and flushes are much lower than pairing but would only add to 0.7798. So the answer to the title is at most 0.2201.
Any help would be appreciated!
I would have thought the probability, for two players with two hole cards each and five community cards, of neither player having even a pair (not counting straights or flushes) was
$$\left(\frac{52}{52}\times \frac{48}{51}\times \frac{44}{50}\times \frac{40}{49} \times \frac{36}{48}\right)\times\left( \frac{32}{47}\times \frac{28}{46}\right) \times \left(\frac{6}{45}\times\frac{27}{44}+\frac{24}{45}\times\frac{26}{44}\right) \\ \,\\ \approx 0.0834$$
based on the community cards being all different values, the first hand also being different values, and the second hand's first card either matching one of the first hand's values or not but still not matching any of the community cards' values.