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I could probably calculate this myself, but I' probably screw it up. If I am in a 10 handed game and I don't have a pocket pair, what is the probability of at least one other player having a pair? What about in a 6 handed game?
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Here is the solution for 8-handed from spectator's point of view (easily adapted to 9 or 5 opponents).
Here is the solution when you hold 33 and want the probability of an opponent holding a pair > 33.