This may not be 100% correct, but should be close. Odds of getting a poket pair is 6% as I recall, or 1 in 17 hands. The chance of 2 (or more) people at a full ring (9 players) is around 3.3% or 1 in 30 hands. So, not all that uncommon. Been a while, but there are many odds calculators out there.

Actually, If you have been dealt a PP, the chances of anybody else having a HIGHER PP is (7n - (xn/2))%. Where X is the value of your cards and N is the number of players left to act.

ex you have a pair of 8's, with 4 people left to act, theres a 12% chance that somebody has a pair higher than that.

pair of 2's with 9 people to act, thats a 54% chance that somebody else has a higher PP.

If you dont believe this, I posted the proof for this here and on 2+2 a few years ago, where it was vaildated on both sites. It is an approximation, doable at the table (usefulness comes intop question though). You can run the exact math yourself if you want, this formula is about .5%-1% off.

Point is, seeing 2 pairs at a table is not uncommon at all. In fact, you would see a lot more of it at the better tables if players werent so smart. It's the tables with stupid players that can't let go of a pair that you see this often. Gotta remember, most of the time, if your holding less than 8's, and you don't hit your set on the flop, most likely you should fold.

Without breaking out the calc, or looking it up, that looks about right Marm, close enough anyway. KK with 1 person to act, 0.5% (1 in 200) they have AA. 22 with 9 left, more that half the time your beat going in. See a cheap flop hoping for a set, but even then there's a chance someone flopped a higher set.

Edit: If it seems like it at first glance; if you really look at it, my numbers above dont disagree with Marms; they are just taking a different approach to the same question.

Didnt say you could trust it, just that it's within normal variance :P