For these types of problems, the following approximation is always within a few tenths. The probability of a single opponent having AA is 6/(50*49/2) = 6/1225. The probability that someone has AA with 9 opponents is 1 minus the probability of no one having it, and this is approximately 1-(1-6/1225)^9 = 4.3% or 1 in 23.1. This is approximate since we are treating the hands as independent, and they are not, but the approximation is excellent.
In this case, we could also just multiply the probability of one person having KK by 9 to get 9*(6/1225) = 4.4% or 1 in 22.7. As we shall see, this is even closer to the exact answer which turns out to be 1 in 22.77. The only reason this is approximate is that it double counts the cases of two people having KK, which would have to be subtracted from this to get the exact answer, but that probability is very small. It has nothing to do with the fact that the opponents cannot "simultaneously pick", and someone could get just an A, as someone incorrectly stated. That is an issue for the first approximation where we are assuming independence. For the second approximation, we are summing probabilities which only assume mutually exclusive events. Note for example that if we had AA, the probability that a particular other player has AA is 1/1225, and the probability that one of 9 players has AA is EXACTLY 9/1225 because it is not possible for more than one other player to have AA. The probability of each player having AA is 1/1225, the same as if they "simultaneously picked".
Whether this second approximation is always closer than the first approximation is an area I would have to research further. I have always used the first approximation, and it is always sufficiently close.
A short while ago, I worked out the exact answer for many of these types of problems for the WPT TV show. These used 9 or 4 total players. You can see these results with an explanation of how they are derived at:
http://www.twoplustwo.com/forums/sho...rue#Post212797
The post "more exact answers" makes some corrections to the formulas explained earlier in the thread. The AA vs. KK problem was actually solved exactly several different ways. It is probably never necessary to be this precise (except for TV).
Here is the exact formula for AA vs. KK 10 handed:
[ 6*9*P(48,16)/2^8-6*C(9,2)*P(46,14)/2^7 ] / [ (P(50,18)/2^9 ) ] = 1 in 22.77.