1) If you have pocket kings, what are the odds you'll run into pocket aces in a 9 handed game? (And in a 4 handed game?). Is it just the negative parlay of 1:220 to the 8th power (and thus about 1/25)?
Your approximation using 220:1 would actually give 1-(220/221)^8 = 1/28. You really want to use 204:1 since you having kings means there are only 50*49/2 = 1225 remaining hands possible, so the odds of AA are 1225/6 = 204:1. Then 1-(203/204)^8 = 1 in 25.9. The exact answer does turn out to be 1 in 25.5 as you said. The exact answer is more difficult since the player's hands are not independent, that is, certain hands make AA more or less likely. We must also consider the possibility of more than one player having AA. That could be ignored in this problem because it is small, but we cannot ignore it in some of your other problems. This is how the exact answer must be computed:
[ 6*8*P(48,14)/2^7 - C(8,2)*P(46,12)/2^6 ] / [ P(50,16)/2^8 ]
= 1 in 25.5.
The denominator is the number of ways to deal hands to the remaining 8 players. It is the number of ways to deal 16 cards out of 50, and then we divide by 2^8 since we don't care about the order of the 2 cards in each player's hand. The first term in the numerator says there are 6 ways a player can have AA, there are 8 players that can have AA, and there are P(48,17)/2^7 ways to deal the remaining hands to 7 players. This would double count cases where two players get AA, so the second term in the numerator subtracts this off. This is so small it could be ignored in this problem.
For 4 players this becomes:
[ 6*3*P(48,4)/2^2 - C(3,2)*P(46,2)/2^1 ] / [ P(50,6)/2^3 ] = 1 in 68.1
4) AK vs. AA or KK
This problem is similar to 1 except there are now 9 ways that two players could have AA or KK. Note that no more than two players can have these hands, and one must have AA while the other has KK. There are still 6 ways for a player to have AA or KK.
9 handed:
[ 6*8*P(48,14)/2^7 - 9*C(8,2)*P(46,12)/2^6) / [ P(50,16)/2^8 ]
= 1 in 25.6
4 handed:
[ 6*3*P(48,4)/2^2 - 9*C(3,2)*P(46,2)/2^1 ] / [ P(50,6)/2^3 ]
= 1 in 68.1
The other problems are done similarly, but they will be more time consuming because there are more ways that multiple players can have better hands that we must consider. This means more terms to be added together. The worst case is problem 3 where up to 6 players could have a better hand. This would require 6 terms to be computed, however, some of these could be ignored. Basically you compute terms until they become sufficiently small. The second term is subtracted, the 3rd is added, and so on alternately. This is called the inclusion-exclusion principle.