Yeah, that seems about right.

Well, there are 4*4 = 16 combinations of AK (4 aces, each with 4 different kings), and to calculate the probability of AK, we simply use the following formula:

P(AK) = 16/(52 choose 2) = 0.012

P(someone else has AK) = P(AK) * number of opponents

so for FR, this is 0.096 = 9.6%

to get the probability that someone has a range of {AK,QQ+} while you don't have any A, K or Q, you simply calculate:

P(AA) + P(KK) + P(QQ) + P(AK) =

3.6% + 3.6% + 3.6% + 9.6% =

**20.4%**
Of course it gets a bit more complicated if you for example hold an A, which blocks some aces from AA and AK... In that case there are 3 combinations of AA left (3 choose 2) and 12 combinations of AK (3 aces times 4 kings).

In that case, the probabilities for AK and AA are the following:

P(AA, 1A blocked) = 3/(52 choose 2) = 0.0022 (half less than if you didn't have an A)

P(AK, 1A blocked) = 12/(52 choose 2) = 0.0090

so the probability of any of the opponents having something from the range of {AK, QQ+} while you hold one A would be:

P = 1.8% + 3.6% + 3.6% + 7.2% = 16.2%

You should be able to calculate similar stuff by yourself now, although it might get a bit more complicated if you start wondering about what the probabilities are that two people have something from that range at a time, etc..

-SF