[AaronBrown]

Another way to get the same answer, that may be easier to visualize, is to consider the number of ways to form A66. If the other player's hand has no Aces or sixes in it, there are 24. If the other player's hand contains an Ace, there are only 18. So the probability that she holds AK is 18/(18+24+24+24) = 3/15.

I like this method better, because it's more natural to incorporate subsequent information. But Reverend Thomas Bayes proved that you get the same answer either way.

Of course, this is affected by what you hold if you hold a 6 or any card the other player might hold.