What are the chances.... (Feb 5th)

So everyone knows if you have something like QQxx and the board is KQx and someone, who very well could have KK shows strength, you should probably throw away your set of Queens (Scenario 1).

However, if you have QQKx in the same situation, it's a lot harder to give credit to someone having KK, because there is only one way to have the two case queens (Scenario 2). I've been playing a lot recently, but I've paid two people off where they have KK. Also, as extra credit, is it more likely to hit the case queen for quads than the chance that someone has KK. I'm going to go with hitting quads as the more likely scenario.

To make the analysis simple let's say that you are playing heads up.

Scenario 1
You have QQxx (where xx are not kings)
Board is KQx

So there are 45 unknown cards and 3 kings unaccounted for (so 42 other cards).
What is the chance that your single opponent was dealt KKxx.
C (3,2) * C(42,2) = 2,709 was for him to have KKxx
C (45,4) = 148,995 possible 4 card combinations with 45 cards left in deck

So I calculated 1.73%, which is close to right answer.
(See here for a link to better calculations: http://forumserver.twoplustwo.com/showflat.php?Cat=0&Number=9063855&an=0&page=0#Post9063855)

Scenario 2
You have QQKx (where xx are not kings)
Board is KQx

So there are 45 unknown cards and 2 kings unaccounted for (and 43 other cards).
What is the chance that your single opponent was dealt KKxx.
C (2,2) * C(43,2) = 903 ways for him to have KKxx
C (45,4) = 148,995 possible 4 card combinations with 45 cards left in deck.

Answer: 0.61% chance that he has the two case Kings.

Bonus question. If the above Scenario 2 happens to you, then you actually have a 4.9% chance of hitting your one outer for Quads Queens. (http://www.propokertools.com/simulator/simulate.jsp?g=oh&b=KhQd7s&h1=QsQhKd2c&h2=KsKc8h3s&h3=&h4=&h5=)