Interesting Problem

You're at a bar with 3 friends.

Your beer comes on a beer mat (4 in total).

There are three different beer mats (same company, different text on them), so one is double (2+1+1).

How likely is it that there are only 3 different beer mats?

I claimed it would be more likely that there were 6 or 7 different beer mats if the only thing you know is that you got 2+1+1 when "drawing" 4.

We got really stuck at the maths though....

>>4
Listen to this man.  If n mats exist, then the probability of the result described by op is (6*n*(n-1)*(n-2))/n^4.  The 6 comes from (4 choose 2) ways of pairing members of your party for the double.

But I'm not sure a solution exists to the problem posed by the op.  Don't we need more information about the probability of there being n mats?  If we assume there can be any number n of mats where n is a natural number greater than 2, the independent probability of there being n mats is 0, in which case the conditional probability is undefined.