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....

if there are three beers, the chance of this outcome would be 1*2/3*1/3*(1/3*3) = 2/9 = 18/81
If there are four beers,
1*3/4*2/4*(1/4*3) = 18/64 > 18/81
five  <=== highest probability
1*4/5*3/5*(1/5*3) = 36/125 > 36/128
six
1*5/6*4/6*(1/6*3) = 60/216 < 36/125