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....

Well that's certainly wrong, considering there's always 6 ways to choose the double out of 4.

So we have for three beer mats:

6*(3*2*1) / 3^4 = 4/9 = 0.4444

For 4:

6*(4*3*2) / 4^4 = 9/16 = 0.5625

For 5:

6*(5*4*3) / 5^4 = 72/125 = 0.5760

For 6:

6*(6*5*4) / 6^4 = 5/9 = 0.5555

For 7:

6*(7*6*5) / 7^4 = 108/343 = 0.5248

For 8:

6*(8*7*6) / 8^4 = 63/128 = 0.4922

The question is though...

What are the chances there are exactly X different beer mats? (eg. in our example, 3).

Calculating the above isn't exactly difficult.