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

>>2
Not what he's asking.  The question is, what is the probability of a certain # of mats existing (call the event B) knowing which mats you and your friends received (event A).  So he wants P(B given A), but you're giving him P(A given B), and that my friend is the conditional probability fallacy.

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.

>>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.

>>5

OP here

Yeah, in the end we considered this possibility, hmm...

I guess if you limit the mats to a reasonable number like 100 the problem just turns into a huge summation party? :p