6 cards, labeled 1,2, ...6, are randomly put in 3 different envelopes, at least one card in each.
Evaluate the probability of 2,2,2 distribution.
It took me until this morning to come around to Steven Lord's way of thinking, and per usual Charlie got there ahead of me. I agree that the answer depends on the method used to make sure that there is one card in each envelope.
If you start by putting one card in each envelope and then randomly distribute the rest, then the probability is 2/9.
If you randomly distribute all 6 cards without attempting to ensure that that each envelope has a card, and then notice after the fact that no envelope is empty, then the probability is 1/6.
And if you place cards one at a time, randomly, while keeping an eye on the envelopes, and both force the 5th card to go in an empty envelope if two envelopes are still empty, and also force the 6th card to go in the empty envelope if one envelope is empty, then the probability is even lower.
I convinced myself of this while trying to find Steven Lord's "error". I tried the simpler case with 4 cards and 2 envelopes. There are 16 ways to randomly distribute the cards without forcing an envelope to be non-empty. Of these, 2 have an empty envelope (they are distributed (4,0)), 8 are (3,1) and 6 are (2,2). Thus, the conditional probability of (2,2) given that the distribution not (4,0) is 6/14. Whereas the "easy-peasy" solution is 1/2.
Not Easy-peasy
