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.

BTW: Why do the cards need to be numbered??

I sense there is some built-in ambiguity here in the problem, simply to cause trouble ;-)

There are two ways to interpret this problem. Way #1 and Way #2:

The guarantee that there is at least 1 card in each envelope is fulfilled:

#1) before filling. In such case, we fill each envelope with a card and then the probability that the remaining three cards are distributed evenly is: 2/3 x 1/3 = **2/9.** This is the probability that the fifth card found a different home than the 4th (2/3) and that the 6th card found the envelope with just one card (1/3).

#2) after filling: In such case, the envelopes are filled. Then, it is noticed that there are one or more cards in each envelope.

Well, random stuffing gives the t=trinomial coefficient n!/(i! j! k!) times then number of ways of achieving the i,j,k distribution. E.g., 6,0,0 may be achieved 3 ways [(6,0,0),(0,6,0), (0,0,6)] while (2,2,2) may be achieved only one way. (I call the ways achieved the "degeneracy" d.)

i,j,k d t=6!/(i! j! k!) d x t

---------------------------------------

600 3 1 3

510 6 6 36

411 3 30 90

420 6 15 90

330 3 20 60

321 6 60 360

222 1 90 90

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3^6 = 729

So, assuming no empty envelopes:

The likelihood of 222 is

p(222)/[ p(411)+p(321)+p(222) ]

= 90/(90+360+90) = 1/6

So, the answer is 2/9 = 0.222... or 1/6 = 0.1666...

depending on how you look at it.

*Edited on ***January 26, 2019, 3:56 am**