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
------------------------------------------
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
two different answers
