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Uniform distribution (Posted on 2019-01-25) Difficulty: 2 of 5
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.

No Solution Yet Submitted by Ady TZIDON    
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two different answers | Comment 2 of 9 |

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
  Posted by Steven Lord on 2019-01-25 21:24:44

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