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.
(In reply to
re: two different answers s.b. one by Ady TZIDON)
Thank you sincerely Ady, for catching my typos. I have corrected my post. I give my replies to yours in order:
(In reply to two different answers by Steven Lord)a. 5,2,0 s.b. 510 and 6!=120; Please redo your table.
Yes - I meant (and in fact I used 5,1,0) and now this is fixed. Thanks! And it is 3^6 = 729 = the total number of different ways to fill the envelopes. Again, corrected - thanks!
b. You should get the same answer i.e. 2/9
For the two cases I depicted, I think not.
Suppose (this is my case #2): the the envelopes were stuffed and then someone noticed there was at least one card in each envelope. BTW, this one or-more card per envelope scenario happens
[P(411)+P(321)+P(222)]/3^6 = 20/27 = 0.740740740... of the time.
Then, the puzzle becomes a conditional probability question, I think:
p(222 | no zeroes) and my answer p=1/6 obtains.
No?
c. DISTINCT cards or IDENTICAL cards should yield the same result.
Exactly. That was my joke, right? There's no point in mentioning that the cards were numbered, right? You see, I am light-hearted about these puzzles - if they are not fun, then why do them (aside from staving-off senility)?
With my best regards for giving us so many fun puzzles - Steve
Edited on January 26, 2019, 4:29 am
re(2): two different answers s.b. one
