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A postman brought N letters to a house with two letter-boxes. Since the two boxes were empty, he puts 1 mail in each of the two mail boxes. Then he chooses one of boxes with probability proportional to number of letters present in that box, and puts the 3rd letter in it. He does this for all subsequent letters. What is the expected number of letters in the box with lower letters?

I referred the solution for the above problem. Suppose I have a stack of 2 black cards and one red card. Initially I put red between 2 black cards. Now I add black cards randomly between any two cards (so, initially it is either above or below red). Note that the probability that I add the card above the red card, when x-1 is the number of cards above red and y-1 is the number of cards below red is x/(x+y). Let the problem be: if red card is dividing the black cards into two sets, what is the expected number of black cards in the smaller section. So, we see that the two problems are equivalent.

Now this way, we are getting all possible combinations in which one red and n black cards can be mixed, we see that the probability that the red card is at height h is independent of h. So, the probability that the smallest box contains n/2 letter or 1 letter (or any number of letters between 1 and n/2) are all same. So, expected number of letters in the smaller box is asymptotically n/4

I do not find the highlighted line intutive :/

Request help to understand this approach and also welcome any more intuitive approaches. Thanks

DeeDee
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1 Answers1

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Think of the red card as being numbered $1$, and each subsequent black card as being numbered $2,3,4\dots$. The initial black cards on top and below are un-numbered (they are not needed for this explanation to work).

After the first black card is inserted, there are two equally likely orderings of the cards; either $(1,2)$, or $(2,1)$. For each of these orderings, there are three possible places to insert the next black card, leading to six equally likely orderings corresponding to all permutations of $(1,2,3)$ for the next stack. in general, after black card numbered $n$ has been inserted, all possible permutations of $(1,\dots,n)$ are equally likely orderings of the deck (you can prove this by induction, but it should also be intuitively clear). In particular, this implies the height of the red card is equally likely to be any height; does this extra explanation help you see why?

Mike Earnest
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  • My question is when the postman chooses the mailbox with the probability of proportion of letters, after the first letter, say if he puts it in mailbox X instead of Y, now X has P = 2/3 and the symmetry/uniform randomness is broken in a sense. How does this converge to your explanation? Is it because either mailbox is equally likely to be chosen when Px = Py = 1/2? – Robert Melikyan Nov 01 '22 at 15:19