Polya's Urn : proportion Probability There are two urns with one ball each. Each of subsequent n-2 balls is placed into one of these urns, with probability proportional to the number of balls already in that urn. What is the expected number of balls in the urn with fewer balls?
I tried the following : Let En denote the expected balls in the urn contains smaller balls.
Then En+1 = En(1-En/n) + (En + 1)(En/n)..
which yields En = n/3.
However this obviously is incorrect as there are several leaps of faith and also the case when the two urns become equal is ignored.
P.S I came across following links but unable to follow the arguments :
http://gurmeet.net/puzzles/polyas-urn-problem/
 A: I will try to open up your mind for the nice explanation that is linked.
Place a marked card. Then place a second card on the left of the marked card or on the right of the marked card. Then place a third card on the left of both cards that are allready placed, in between them, or on the right of them. Note that for the third card there are $3$ spots where it can be placed. Continue this, realizing that for the $i$-th card there are $i$ spots where it can be placed, and let it be that each spot has the same probability to be the elected one. 
In total $n-1$ cards are placed (this number includes the marked card), so $n-1$ spots are filled up with cards. Essential question:  Do you understand that each of these spots has equal probability to be the one that is filled with the marked card?
Relate the two urns with $2$ balls in it - placed side by side - with the marked card. Relate the second card with the first ball that is added to one of the urns. A card on the left of the marked card corresponds with a ball in - let's say -the urn at the left side. Likewise a card on the right of the marked card corresponds with a ball in the urn at the right side. If the card is placed is on the left side of the marked card then from that moment on there are $2$ spots on the left side where a next card can be placed  and $1$ spot on the right side. So the probability that the next card will also be placed on the left side of the marked card is $\frac23$. This agrees with the prescription of probability that the next ball will be placed in the urn that contains $2$ or the $3$ balls. This resemblance will also be there for later placings. 
This result in equal probabilities of the configurations $(1,n-1),\dots,(n-1,1)$. This because each of these configurations corresponds with a spot where the marked card is placed.
A: If $n$ is even number, the final distribution if the balls in urns is 
$ (1,n-1), (2,n-2),\ldots, (\frac{n}{2}-1, \frac{n}{2}+1), (\frac{n}{2}, \frac{n}{2}), (\frac{n}{2}+1, \frac{n}{2}-1), \ldots, (n-2,2), (n-1,1)$
with probability $\frac{1}{n-1}$. Then the expectation of the balls in the urn with fewer balls is $\frac{2}{n-1}\left(1+2+\ldots+ (\frac{n}{2}-1)\right) + \frac{1}{n-1}\times \frac{n}{2} = \frac{n^2}{4(n-1)}$
Similarly, if $n$ is an odd number, we get $\frac{n+1}{4}.$
