Probability of drawing white ball after transferring to new urn n times I am in a probability theory course and could not find the solution to this question anywhere. The assignment is already turned in, and I am asking this for my knowledge and for others who are also curious. (Sorry for the messy math typing. Doing this off of a phone.)
There are $n$ urns in a line that all have $a$ white balls and $b$ black balls. Now, a ball is selected at random from the first urn and transferred to the second urn. Then a ball is transferred from the second urn to the third urn. This process continues until the $n$th urn. 
What is the probability P(W) that you will choose a white ball from the nth urn?
One way of thinking that I came to was:
$$P(W) = P(W_{n-1})P(W_n|W_{n-1}) + P(B_{n-1})P(W_n|B_{n-1})$$
In this method I know that n = 2 works, but unsure if n= 3, 4, etc works as well.
And another possible one that I thought of as I was typing is:
You are left with either $\frac{a+1}{(a+1) + b}$ or $\frac{a}{a + (b+1)}$ so placing a weight on these two can lead to the probability of a white ball being chosen.
 A: Claim: For all $n$, we have $P(W_n)=\frac{a}{a+b}$


*

*$P(W_1)=\frac{a}{a+b}$

*Suppose that $P(W_{n-1})=\frac{a}{a+b}$, with $n>1$. Using the formula you wrote down, 
$\begin{align}P(W_{n})&=P(W_{n-1})P(W_n|W_{n-1})+P(B_{n-1})P(W_n|B_{n-1})\\
&=\frac{a}{a+b}\frac{a+1}{a+b+1}+\frac{b}{a+b}\frac{a}{a+b+1}\\
&=\frac{a(a+b+1)}{(a+b)(a+b+1)}\\
&=\frac{a}{a+b}
\end{align}$
By induction, the claim is true.
A: A more general statement is that if you have an urn known to contain
$N$ balls, and if the expected number of white balls in the urn is $x$,
then the probability of drawing a white ball on the first draw is $\frac xN.$
To show why this is true, suppose you randomly distribute the contents
of the urn into $N$ previously empty urns, one ball per urn.
Now each of the $N$ new urns contains one ball, and the expected number of
white balls in each urn is $x_1$ (the same value for each urn, by symmetry).
The expectation of a sum is the sum of expectations of the terms, so
the expected number of white balls in all $N$ urns together is $N x_1$;
but this is also the expected number of white balls in the original urn.
Therefore
$$x_1 = \frac xN.$$
But $x_1$ is the probability that the (single) ball in one of the urns
is white, which is the probability that a single ball drawn from the
original urn will be white.
That's a lot of trouble to answer this one question (especially since the
answer isn't even finished yet), but once you know the general statement
is true you can apply it in a lot of other places, not just here.
Back to the original problem, the expected number of white balls in the second urn is initially $a$. When we transfer one ball from the first urn to the
second, the expected number of white balls transfered is
$\frac{a}{a+b}.$
The expected number of white balls in the second urn after the transfer is
therefore
$$a + \frac{a}{a+b} = \frac{a(a + b + 1)}{a + b}.$$
Since there are $N = a + b + 1$ balls in the second urn now,
the probability of drawing a white ball is
$$\frac{\left(\frac{a(a + b + 1)}{a + b}\right)}{a + b + 1} = \frac{a}{a + b}.$$
Rewrite the transfer from urn number $1$ to urn number $2$
as a transfer of one ball from urn number $k$ to
urn number $k+1$, assuming that the ball chosen from urn $k$ is
white with probability $\frac{a}{a + b},$ and you can prove by induction
that the probability of drawing a white ball from the $n$th urn
is also $\frac{a}{a + b}.$
