Probability that the ball drawn from $n$th urn is white There are $n$ urns each having $a$ white and $b$ black balls. One ball is taken from urn 1 and is transferred to urn 2. Then one ball is taken from urn 2 and transferred to urn 3 and so on. Find the probability that the ball drawn from $n$th urn is white.
I get the intuition that the answer should be $\frac{a}{a+b}$, but I'm unable to prove it.
 A: When you go from urn 1 to urn 2, the white probability in urn 2 is now $\frac{a+\frac{a}{a+b}}{a+b+1} = \frac{(a+b)a+a}{(a+b+1)(a+b)}=\frac{a}{a+b}$  Therefore as you continue, the probabilities in each urn remain the same after each transfer, leading to your final answer.
A: Wrap the balls in a paper and put the number of the original urn to the outside of the ball. When you draw the ball from the last urn, first look at the number on the ball, and then open the package and look at the color.
The color of the ball doesn't have any effect in what happens, so after we have looked at the number, no matter what the number was, the probability that the ball is white is $\frac{a}{a+b}$. Therefore, the probability is $\frac{a}{a+b}$ even if we don't look at the number.

Mathematically, let $X$ be the number of the urn where the last ball originally comes from, and let $Y$ be the color. We know that for all $x$, $\mathcal{P}(Y=\text{white}|X=x)=\frac{a}{a+b}$, so we know that $\mathcal{P}(Y=\text{white})=\frac{a}{a+b}$. 
A: In the average of many trials, you can consider that, when you draw a ball from the first urn and move it to the second one, it is part white and part black, in fractions $\frac{a}{a+b}$ and $\frac{b}{a+b}$ respectively. This doesn't make sense in terms of one trial, but as an average, it does.
Thus, adding such a fractional ball to urn #2 does not change the probability of what is drawn from that urn. This argument extends to any number of urns.
