Adding balls with probabilities according to existing balls I came across the following probability problem:
Start with $1$ black ball and $1$ white ball in a box. At each step, we will put in a new ball. If there are $a$ black balls and $b$ white balls, we put in a black ball with probability $\dfrac{a}{a+b}$ and a white ball with probability $\dfrac{b}{a+b}$. We do this until there are $n$ balls ($n\geq 2$). Prove that the probabilities that there are $1,2,\ldots,n-1$ black balls are all equal.
This problem is trivial by induction on $n$, the total number of balls. I wonder, however, if there is an intuitive way to interpret the result, without the use of induction?
 A: It's actually obvious-think of the number of ways of choosing "x" objects out of a given collection of two kinds of objects (let it have a objects of type 1 and b objects of type 2). this can be done in x+1 ways (the problem reduces to the number of ways of assigning a number between 0 and x to the number of objects chosen from type 1). 
hence, the number of ways of filling the box in your problem is n-1 ( the number of ways of choosing n-2 objects from a given collection of white and black balls). Clearly, out of these n-1 ways, there is one each for a given number of chosen black balls between 0 and n-2. Since there are already one white and one black ball in the box, there is 1 black ball when 0 is chosen and n-1 when n-2 are chosen. hence  the probabilities that there are 1,2,…,n−1 black balls are all equal to 1 in n-1.
A: Choose some $1\leqslant k\leqslant n-1$.


*

*There are $k$ white balls after $n-2$ steps if one chooses $k-1$ times to add a white ball and $n-k-1$ times to add a black ball.

*For every set of $k-1$ times at which one chooses to add a white ball, the probability of an individual path from $(1,1)$ balls to $(k,n-k)$ balls is some product of $\dfrac{a}{c}$ and of $\dfrac{b}{c}$ where $a$ takes every value from $1$ to $k-1$ (when one adds the white balls), $b$ takes every value from $1$ to $n-k-1$ (when one adds the black balls), and $c$ takes every value from $2$ to $n-1$ (since the number of balls before one adds a ball, white or black, is $2$ then $3$ then... $n-1$). Thus the probability of each individual path is $\dfrac{(k-1)!(n-k-1)!}{(n-1)!}$.

*There are as many such paths are there are ways of choosing the $k-1$ times when one adds a white balls, that is, $\displaystyle{n-2\choose k-1}$. 


Thus, the probability to have $k$ white balls when there are $n$ balls is 
$$
\frac{(k-1)!(n-k-1)!}{(n-1)!}\cdot{n-2\choose k-1}=\frac1{n-1}.
$$
