Question: 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?

Answer: 
You solve $18 u + 14 v = 2 = \gcd(18,14).$ Solve $2 w + 63 z = 1.$ Combine to get $18 x + 14 y + 63 z = 1.$ Then multiply all by $5.$
[For the following paragraphs, please refer to the figure at the end of the last paragraph ([the figure is also available in PDF](http://www.aespen.ca/AEnswers/LgXnx1444707176.pdf)).]

The manipulations performed from steps (0) to (17) were designed to create the linear system of equations (0a), (5a), (11a) and (17a). The manipulations end when the absolute value of a coefficient of the latest equation added is 1 (see (17)).

Equation (0a) is given. It is possible to infer equations (5a), (11a) and (17a) from (0), (5) and (11) respectively without performing manipulations (0) to (17) directly. In every case, select the smallest absolute value coefficient, generate the next equation by replacing every coefficient with the remainder of the coefficient divided by the selected coefficient (smallest absolute value coefficient) – do the same with the right-hand constant – and add the new variable whose coefficient is the smallest absolute value coefficient. If the new equation has a greatest common divisor greater than one, divide the equation by the greatest common divisor (it may be necessary to divide this greatest common divisor from previous equations). Stop when the absolute value of a coefficient of the latest equation added is 1.
Then proceed to solve the linear systems of equations.

[![enter image description here](https://i.stack.imgur.com/1VTjZ.jpg)](https://i.stack.imgur.com/1VTjZ.jpg)