Colour change in drawing balls (Expectation same) You are given an urn with 100 balls (50 black and 50 white). You pick balls from urn one by one without replacements until all the balls are out. A black followed by a white or a white followed by a black is "a colour change". Calculate the expected number of colour changes if the balls are being picked randomly from the urn.
The solutions for this puzzle goes as:
There are 99 consecutive pairs. Let $X_i$ be a random variable taking value 1 if $i$th pair has a colour change and zero otherwise.
We have to find expected value of $E[X_1 + X_2 + ... + X_{99}]$
Since all $X_i$ are equivalent, the answer is $99\, E[X_1]$
$E[X_1] = (50/100)\, (50/99)+(50/100)\, (50/99) = 50/99$
What is the intuition or proof behind all the $X_i$ being equivalent ?
 A: Your process of picking out the balls one by one without replacement can be thought of as giving all balls a number in $\{1,\dots,100\}$.
We have $X_i=1$ if the balls $i$ and $i+1$ have different colors. 
Ball $i$ has a specific color (black or white). 
Then $49$ of the other $99$ balls do have that specific color and $50$ of them do not have that specific color. 
That means that all these $99$ balls have a chance of $\frac{50}{99}$ to have not that specific color.
Ball $i+1$ is one of these balls, so we conclude that $P(X_i=1)=\frac{50}{99}$.
This reasoning is valid for every $i\in\{1,\dots,99\}$.
Moreover it is valid for distinct balls that are not necessarily consecutive.
A: *

*If you put the balls randomly in a circle with all possible patterns equally likely,  by symmetry the probability of a colour change between a pair of points does not depend on where you are in the circle

*Putting the balls randomly in a line has the same probability distribution as putting the balls randomly in a circle then breaking the circle at random then straightening it into a line: all possible patterns are equally likely 

*Therefore, in a random line of balls, the probability of a colour change between a pair of points does not depend on where you are in the line
A: As an additonal note regarding this problem, it is possible to compute the complete probability distribution of the number of colour changes.
The following result will be used.
The number of solutions to the equation
$$
x_1 + \dotsb + x_r = s
$$
in non-negative integers is $\binom{s + r - 1}{r - 1}$.
Suppose that the urn contains $m$ black balls and $m$ white balls, and denote by $X$ the number of colour changes.
First, note that the total number of sequences is $\binom{2m}{m}$.
The number of sequences for which there are $k$ colour changes can be computed as follows.

*

*Suppose that $k$ is odd, say $k = 2j - 1$ for some positive integer $j$. Assume, for concreteness, that the first colour change is white to black. Then there are $j$ white-to-black colour changes and $j - 1$ black-to-white colour changes. Fixing the relative positions of $j$ black balls and $j$ white balls, one may use the result above to deduce that the remaining $m - j$ black balls can be distributed within the sequence in
$$
\binom{m - j + j - 1}{j - 1} = \binom{m - 1}{j - 1}
$$
ways. Similarly, the remaining $m - j$ white balls can be distributed within the sequence in
$$
\binom{m - 1}{j - 1}
$$
ways. It follows that the number of sequences which exhibit $k = 2j - 1$ colour changes is
$$
2 \cdot \binom{m - 1}{j - 1}^2\text{.}
$$


*Using a similar argument to the case where $k$ is odd, one may show that the number of sequences which exhibit $k = 2j$ colour changes is
$$
2 \cdot \frac{m - j}{j} \cdot \binom{m - 1}{j - 1}^2\text{.}
$$
This fully describes the probability distribution for the number of colour changes in a sequence of $m$ black balls and $m$ white balls.
One has
$$
\begin{equation*}
\begin{split}
\mathbb{E}X &= \frac{\sum_{j = 1}^{m} 2 \cdot (2j - 1) \cdot \binom{m - 1}{j - 1}^2 + \sum_{j = 1}^{m} 2 \cdot 2j \cdot \frac{m - j}{j} \cdot \binom{m - 1}{j - 1}^2}{\binom{2m}{m}}\\
&= \frac{(4m - 2)\cdot\sum_{j = 0}^{m - 1} \binom{m - 1}{j}^2}{\binom{2m}{m}}\\
&= \frac{(4m - 2) \cdot \binom{2m - 2}{m - 1}}{\binom{2m}{m}}\\
&= m
\end{split}
\end{equation*}
$$
as expected.
With some more binomial coefficient gymanstics, one may also compute (for example) the variance of $X$.
