What is the reason that $p_i$ does not depend on $i$? Consider the following formula :
$$p_i=\sum_{j=1}^n\frac{\binom{i-1}{j-1}\binom{n^2-i}{n-j}}{\binom{n^2}{n}},$$
where $i,n$ are positive integer and $i=1,\ldots,n^2$.
Suppose there are $n$ boxes (the boxes are labeled with $1,2,\ldots,n$) each containing $n$ balls. The balls in each box are ordered according to their size. The $i$th $(i=1,2,\ldots,n^2)$ smallest ball among all $n^2$ balls is the $j$th smallest ball  $(j=1,2,\ldots,n)$ of a box if and only if  (1) the first $(i-1)$ values contain exactly $(j-1)$ 1's. This can be done in $\binom{i-1}{j-1}$ ways, $(2)$ the $i$th value is a $1$, (3) And therefore there are exactly $(n-j)$ 1's distributed among the remaining $(n^2-i)$ values. This can be done in $\binom{n^2-i}{n-j}$ ways. 
The total number of ways in which the $n^2$ balls can be randomly allocated to $n$ boxes is $\binom{n^2}{n}$. 
Then the probability that the $i$th smallest ball among all $n^2$ balls is the  $j$th smallest ball of a box is 
$$\frac{\binom{i-1}{j-1}\binom{n^2-i}{n-j}}{\binom{n^2}{n}}.$$
We select a ball if it is the $1$st smallest ball of the $1$st box, or  $2$nd smallest ball of the $2$nd box, thus $n$th smallest ball of the $n$th box. It is not possible that the $i$th smallest ball overall was in two boxes, but it had to be in some box. Therefore the event that "the $i$th smallest ball was in box $j$", is an exhaustive partition of the possibilities. Consequently there chances add up. Hence the probability that the $i$th smallest ball is selected is 
$$p_i=\sum_{j=1}^n\frac{\binom{i-1}{j-1}\binom{n^2-i}{n-j}}{\binom{n^2}{n}}.$$
I expected different value of $p_i$ for different $i$. But I was astonished that the $p_i$ is same for all $i$.  
It seems 
$$p_i=\sum_{j=1}^n\frac{\binom{i-1}{j-1}\binom{n^2-i}{n-j}}{\binom{n^2}{n}}=\frac{1}{n}.$$
I checked it by calculating $p_i$ for different $n$ such as $2$ or $3$.
What is the  reason that $p_i$ does not depend on $i$?
Why this complex formula is nothing but a uniform probability? 
 A: Here is an algebraic verification. Denoting with $[z^k]$ the coefficient of $z^k$ in a series we can write e.g.
\begin{align*}
[z^k](1+z)^n=\binom{n}{k}
\end{align*}

We obtain
  \begin{align*}
\color{blue}{\sum_{j=1}^n}&\color{blue}{\binom{i-1}{j-1}\binom{n^2-i}{n-j}}\\
&=\sum_{j=0}^\infty [z^{j-1}](1+z)^{i-1}[u^{n-j}](1+u)^{n^2-i}\tag{1}\\
&=[u^n](1+u)^{n^2-i}\sum_{j=0}^\infty u^j[z^j]z(1+z)^{i-1}\tag{2}\\
&=[u^n](1+u)^{n^2-i}u(1+u)^{i-1}\tag{3}\\
&=[u^{n-1}](1+u)^{n^2-1}\tag{4}\\
&\color{blue}{=\binom{n^2-1}{n-1}=\frac{1}{n}\binom{n^2}{n}}
\end{align*}

Comment:


*

*In (1) we apply the coefficient of operator twice and set the upper limit of the series to $\infty$ without changing anything since we are adding zeros only.

*In (2) we use the linearity of the coefficient of operator and apply the rule $[z^{p-q}]A(z)=[z^p]z^{q}A(z)$.

*In (3) we apply the substitution rule of the coefficient of operator with $z:=u$
\begin{align*}
A(u)=\sum_{j=0}^\infty a_j u^j=\sum_{j=0}^\infty u^j [z^j]A(z)
\end{align*}

*In (4) we do some simplifications and are ready to select the coefficient of $[u^{n-1}]$.
A: $p_i = \sum^{n}_{j=1} \dfrac{\binom{i-1}{j-1}\binom{n^2-i}{n-j}}{\binom{n^2}{n}} =  \sum^{n}_{j=1} \dfrac{\binom{i-1}{j-1}\binom{n^2-i}{n-j}}{\binom{n^2-1}{n-1}\times\dfrac{n^2}{n}} = \dfrac{1}{n}\sum^{n}_{j=1} \dfrac{\binom{i-1}{j-1}\binom{n^2-i}{n-j}}{\binom{n^2-1}{n-1}}$.
The formula inside the sum can be interpreted the following way:
Let's say I have $n^2-1$ balls from which $i-1$ are red and $n^2-i$ are blue. If I select $n-1$ balls from this set, the probabilty of getting $j-1$ red balls and $n-j$ blue balls is $\dfrac{\binom{i-1}{j-1}\binom{n^2-i}{n-j}}{\binom{n^2-1}{n-1}}$.
Now we sum probabilities over all possible outcomes, which is equal to $1$, hence $p_i = \dfrac{1}{n}$.
