Expectation of minimum of two random numbers Let's assume we picked chose two numbers in the range $[1,n]$ where $n\in \mathbb N$ independently, such that each one of them is distributed randomly. 
How can I find the expectation of the minimum of these two numbers?
I know that we have $2n-1$ possibilities of getting $1$ as minimum, $2(n-1)-1$ possibilities of getting $2$ as minimum, and so on!
But how do I formalize this and calculate the expectation?

For $1 \le k \le n$, let $p_k$ be the probability that the minimum of the two numbers is $k$, and let $e$ be the expected minimum value.
\begin{align*}
\text{Then}\;\;p_k &= 
2
\left(
{\small{\frac{1}{n}}}
\right)
\left(
{\small{\frac{n-k+1}{n}}}
\right)
-
{\small{\frac{1}{n^2}}}
\\[6pt]
&=\frac{2n-2k+1}{n^2}\\[8pt]
\text{Hence}\;\;e &=\sum_{k=1}^n kp_k\\[4pt]
&=\sum_{k=1}^n k\left(\frac{2n-2k+1}{n^2}\right)\\[4pt]
&=\frac{1}{n^2}\left(\sum_{k=1}^n  k(2n-2k+1)\right)\\[4pt]
&=\frac{1}{n^2}\left(\sum_{k=1}^n (2n+1)k-2k^2)\right)\\[4pt]
&=\frac{1}{n^2}\left(\left(\sum_{k=1}^n (2n+1)k\right)-\left(\sum_{k=1}^n 2k^2)\right)\right)\\[4pt]
&=\frac{1}{n^2}\left(
\left((2n+1)\sum_{k=1}^n k\right)
-
\left(2\sum_{k=1}^n k^2)\right)
\right)\\[4pt]
&=\frac{1}{n^2}
\left(
(2n+1)
\left({\small{\frac{n(n+1)}{2}}}\right)
-2
\left({\small{\frac{n(n+1)(2n+1)}{6}}}\right)
\right)
\\[4pt]
&=\frac{(n+1)(2n+1)}{6n}\\[4pt]
\end{align*}
