What do you expect?

Here, we'll give a general statement: The expected value of the smallest element in an r-subset of an n-element set is given by $\frac {n+1}{r+1}$ .

Proof: The smallest element of an $r$ -subset can be anything from $1$ to $n-r+1$ . For any such integer $k$ in $[1,n-r+1]$ , we count the number of $r$ -subsets having $k$ as the least element. But this is easy; we simply need to choose $r-1$ other elements from the set ${k+1,...,n}$ , which can be done in $n-k \choose r-1$ ways.

Thus, the expected value to be found out is

$E = \frac {\displaystyle \sum_{k=1}^{n-r+1} k {n-k \choose r-1}}{{n \choose r}}$ . Let the numerator of this fraction be $N$ .

Now, notice that this step is true: $N =\displaystyle \sum_{k=1}^{n-r+1}{n-k \choose r-1}+\sum_{k=2}^{n-r+1}{n-k \choose r-1}+...+\sum_{k=n-r+1}^{n-r+1}{n-k \choose r-1}$ .

Thus $N ={n\choose r}+{n-1\choose r}+...+{r\choose r}$

$= {n+1 \choose r+1}$ (Try proving this step!)

Therefore $E= \frac {{n+1 \choose r+1}}{{n \choose r}} = \frac {n+1}{r+1}$ .

The answer to the given question is, therefore, $\frac {2016}{63} = 32$ . Hence $a=32, b=1 , a+b =33$ .