why does exact binomial confidence intervals have wider than nominal coverage? The exact $(1-\alpha)$ level confidence interval lower limit is given by 
$$
\sum_{k=y}^{n} {n \choose k} {p_L}^k(1-p_L)^{n-k}=\alpha/2
$$
and the upper limit analogously.
Why does the resulting C.I. has greater coverage than the nominal $(1-\alpha)$? I thought that the above equation has a valid solution for $p_L$ at any $\alpha$ and so should provide the correct coverage.

Because the distribution is discrete.  One might get
$$
\sum_{k=y}^n \cdots \ge \frac\alpha2 > \sum_{k=y+1}^n \cdots
$$
but there's nothing between $y$ and $y+1$ that you can put there to make it exactly $\alpha/2$, so you err on the side of caution and use the one that makes it more than $\alpha/2$.
