Solution:

Answer: $\frac{2013^{2013}-2012^{2013}}{2013^{2012}}$

For each $n \in \{1,2, \ldots, 2013\}$, let $X_{n}=1$ if $n$ appears in $\left\{a_{1}, a_{2}, \ldots, a_{2013}\right\}$ and $0$ otherwise. Defined this way, $\mathrm{E}\left[X_{n}\right]$ is the probability that $n$ appears in $\left\{a_{1}, a_{2}, \ldots, a_{2013}\right\}$.

Since each $a_{i}$ ($1 \leq i \leq 2013$) is not $n$ with probability $\frac{2012}{2013}$, the probability that $n$ is none of the $a_{i}$'s is $\left(\frac{2012}{2013}\right)^{2013}$, so $\mathrm{E}\left[X_{n}\right]$, the probability that $n$ is one of the $a_{i}$'s, is $1-\left(\frac{2012}{2013}\right)^{2013}$.

The expected number of distinct values in $\left\{a_{1}, \ldots, a_{2013}\right\}$ is the expected number of $n \in \{1,2, \ldots, 2013\}$ such that $X_{n}=1$, that is, the expected value of $X_{1}+X_{2}+\cdots+X_{2013}$.

By linearity of expectation,
$$\mathrm{E}\left[X_{1}+X_{2}+\cdots+X_{2013}\right]=\mathrm{E}\left[X_{1}\right]+\mathrm{E}\left[X_{2}\right]+\cdots+\mathrm{E}\left[X_{2013}\right]=2013\left(1-\left(\frac{2012}{2013}\right)^{2013}\right)=\frac{2013^{2013}-2012^{2013}}{2013^{2012}}.$$