Imprecise statement of the question
You are given a very large positive integer $n$ and a set $X$ with $n$ elements. You pick at random a map $f:X\to X$, and you get $1/n$ dollar for each element of $X$ you hit with $f$ (that is, for each element $y\in X$ which is of the form $f(x)$).
What is, approximately, your expected return?
Precise statement of the question
Let $n$ be a positive integer and $X$ a set with $n$ elements. Set $$ a_n:=n^{-n-1}\sum_{f:X\to X}|f(X)|, $$ where the sum runs over all the maps $f:X\to X$, and $|f(X)|$ is the number of elements in the image $f(X)$ of $X$. This defines a sequence of rational numbers in the interval $(0,1)$.
Does the limit $$\lim_{n\to+\infty}a_n$$ exist? If it does, what is its value?
Some observations
The question can be expressed in terms of Stirling numbers of the second kind as follows.
Let again $X$ be our set with $n$ elements, and let $k$ be an integer with $1\le k\le n$.
To choose a map $f:X\to X$ with $|f(X)|=k$, we may first choose a subset $f(X)$ of size $k$ of $X$, and then choose a surjection $X\to f(X)$, $x\mapsto f(x)$.
There are $\binom{n}{k}$ options for the first choice, and $k!\genfrac\{\}{0pt}{2}{n}{k}$ for the second, where $\genfrac\{\}{0pt}{2}{n}{k}$ is the Stirling numbers of the second kind attached to the couple $(n,k)$, so that there are $$ k!\ \binom nk\ \begin{Bmatrix}n\\ k\end{Bmatrix}=\frac{n!}{(n-k)!}\ \begin{Bmatrix}n\\ k\end{Bmatrix} $$ maps $f:X\to X$ with $|f(X)|=k$, and we get $$ a_n:=\frac{n!}{n^{n+1}}\sum_{k=1}^n\ \frac k{(n-k)!}\ \begin{Bmatrix}n\\ k\end{Bmatrix}. $$ The numbers $a_2,a_3,\ldots,a_7$ are approximately equal to $$ 0.75,\ 0.7037037037,\ 0.68359375,\ 0.67232,\ 0.66510202332,\ 0.660083. $$ I used WolframAlpha to compute them, as in this link.
An obvious guess is that we have $a_n\ge\frac12$ for all $n\ge1$, and that the sequence decreases. This would imply the existence of a limit. A slightly less obvious guess is that the limit is $\frac12$.
You hit a given element in $X$ with a probability of $$a_n=1-\left(\frac{n-1}{n}\right)^n.$$ The expected size of the image is thus $na_n$ and the expected winning is $a_n$. This is just linearity of expectation.
Therefore $$\lim_{n\to\infty}a_n=1-e^{-1}.$$
We can get an asymptotic expansion: $$n\ln\frac{n-1}n=-1-\frac1{2n}-\frac1{3n^2}+O(n^{-3})$$ so that $$\left(\frac{n-1}n\right)^n=e\exp\left(-\frac1{2n}-\frac1{3n^2}+O(n^{-3})\right) =e\left(1-\frac1{2n}-\frac{5}{24n^2}+O(n^{-3})\right)$$ so that $$a_n=1-e+\frac{e}{2n}+\frac{5e}{24n^2}+O(n^{-3}).$$