probability distribution of coverage of a set after $X$ independently, randomly selected members of the set I have a set of numbers where I am randomly and independently selecting elements within a set .  After a number of these random element selections I want to know the coverage of the elements in the set. Coverage being how many elements from the set have been selected at least once divided by the total number of elements in the set.
To restate this: what is the probability distribution of the different coverage values on a set after $X$ randomly, independently selected elements of the set?
 A: The expected proportion of elements covered, $E\left(\frac{m}{n}\right)$, 
has a simple limiting form as $n \rightarrow \infty$ with the sampling rate $ r / n $ fixed.
Note that  $\lim_{n \rightarrow \infty} \left(1-\frac{1}{n}\right)^n = e^{-1}$, and rewrite:
$$\lim_{n \rightarrow \infty} E\left(\frac{m}{n}\right) = 1 - e^{-\frac{r}{n}}$$
so that for example sampling $r=n$ times is expected to cover about 63% of the set.  This is a reasonable approximation even for $n > 100$.
A: Derivation of $\operatorname E [M]$ with a classic use of indicator variables and total expectation:
We sample from set $X = \{1, \dots, n \}$. Let $X_i$ be $1$ if member $i$ is in our sample, $0$ otherwise.
For a sample of size $x$, the probability that none of the values are $i$ is $\left(\frac{n-1}{n}\right)^x$. Thus the probability that the sample includes $i$, and therefore $X_i = 1$, is $1-\left(\frac{n-1}{n}\right)^x$.
We have $$\operatorname E[M] = \operatorname E[X_1 + \dots + X_n] = \operatorname E[X_1] + \dots + \operatorname E[X_n] = n \left(1-\left(\frac{n-1}{n}\right)^x\right)$$
Variance is calculated similarly, but we have to consider separately $\operatorname E[X_i^2] $ and $\operatorname E[X_i X_j]$ for $ i \ne j$.
A: If there are $n$ elements of the set then the probability that $M=m$ have been selected after a sample of $x$ (with replacement) is  
$$\frac{S_2(x,m) \; n!}{n^x \; (n-m)!} $$
where $S_2(x,m)$ is a Stirling number of the second kind. 
The expected value of $M$ is: $n \left(1- \left(1-\dfrac{1}{n}\right)^x \right)$. 
The variance is: $n\left(1-\dfrac{1}{n}\right)^x  + n^2 \left(1-\dfrac{1}{n}\right)\left(1-\dfrac{2}{n}\right)^x - n^2\left(1-\dfrac{1}{n}\right)^{2x}. $
A: You can build a recurrence relation to construct the probability distribution using dynamic programming.
We define the recurrence p(m, x) as the probability of selecting m unique elements after x picks from a set of size n.
After 0 picks, we must have selected exactly 0 unique elements.
p(m=0, x=0) = 1
p(m!=0, x=0) = 0
Say the xth pick results in m unique elements selected.  Then either this last pick was already previously chosen (with probability m/n from the state (m, x-1)), or this pick adds a new mth unique value (with probability 1 - (m-1)/n) from the state (m-1, x-1))
So we set the recurrence to:
p(m,x) = p(m-1, x-1) * (1 - (m-1)/n) + p(m, x-1) * (m/n)
