Probability of sampling all elements of an unknown set Suppose that you are drawing elements randomly and uniformly from a collection of unknown size. What is the probability that after drawing $t$ samples from this collection that you have seen all of the samples at least once? The set of sample you have already drawn is denoted as set $A$ and that the number of times a particular sample $a$ has been drawn is given by $f(a)$.
As an additional bound if necessary, you may also assume that the maximum size of the collection could be is given by $m$ (though it may be smaller).
I thought I had a solution and it appeared that the solution was unbounded without $m$, but I don't think it works after doing some experimentation.
$$
\sum_{k=1}^m (\frac{k}{|A| + k})^t
$$
 A: I will give you a partial answer and you can perhaps continue this line of reasoning on your own.
The main observation to make is how many unique elements you have in your $t$ samples. If, for example, you take $10$ samples and they are all the same, you intuitively understand that it's quite probable the set is just one element.
Let's denote the state of taking $t$ samples with $u$ unique values in them as $(t,u)$ where $1\le u \le t$. Also let's name the unknown size of our set as $n$. Let's also assume that the upper bound for $n$ is $m$
What you want to find is $P(n=u|(t,u))$. Employing Bayes' rule we get:
$$P(n=u|(t,u)) = \frac{P((t,u)|n=u)\cdot P(n=u)}{P(t,u))} =\frac{P((t,u)|n=u)\cdot P(n=u)}{\sum_{i=u}^{i=m} P((t,u)|n=i)\cdot P(n=i) }$$
First note that there is an issue with the prior probabilities $P(n=i)$. We should have some information about them. If we don't, we can assume them to be equiprobable, but that is an assumption. The problem statement needs to provide you with this information. Furthermore, we immediately see the problem if $m$ is unbounded, as these prior probabilities will become $0$ (under our equiprobable assumption). For all practical purposes though we can set $m$ to be any very large finite number and we'll be ready to roll.
Let's investigate the case where $u=1$
$$P(n=1|(t,1)) =\frac{P((t,1)|n=1)\cdot P(n=1)}{\sum_{i=1}^{i=m} P((t,1)|n=i)\cdot P(n=i) }$$
Notice that $P((t,1)|n=1)=1$. Also $P(n=i)=\frac{1}{m}$ assuming that the prior probabilities are equiprobable. The only thing we need to figure out is $P((t,1)|n=i)$ where $i>1$. What's the probability that we've drawn the same element again and again $t$ times, given that we have $n=i$ distinct elements?
Assuming a uniform random selection, it is simply $\left(\frac{1}{i}\right)^{t-1}$. So we get:
$$P(n=1|(t,1)) =\frac{\frac{1}{m}}{\sum_{i=1}^{i=m} \left(\frac{1}{i}\right)^{t-1}\cdot \frac{1}{m} } = \frac{1}{\sum_{i=1}^{i=m} \left(\frac{1}{i}\right)^{t-1}}$$
Hmm, I am now thinking whether we can generalise the solution for $m=\infty$ given that the priors are cancelling out. Maybe yes, but I am not sure about this step. If we accept $m=\infty$ the solution can be written in a simplified form: $P(n=1|(t,1)) = \frac{1}{\zeta(t-1)}$, where $\zeta()$ is the Riemann-zeta function and $t>2$. For $t=2$ this probability seems to be $0$ when $m=\infty$ which seems counterintuitive to me. Of course for a finite value of $m$ the probability is non zero (for example, for $m=100$ $P(n=1|(2,1)) \approx 0.193$ 
Here's what the probability $P(n=1|(t,1))$ equals for some values of $t>2$ for $m=100$ or $m=\infty$
$$
\begin{array}{c|c|c}
t & m=100 & m=\infty \\
\hline
3 & 0.612 & 0.608 \\ 
4 & 0.832 & 0.832 \\
5 & 0.924 & 0.924 \\
10 & 0.998 & 0.998 \\
20 & 0.99999 & 0.99999 \\
\end{array}
$$
As you notice the differences are negligible between the different $m$ for $t>3$.
You could perhaps try to work out what $P(n=t|(t,t))$ is equal to, and from there you can try the general case $P(n=u|(t,u))$. For the general case, I believe you'll need $f(a)$ (the function that gives you the repetitions for each element of your sample), but I have not tried it myself. 
A: If the size of the collection is $n$, the probability that in $t$ random extractions with replacement all elements have been drawn is
$$ P(E(t) \mid n ) = \frac{{t \brace n} n!}{n^t} \tag{1}$$ 
where ${t \brace n} $ is the Stirling number of the second kind.
If you don't know $n$, then you should at least assume an apriori $p(n)$, and then
$$ P(E(t)) = \sum_n P(E(t) \mid n ) \, p(n) \tag{2}$$
For example, if you assume a uniform distribution between $1$ and  $m$, then
$$ P(E(t)) = \frac{1}{m} \sum_{n=1}^m P(E(t) \mid n ) =  \frac{1}{m} \sum_{n=1}^m \frac{{t \brace n} n!}{n^t} \tag{3}$$
