Likelihood of sampling each types from a population comprising many types? I have a total population size of 10,000 comprising 50 unique types which I assume are equally represented, i.e. their are 200 individuals per unique type in my population. 
 - How many samples would I need to take (without replacement) to ensure that I was x% (100%) confident of obtaining at least one representative of each unique type?
 - How many unique types could I expect to obtain if I only sampled 500 individuals (without replacement)?
 A: A partial answer (for the first part of the question):
Let $n$ be the size of the sample. Then there are $\left(\begin{matrix} 10000 \\ n \end{matrix}\right)$ possible samples, all equally likely.
Let $E_1$ be the event that type $1$ is unrepresented. There are $\left(\begin{matrix} 9800 \\ n \end{matrix}\right)$ possible samples in which type $1$ is unrepresented, so:
$$P(E_1) = \frac{\left(\begin{matrix} 9800 \\ n \end{matrix}\right)}{\left(\begin{matrix} 10000 \\ n \end{matrix}\right)}$$
Call this probability $p_1$. Similarly for all $2\le k\le 50$ we have $P(E_k) = p_1$ where $E_k$ is the event that type $k$ is unrepresented.
Now we are interested in the probability that at least one type is unrepresented (since that is $1$ minus the probability that every type is represented). Therefore would like to know:
$$P(E_1 \cup E_2 \cup … \cup E_{50})$$ 
For this we can use a generalisation of the formula $P(A\cup B) = P(A)+P(B)-P(A\cap B)$. This generalisation is the inclusion-exclusion principle as applied to probability, see the following article:
https://en.wikipedia.org/wiki/Inclusion-exclusion_principle#In_probability
In this case we can use the fact that the probability that any two types are both unrepresented, for example $P(E_1 \cap E_2)$, is equal to
$$p_2 = \frac{\left(\begin{matrix} 9600 \\ n \end{matrix}\right)}{\left(\begin{matrix} 10000 \\ n \end{matrix}\right)}$$
And in general, the probability that any $t$ types are all unrepresented is
$$p_t = \frac{\left(\begin{matrix} 10000-200t \\ n \end{matrix}\right)}{\left(\begin{matrix} 10000 \\ n \end{matrix}\right)}$$
Therefore we can use the special case of the inclusion-exclusion principle in which the probability of the intersection depends only on the cardinality of the intersection (i.e. on what we are calling $t$). In that special case the formula for probability of the union is:
$$P(E_1 \cup E_2 \cup … \cup E_{50}) = \sum_{t=1}^{50}(-1)^{t-1}\left(\begin{matrix} 50 \\ t \end{matrix}\right)p_t$$ 
Therefore the probability of obtaining at least one representative of each type is
$$1-\sum_{t=1}^{50}(-1)^{t-1}\left(\begin{matrix} 50 \\ t \end{matrix}\right)p_t$$ 
Where
$$p_t = \frac{\left(\begin{matrix} 10000-200t \\ n \end{matrix}\right)}{\left(\begin{matrix} 10000 \\ n \end{matrix}\right)}$$
We could then use trial and improvement to find the value of $n$ that would ensure $x$ percent confidence. 
