Probability that a size $d$ sample will contain all $k$ colours present I tried looking for this question, but couldn't find it exactly... apologies if this is a repeat!
Imagine an urn with $m$ balls. Each ball has a different colour and there are $k$ colours (obviously, $m \geqslant k)$.
The distribution of colours is known. Specifically, there are $\alpha_{1}$ balls of colour $1$, $\alpha_{2}$ balls of colour $2$,..., and $\alpha_{k}$ balls of colour $k$. Therefore it must be true that $\alpha_{1} + \alpha_{2} + \cdots + \alpha_{k} = m$.
If I have $d$ choices, without replacement, what is the probability that all $k$ colours will be selected? Note that $d > k$.
Solving problems of this nature is simple with examples, but is there a general formula for calculating the probability of selecting all $k$ colours from a sample space of size $d$?
Thanks.
 A: You're after the multivariate hypergeometric distribution.
Summing over the appropriate PMF cases (or better yet the $k$ cases CDF of the possible zero cases) will get you the information you want.
A: Well, yes, more information is needed. Imagine the case where there are 1000 balls, with colours red and blue. The probabilities will be very different depending on the proportion of red to blue balls. Choosing say d=10 balls when there are 999 red balls, 1 blue gives only a 0.01 chance of getting both colours - if the urn holds 500 red and 500 blue then drawing 10 balls gives you very high chance, 0.998, of getting both colours.

(unless by "arbitrary" you mean that the number of balls of each colour is equal - but that isn't what the word means)
A: By the inclusion-exclusion principle, the probability that all colors are present after $d$ draws is $$\sum_{A\subseteq k}(-1)^{|A|}{\text{prob(all colors in set $A$ are missed after $d$ draws)}}.$$
For each set $A$ of colors, the probability that all colors in set $A$ are missed after $d$ draws is $$\frac{\left(\sum_{i\notin A}\alpha_i\right)_d}{(m)_d},$$
where $(j)_k$ denotes the Pochhammer symbol for a decreasing factorial, namely $j(j-1)(j-2)\cdots(j-k+1)$. 
So, your desired probability that all colors are present after $d$ draws is
$$\sum_{A\subseteq k}(-1)^{|A|}\frac{\left(\sum_{i\notin A}\alpha_i\right)_d}{(m)_d}.$$
