I'm not a mathematician but I have some applied statistics background. I have some data for which I want to calculate some expected values.
I have a box with 20 balls of 5 different colours, of varying frequencies (say 1 red, 2 blue, 4 yellow, 6 orange, 7 green). If I select 1 ball, I will have one colour. I'd like to know how many colours I should expect to have, on average, as I then pick out successive balls, i.e. how many colours would I expect to have if I pick another ball, and another ball, all the way up to 20 when I will obviously have 5 colours. I imagine the values from 2 - 19 will be fractions, as this is average expectation.
Either this has a really simple answer, or it's really complicated. I'd like to be able to generalise the solution to a box of any number of balls with any number of colours.
By linearity of expectation, the expected number of colours is the sum of the probabilities of each colour being selected. If you have $n_i$ balls of colour $i$ and a total of $n$ balls and you draw $k$ balls without replacement, this is
$$ \sum_i\left(1-\frac{\binom{n-n_i}k}{\binom nk}\right)\;. $$