Suppose I have $12$ counters consisting of $4$ identical copies of the letters $A$, $B$ and $C$. I now pick $5$ counters arbitrarily, each counter being equally likely to be chosen.
If I wanted to calculate the expected number of distinct letters, I am thinking of going about doing the following computation.
$$E \left [x \right ] = (1\times P_1) + (2\times P_2) + (3 \times P_3)$$
where $P_i$ is the probability of having $i$ distinct letters, where $i=1,2,3$.
However, this seems rather cumbersome. Is there perhaps a faster way of computing the expected number of distinct letters, assuming my method above is even correct?