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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?

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However, this seems rather cumbersome.

No, in your case, it's simple. $P_1=0$, $P_3=1-P_2$, $P_2=P_{2A}+P_{2B}+P_{2C}$ where $P_{2A}$ is probability of not having $A$ in picked counters, $P_{2A}=P_{2B}=P_{2C}$ (since letters are symmetrical) and finally $P_{2A}=\pmatrix{8 \\ 5} / \pmatrix{12 \\ 5} = {7 \over 99}$.

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