I have been thinking on this one for a while, and I'm not quite sure the best way to approach it. I feel I'm missing something obvious, but I'm just a novice in probability, and getting a bit of a push along the way would help.
Suppose we have a deck of cards with 3 sets of the values 1 through 15 (45 cards total). If we draw 9 cards from the deck without replacement, what is the expected number of unique values drawn?
For the sake of brevity, let us call the number of unique values n. I have been attempting to solve this problem by way of calculating P(X=n) for each n between 3 and 9. There are trivial cases, for instance where the number of unique values is 9 or 3, but other values quickly gives way to a lot of combinatorics for the repeated values, and I feel that this is more work than is truly required to find E[n].
Is there something that I am missing that simplifies this calculation, such as a property of normal distributions I have overlooked, or perhaps a fast method to do this iterated expectation? Or is this lengthy method the only valid approach?
Hints: The linearity of expectation implies that $E[X+Y]=E[X]+E[Y]$
Define the indicator random variables $X_i = \begin{cases} 1&\text{if}~i~\text{was among the numbers drawn}\\0&\text{otherwise}\end{cases}$
Let $X=\sum\limits_{i=1}^{15} X_i$
What does $X$ represent then in regards to your problem?