I have made a calculation and now I do not understand what I did there.
It is about the following question: Imagine you have n cards of which there are 2 aces, what is the expectation value to get the first ace and what is the expectation value to get the 2nd ace? So you take one card in each step and the question is: What is the expectation value to get the first ace and what is the expectation value to get the second ace?
My calculation was: $$E(X)=\sum_{j=1}^{n-1} j\frac{(n-j)}{\frac{n}{2}(n-1)} = \frac{1}{3}(n+1)$$ and $$E(Y)=\sum_{j=2}^{n} j\frac{(j-1)}{\frac{n}{2}(n-1)} = \frac{2}{3}(n+1)$$
I noticed that in the demoniator we have $\frac{n}{2}(n-1)$ which is just the binomial coefficient for taking 2 out of n, but I don't know at which probability space I was looking and how I got to these two sums. Can anybody help me here? ( So it is perfectly clear how to evaluate those sums and get the right-hand side, but I have troubles to understand how I got those two sums).
You are asking for the expected position of the first or second ace in a shuffled deck of $n$ cards containing two aces. There are $n \choose 2$ ways to choose $2$ of the $n$ cards to be aces. Of these, if $1 \le j \le n-1$, $n-j$ have the first ace in position $j$ (the second being in any of the $n-j$ positions $j+1, j+2, \ldots, n$). Similarly, if $2 \le j \le n$, $j-1$ have the second ace in position $j$ (the first being in any of the $j-1$ positions $1, 2, \ldots, j-1$).