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In probability, assume I have a set of n labeled balls [1..n]. I pick two balls randomly and uniformly without returning them.

Assume I have two random variables X and Y such as: X represents the selection of the first ball and Y represents the selection of the second ball.

now, I know that the expectancy of Y (E[y])) is (N+1)/2, however I cannot seem to prove it. Using the law of total expectation how can I show this?

Thank you

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A simple way to do it is to note that $\Pr(Y=k)=\frac{1}{n}$ for $1\le k\le n$. Thus $$E(Y)=\frac{1}{n}\left(1+2+\cdots+n\right)=\frac{1}{n}\cdot \frac{n(n+1)}{2}.$$ If we really want to, we can condition on $X$. We have $\Pr(X=k)=\frac{1}{n}$. If $X=k$, then for $Y$ we are picking with proabability $\frac{1}{n-1}$ from everybody but $k$. Thus $$E(Y|X=k)=\frac{1}{n-1}\left(\frac{n(n+1)}{2}-k\right).$$ Sum from $1$ to $n$, and divide by $n$. We get $$E(Y)=E(E(Y|X))=\frac{1}{n}\cdot \frac{1}{n-1}\left(n\frac{n(n+1)}{2}-\frac{n(n+1)}{2}\right).$$ This simplifies to $\frac{n+1}{2}$.

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