Removing balls from an urn in pairs: expected number of pairs in which both are red

Question

I am sorry that I cannot make the title more clear. The following is from Sheldon M. Ross: Introduction to Probability Models (11th Edition). I am able to reach the desired answer but actually I don't fully understand it. Please help.

Question:

An urn contains $2n$ balls, of which $r$ are red. The balls are randomly removed in $n$ successive pairs. Let $X$ denote the number of pairs in which both balls are red. Find $E[X]$ and Var$(X)$.

I then denote $X_i$ be a random variable which equals to $1$ if the $i$-th pair which $2$ red balls and $0$ if otherwise. Then, we have $$ E[X] = \sum_{i=1}^n {E[X_i]} = \sum_{i=1}^n{\frac{r(r-1)}{2n(2n-1)}}=\frac{r(r-1)}{2(2n-1)}$$ which is the correct answer.

But then I am not satisfied by the answer: why all $E[X_i], i=1,...,n$, are of the same value? Say, I am calculating $E[X_2]$. Then I should have $$E[X_2] = (1)P(X_2 = 1) +(0)P(X_2 = 0).$$ But shouldn't $P(X_2 = 1)$ depend on the outcome of the first pair?

On the other hand, I calculate Var$(X)$. The suggested solution gives the following: $$Var(X) = \sum_{i=1}^n Var(X_i) + 2 \sum \sum_{i<j} Cov(X_i,X_j)=(n)Var(X_1)+(n)(n-1)Cov(X_1,X_2).$$ I am not sure about the last equality, especially the existence of $n(n-1)$.

Thanks in advance!

Answer 1

As Didier pointed out in a comment, you can see that the probabilities are the same for each pair by noting that each pair contains two uniformly chosen balls, that is, since the balls are by definition identical (except for the colour, which doesn't influence their selection), in a given draw every pair of balls is equally likely to be drawn, and this determines all probabilities related to a single draw.

Regarding the variance, your calculation is correct (the factor $\binom n2$ is the number of pairs in the summation over the covariances), but I find it easier to calculate variances using expectation values, like this:

\begin{align} \operatorname{Var}(X)&=E[X^2]-E[X]^2\\ &=nE[X_1^2]+n(n-1)E[X_1X_2]-(nE[X_1])^2\\ &=nE[X_1]+n(n-1)E[X_1X_2]-(nE[X_1])^2\\ &=n\frac{\binom r2}{\binom{2n}2}+n(n-1)\frac{\binom r4}{\binom{2n}4}-n^2\left(\frac{\binom r2}{\binom{2n}2}\right)^2\;. \end{align}