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

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 $n(n-1)$.

• Because each pair is made of two uniformly chosen balls, whether this is the first pair to be removed or any other pair.
– Did
Oct 5 '14 at 12:43

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:
• How did you go from first line to the second? I mean $E[X^2] = E[(\sum_{i=1}^n X_i)^2]$. I don't fully understand why all the $X_i$ are identical, I mean if you picked red balls then it should the future $X_i$. But even taking that as granted, how does $X_1$ and $X_2$ appear? Sep 24 '19 at 14:08
• @Aditya: Yes, if you picked red balls, it influences the future $X_i$, but that just means that the conditional quantities (probabilities and expectations) are different. The unconditional quantities (when we haven't picked any red balls yet) are the same. I tried to explain this in the first paragraph; if it's still not clear, please indicate what part of that explanation is unclear. About $X_1$ and $X_2$: The squared sum contains $n$ terms of the form $X_i^2$ and $n(n-1)$ terms of the form $X_iX_j$ with $i\ne j$, and these are all equal in expectation to $X_1^2$ and $X_1X_2$, respectively. Sep 28 '19 at 11:53