This question is asked as an extension of:
Expectation of number of trials before success in an urn problem without replacement
(Note: I am not the author of the original question.)
We have $b$ blue balls and $r$ red balls in an urn. Sampling the urn sequentially and without replacement, we remove red balls until we select a blue ball. In the previous question, Byron Schmuland (and others) calculated that the expected number of balls drawn before drawing a blue ball should be:
$\mathbb{E}(\mbox{number of balls drawn})=1+\sum_{i=1}^r \mathbb{E}(Z_i)=1+r\left({1\over b+1}\right)$
And therefore, we have that:
$\mathbb{E}(\mbox{number of red balls drawn})=1+\sum_{i=1}^r \mathbb{E}(Z_i)-1=1+r\left({1\over b+1}\right)-1$
My question is:
What is the variance for the number of red balls drawn, i.e. $Var[\mbox{number of red balls drawn}]$?
Best Answer
Edit: The indicator random variables have been radically changed, so that one gets a very quick computation of the mean and variance.
Let $Y$ be the number of reds drawn before the first blue. Suppose that the red balls have labels $1, 2, 3,\dots,r$. Let $X_i=1$ if red ball with label $i$ is drawn before the first blue is drawn, and let $X_i=0$ otherwise.
Then $Y=X_1+\cdots+X_r$. Note that the number of draws up to an including the first blue is $Y+1$. But $Y+1$ and $Y$ have the same variance.
To calculate the variance of $X_i$, we first calculate the mean. By linearity of expectation we have $$E(Y)=E(X_1)+E(X_2)+\cdots+E(X_r).$$ By symmetry, all the $E(X_i)$ are the same. The probability red with label $i$ comes before any of the $b$ blue is $\frac{1}{b+1}$. It follows that $E(Y)=\frac{r}{b+1}$.
To calculate the variance of $Y$, calculate $E(X_1+\cdots +X_{r})^2$ and subtract the square of $E(Y)$, which we know.
To find $E(X_1+\cdots+X_r)^2$, expand the square and use the linearity of expectation. We know the expectation of $\sum X_i^2$, since $X_i^2=X_i$. So we need the expectations of the cross terms.
For $i\ne j$, $X_iX_j=1$ if both red ball $i$ and red ball $j$ come before any blue. This has probability $\frac{2}{(b+2)(b+1)}$. Multiply by $2\binom{r}{2}$ to get the sum of the cross terms.