Variance for the number of trials before success in an urn problem without replacement 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. $Var[\mbox{number of red balls drawn}]$?
 A: 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.
A: The probability of drawing $n$ red balls then a blue ball is
$$
\frac{r}{b+r}\frac{r-1}{b+r-1}\frac{r-2}{b+r-2}\cdots\frac{r-n+1}{b+r-n+1}\frac{b}{b+r-n}
=\frac{\binom{\vphantom{b+}r}{n}}{\binom{b+r-1}{n}}\frac{b}{b+r}\qquad\tag{1}
$$
Since a blue ball must be drawn eventually, $(1)$ indicates that
$$
\sum_{n=0}^r\frac{\binom{\vphantom{b+}r}{n}}{\binom{b+r-1}{n}}=\frac{b+r}{b}\tag{2}
$$
which can be proven by induction on $r$.

$(2)$ is true for $r=0$. Assume $(2)$ is true for $r-1$, then
  $$
\begin{align}
\sum_{n=0}^r\frac{\binom{\vphantom{b+}r}{n}}{\binom{b+r-1}{n}}
&=1+\sum_{n=1}^r\frac{\frac{\vphantom{b+}r}{n}\binom{r-1}{n-1}}{\frac{b+r-1}{n}\binom{b+r-2}{n-1}}\\
&=1+\frac{r}{b+r-1}\sum_{n=0}^{r-1}\frac{\binom{r-1}{n}}{\binom{b+r-2}{n}}\\
&=1+\frac{r}{b+r-1}\frac{b+r-1}{b}\\
&=\frac{b+r}{b}
\end{align}
$$

There is a more direct proof of this identity at the end of this answer.
Using $(2)$, the expected number of red balls drawn is
$$
\begin{align}
\sum_{n=0}^r\frac{\binom{\vphantom{b+}r}{n}}{\binom{b+r-1}{n}}\frac{b}{b+r}n
&=\frac{br}{b+r}\sum_{n=0}^r\frac{\binom{r-1}{n-1}}{\binom{b+r-1}{n}}\\
&=\frac{br}{b+r}\sum_{n=0}^r\frac{\binom{r\vphantom{+1}}{n}-\binom{r-1}{n}}{\binom{b+r-1}{n}}\\
&=\frac{br}{b+r}\left(\frac{b+r}{b}-\frac{b+r}{b+1}\right)\\
&=\frac{r}{b+1}\tag{3}
\end{align}
$$
which agrees with the result in the question.
Using $(2)$ the expected value of $n(n-1)$ is
$$
\begin{align}
\sum_{n=0}^r\frac{\binom{\vphantom{b+}r}{n}}{\binom{b+r-1}{n}}\frac{b}{b+r}n(n-1)
&=\frac{br(r-1)}{b+r}\sum_{n=0}^r\frac{\binom{r-2}{n-2}}{\binom{b+r-1}{n}}\\
&=\frac{br(r-1)}{b+r}\sum_{n=0}^r\frac{\binom{r\vphantom{-1}}{n}-2\binom{r-1}{n}+\binom{r-2}{n}}{\binom{b+r-1}{n}}\\
&=\frac{br(r-1)}{b+r}\left(\frac{b+r}{b}-2\frac{b+r}{b+1}+\frac{b+r}{b+2}\right)\\
&=\frac{2r(r-1)}{(b+1)(b+2)}\tag{4}
\end{align}
$$
Thus, the variance, which is the mean of the squares minus the square of the mean is
$$
\frac{2r(r-1)}{(b+1)(b+2)}+\frac{r}{b+1}-\left(\frac{r}{b+1}\right)^2
=\frac{br(b+r+1)}{(b+1)^2(b+2)}\tag{5}
$$
A: $Z_i$ is the indicator that red ball $i$ is drawn before the first blue ball.  (That all blue balls are drawn after it.)
Among the set of $b+1$ balls that are: "all blue balls plus red ball $i$", each one is equally likely to be drawn first, so the probability for our favoured event happens is $1/(b+1)$.$$\mathsf E(Z_i)=\dfrac 1{b+1}\\\mathsf E(\sum_{i=1}^rZ_i)=\dfrac{r}{b+1}$$
Similarly, by considering the set of $b+2$ balls that are: "all blue balls, plus two distinct red balls $i$ and $j$", we may derive the probability for both these red balls being before the blue balls.  Then use this $\mathsf E(Z_iZ_j)$ to evaluate:
So  $$\begin{align}\mathsf{Var}(Z) &= \mathsf E(Z^2)-\mathsf E(Z)^2\\[1ex] &=\sum_{i=1}^r\mathsf E(Z_i^2)+2\sum_{i=2}^{r}\sum_{j=1}^{i-1}\mathsf E(Z_iZ_j)~-\mathsf E(Z)^2\\[1ex]&=\dfrac r{b+1}+\dfrac{2~r~(r-1)}{(b+2)(b+1)}~-\dfrac{r^2}{(b+1)^2}\\[1ex]&=\dfrac{b r (b + r + 1)}{(b + 1)^2 (b + 2)}\end{align}$$
