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. $Var[\mbox{number of red balls drawn}]$?

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
$$
\frac{2r(r-1)}{(b+1)(b+2)}+\frac{r}{b+1}-\left(\frac{r}{b+1}\right)^2
=\frac{br(b+r+1)}{(b+1)^2(b+2)}\tag{5}
$$
