Derivation of the Negative Hypergeometric distribution's expected value using indicator variables I'm trying to understand how to derive the Negative Hypergeometric's expected value using indicator variables. Note, in the problem below, we are only interested in the expected value before the first success:

An urn contains $w$ white balls and $b$ black balls, which are randomly drawn one by one without replacement. The number of black balls drawn before drawing any white balls has a Negative Hypergeometric distribution. Find this expected value.

This is the solution from the textbook (Introduction to Probability by Blitzstein and Hwang):

Let us prove this using indicator r.v.s. Label the black balls as $1, 2, . . . , b$, and let $I_j$ be the indicator of black ball $j$ being drawn before any white balls have been drawn.
Then $\mathbb{P}(I_j = 1) = \displaystyle\frac{1}{w + 1}$ since, listing out the order in which black ball $j$ and the white balls are drawn (ignoring the other balls), all orders are equally likely by symmetry, and $I_j = 1$ is equivalent to black ball $j$ being first in this list.
So by linearity,
$$\mathbb{E}\left(\sum_{j=1}^b I_j\right)=\sum_{j=1}^b \mathbb{E}(I_j)=\frac{b}{w+1}$$

What I don't get is why the probability of the indicator variable is $\mathbb{P}(I_j = 1) = \displaystyle\frac{1}{w + 1}$.
Shouldn't it be $\displaystyle\frac{b}{w+b}$?
Since there are $b$ black balls and $b+w$ total balls.
But then if I use this, I think the expected value I get is Hypergeometric and not the Negative. Also, in the solution it says to ignore the other balls and I am not quite sure why we can do that.
 A: Call the one particular black ball we're concerned with $B$.  Then the orders in question are: $$ \begin{array}{cccccccccl} W & W & W & \cdots & W & W & W & B & & w!\text{ different orders} \\ W & W & W & \cdots & W & W & B & W & & w!\text{ different orders} \\ W & W & W & \cdots & W & B & W & W & & w!\text{ different orders} \\ \vdots & \vdots & \vdots & & \vdots & \vdots & \vdots & \vdots \end{array} $$
The question is: Why are the events identified by the rows of this array equally probable?
And why can we ignore the other black balls?
And are these really two different questions?
In each row, there are $w!$ different orders in which the white balls can appear, and there are $(b-1)!$ different orders in which the black balls can appear, thus $w!(b-1)!$ different outcomes corresponding to each one row of this array.  The fact that the number of outcomes corresponding to every row is the same for every row is the reason why all the rows are equally probable events.  Since there are $w+1$ rows, each row has probability $1/(w+1)$.
Perhaps one should say $w!(b-1)!$ different orders in each row rather than saying "ignore the other black balls".  Saying "ignore the other black balls" really just means there are equallly many orders in which the other black balls can appear in each row of the array.  In the same way, the orders in which the white balls appear in each row of the array was in effect "ignored" because the number of such orders is the same in every row.
Let's try a concrete example with two balls of each color, called $W_1,W_2,B_1,B_2$, and the particular black ball we're concerned with is $B_1$.  The array above then looks like this:
$$
\begin{array}{ccc}
W & W & B \\
W & B & W \\
B & W & W
\end{array}
$$
So there is a $1/3$ chance that that $B_1$ appears before any white ball.  Now look at the list of all $4!=24$ orders
$$
\begin{array}{c}
\left.\begin{array}{cccc}
W_1 & W_2 & B_1 & B_2 \\
W_1 & W_2 & B_2 & B_1 \\
W_1 & B_2 & W_1 & B_1 \\
B_2 & W_1 & W_2 & B_1 \\
W_2 & W_1 & B_1 & B_2 \\
W_2 & W_1 & B_2 & B_1 \\
W_2 & B_2 & W_1 & B_1 \\
B_2 & W_2 & W_1 & B_1
\end{array} \right\} \text{ This is } W\ W\ B \\[6pt]
\left.\begin{array}{cccc}
W_1 & B_1 & W_2 & B_2 \\
W_1 & B_1 & B_2 & W_2 \\
W_1 & B_2 & B_1 & W_2 \\
B_2 & W_1 & B_1 & W_2 \\
W_2 & B_1 & W_1 & B_2 \\
W_2 & B_1 & B_2 & W_1 \\
W_2 & B_2 & B_1 & W_1 \\
B_2 & W_2 & B_1 & W_1
\end{array} \right\} \text{ This is } W\ B\ W \\[6pt]
\left.\begin{array}{cccc}
B_1 & W_1 & W_2 & B_2 \\
B_1 & W_1 & B_2 & W_2 \\
B_1 & B_2 & W_1 & W_2 \\
B_2 & B_1 & W_2 & W_1 \\
B_1 & W_2 & W_1 & B_2 \\
B_1 & W_2 & B_2 & W_1 \\
B_1 & B_2 & W_2 & W_1 \\
B_2 & B_1 & W_2 & W_1
\end{array} \right\} \text{ This is } B\ W\ W
\end{array}
$$
