[Math] Negative Hypergeometric Distribution expectation

Question

I am reading Introduction to Probability by Blitzstein and Hwang – Expectation. The book states :

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. For example, if we shuffle a deck of cards and deal them one at a time, the number of cards dealt before uncovering the first ace is a negative hypergeometric with $w=4,b=48$.

Finding the expected value of a negative hypergeometric r.v. directly from the definition results in very complicated sums of products. But the answer is very simple-looking: $b/(w+1)$.

Let us prove this using indicator r.v.s. Label the black balls as $1,2,3,\ldots,b$ and let $I_{j}$ be the indicator of the black ball $j$ being drawn before any white balls have been drawn. Then, $P(I_{j}=1)=1/(w+1)$, 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 balls $j$ being first in this list.

I know that, when sampling without replacement, the number of failures(drawing a black ball) until the first success(drawing a white ball) is a negative hypergeometric r.v.

But, why is $P(I_{j}=1)=1/(w+1)$? And why do we ignore the other black balls?

Answer 1

Just to give the question a formal answer (related to BGM's comments and Quasar's responses):

• The probability that black ball $b_i$ comes before all the white balls is $\frac1{w+1}$, since of those $w+1$ balls it is equally likely to come first, second, third, etc.
• The expected number of times that black ball $b_i$ comes before all the white balls is therefore also $\frac1{w+1}$, equal for all $i$
• Using linearity of expectation, the expected total number of black balls coming before all the white balls is then $\frac1{w+1}+\frac1{w+1}+\cdots+\frac1{w+1} = \frac{b}{w+1}$