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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?

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    $\begingroup$ The complete draw without replacement is like a permutation of those balls. You may think the possible realizations is like $B_3W_9W_7B_8W_5\ldots$ etc. First you pick out the particular black ball $j$ and all the white balls, forming a group with $w+1$ balls. All permutation are equally likely, so the probability that black ball $j$ being the no 1 is $1/(w+1)$. And then note that the insertion of other black balls does not affect the relative position of this black ball $j$ to those white balls. $\endgroup$ – BGM Jan 12 '17 at 4:29
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    $\begingroup$ Or think it in the other way round, you first sample the whole bag of balls and form a queue/permutation. Then note that you can swap the black ball $j$ with each of the white ball, and each resulting queue correspond to different relative position of the group of black ball $j$ and those white balls, and they are equally likely. $\endgroup$ – BGM Jan 12 '17 at 4:32
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    $\begingroup$ Related? Intuitive/heuristic explanation of Polya's urn $\endgroup$ – BCLC Jan 12 '17 at 5:06
  • $\begingroup$ @BGM, okay I understand that the probability of black ball $j$ being first in a list of $w$ white balls and $1$ black ball is $1/(w+1)$. $\endgroup$ – Quasar Jan 13 '17 at 2:26
  • $\begingroup$ @BGM, thanks for the clarification. To continue further, by symmetry the above holds for all of the black balls from $1,2,3,\ldots,b$. $\endgroup$ – Quasar Jan 13 '17 at 2:46
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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 $\frac1{w+1}+\frac1{w+1}+\cdots+\frac1{w+1} = \frac{n}{w+1}$
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