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Suppose there is a box with N balls, k white and (N-k) black. We draw two balls, one after the other. Thus, 4 possible cases are possible: (white, white), (white, black), (black, white), (black, black).

  1. (white, white) – in each of two draws, balls are replaced. For N balls (k white, (N-k) black): $$P(white \wedge white) = {\left(\frac{k}{N}\right)}^2$$
  2. (white, black) – the special case, discussed later.
  3. (black, white) – in each of two draws, balls are replaced. For N balls (k white, (N-k) black): $$P(black \wedge white) = \frac{N-k}{N} \frac{k}{N}$$
  4. (black, black) – in each of two draws, balls are replaced. For N balls (k white, (N-k) black) probability of the event is: $$P(black \wedge black) = {\left(\frac{N-k}{N}\right)}^2$$

For the case 2. – (white, black) – black ball is not returned. It works like if white ball would "allocate" next black ball preventing its return to box. (The white ball is still replaced, of course).

Also: white balls are replaced, but the experiment stops if k white balls (i.e. all) have been drawn in any of the 4 configurations.

Question: how many the special (white, black) events are expected to occur? The experiment obviously stopps after whichever comes first:

  • drawing white ball k times (in any of the 4 configurations),
  • moving all (N-k) black balls outside box.
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So if I understand correctly the first time around P(white^black) is also (N-k)k/(N)$^2$ but after the white black sequence there are still k white ball but only N-k-1 black balls. –  Michael Chernick May 31 '12 at 1:42
    
@MichaelChernick: yes exactly –  Calikin May 31 '12 at 12:09
    
This is an interesting problem. I will try to work it out. But given that I am busy someone else will probably beat me to it. –  Michael Chernick May 31 '12 at 14:15
    
@Calkin This is all interesting as far as understanding why the urn works the way it does. But the problem is already well-defined mathematically, So it is just a matter of look at the possibilities and drop blacks for subsequent probability calculations based on the W B pair occurring. –  Michael Chernick May 31 '12 at 19:41

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