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Assume $b$ black balls and $w$ white balls. Throw them randomly and uniformly into $m$ bins. What is the expected number of collisions between black and white balls: (a) in each bin, (b) in all bins.

thanks

John Hughes
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user142868
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    What have you tried so far? Also: if a bin contains one white ball and two black balls, is that one collision or two? – John Hughes Apr 20 '20 at 15:51
  • If a bin contains one white ball and two black balls, then the number of collisions is $2$. – user142868 Apr 20 '20 at 15:52
  • I am thinking the following: $\frac{w}{m} \frac{b}{m}$ for the case of each bin and $\frac{wb}{m}$ for all the bins collectively. – user142868 Apr 20 '20 at 16:02
  • For $b$ you can use linearity with the answer from $a$. If there are four black balls in a bin is that six collisions? – Ross Millikan Apr 20 '20 at 16:15
  • What is a collision? Is the number of collisions just the number of equally-colored balls in a bin minus one (i.e., 1 is no collision, 2 is 1 collision, ..., 42 is 41 collisions)? – vonbrand Apr 20 '20 at 16:16
  • Your answer in the comment is correct. Can you justify it? Think linearity. – Ross Millikan Apr 20 '20 at 16:25
  • If there are $4$ black balls and one white, then the number of collisions is $4$. – user142868 Apr 20 '20 at 16:54

1 Answers1

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By assuming $m$ bins, the probability of any two balls of different color to collide is:

$$\frac{1}{m}.$$

We expect $\frac{w}{m}$ and $\frac{b}{m}$ of white and black balls, repsectively, in each bin. Therefore, the expected number $X$ of collisions in each bin is: $$E[X]=\frac{w \: b}{m^2}.$$ The expected number of collisions $Y$ in all bins is: $$E[Y] = E[X_1+X_2+\ldots+X_m].$$

Using linearity of expectation, $$E[Y] = E[X_1] +E[X_2]+\ldots+E[X_m] = \sum_{j=1}^{m} E[X_j] = m \frac{w \: b}{m^2} = \frac{w \: b}{m}.$$

user142868
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