I am trying to figure out a solution to the following problem:

Let there be two groups of people, Group A and Group B. Group A represents x percent (e.g. 1%) of the world's population, and Group B represents y percent (e.g. 2%) of the world's population. What is the probability that a person from Group A will meet a person from Group B? Assume the following things:

  • The average human being meets z people (e.g. 100,000) in a lifetime.
  • The world's population is kept at a constant k people.
  • Everyone in the world was born and will die at the same time.


I came up with this question myself, but I'm not a mathematician, so please feel free to clean this up if need be. Also, if there is not enough information in the problem to solve it, add assumptions and please indicate the reasons for adding them. The assumptions I wrote are my attempt at making the problem easier. If they are not necessary, and removing any produces a more accurate answer, then I encourage the removal of them.

  • $\begingroup$ Are you asking about a particular member of Group A, or just any member of Group A? $\endgroup$ – Alex Becker Jun 16 '12 at 4:42
  • $\begingroup$ I am asking for the probability that any member from Group A will meet any member from Group B. $\endgroup$ – GRardB Jun 16 '12 at 4:45
  • $\begingroup$ Very good formulation of the problem from non-mathematician. $\endgroup$ – Norbert Jun 16 '12 at 6:53
  • $\begingroup$ Thank you, Norbert. I study computer science (undergrad), so I have a good bit of exposure to math problems, but probably not as much as most people on this site. $\endgroup$ – GRardB Jun 22 '12 at 2:48

Let $a$ be the number of people in group $A$ and $b$ be the number of people in group $B$. Then the chance that you never meet somebody from group $B$ is the chance that all your $z$ acquaintances are outside, $\frac {k-b}k\frac{k-b-1}{k-1} \ldots \frac{k-b-z+1}{k-z+1}=\frac {(k-b)!(k-z)!}{k!(k-b-z)!}$ If the chances that an $A$ individual meets a $B$ individual are independent, the chance they never meet is $\left(\frac {(k-b)!(k-z)!}{k!(k-b-z)!}\right)^a$. Although it is not obvious, this should be symmetric in $a$ and $b$.

  • $\begingroup$ The problem with your answer is that z was specified as number /100000 but the world's population was not specified. So if W=world's population then rz is the number that should replace z in your formula where r=W/100000. $\endgroup$ – Michael R. Chernick Jun 16 '12 at 14:29
  • $\begingroup$ @MichaelChernick: my reading was that world popluation is $k$ and $z$ was a constant. My $a$ is $xk$ and $b$ is $yk$ $\endgroup$ – Ross Millikan Jun 16 '12 at 15:00
  • $\begingroup$ Sorry I some how missed the line where he said that the world's population is held constant k. $\endgroup$ – Michael R. Chernick Jun 16 '12 at 16:37
  • $\begingroup$ Awesome answer! Thanks! $\endgroup$ – GRardB Jun 22 '12 at 2:47

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