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The world population is ~7 billion

Social Classes:

  • x: Top .001% is ~70,000
  • y: Second .01% is ~700,000
  • z: Next .1% is ~7 million
  • r: Rest 99.9% is ~6,993,000,000 (billion)

  • p: population ~7,000,000,000

The odds of 1 person in the Rest grp meeting 1 person from the Top grp is it approx: 1 in 100,000 meets?

What are the odds of 1 person in the Rest grp meeting 2 people from the Top grp? is it: 1 in 100,000^2 or 1 in 10,000,000,000?

What are the odds of 1 person in the Rest grp meeting 2 people from the Top grp on the same year?

(Assuming in a lifetime of 100 years.)

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  • $\begingroup$ How big is the room? How do people enter it? $\endgroup$ – Henry Apr 24 '14 at 18:54
  • $\begingroup$ @Henry assume the room is infinite in size. They enter via means of teleport. Is my math correct? $\endgroup$ – nineyrold Apr 24 '14 at 18:54
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    $\begingroup$ your numbers must be wrong. 7 million is about .1% of 7 billion, not 0.001 %. This off by two zeroes error occurs for all of these, so i'm left questioning which value to use $\endgroup$ – Asimov Apr 24 '14 at 18:58
  • $\begingroup$ @JohnJPershing Thanks for pointing that out. I will correct that. Hang on. $\endgroup$ – nineyrold Apr 24 '14 at 19:00
  • $\begingroup$ @JohnJPershing I changed the question. Can you give your answer. Greatly appropriated. $\endgroup$ – nineyrold Apr 24 '14 at 19:28
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If your room contains all 7 billion people, the the probability of having at least 2 people from the top group, and at least 1 from the bottom group, is 100%.

If your room contains 3 people, chosen at random with replacement, then the probability that they consist of exactly 2 people from category A, and 1 from Category B, is $$3a^2b$$ where $a$ is the fraction of the world in category $A$, and $b$ is the fraction of the world in category $B$. The reason for the 3 is that the person from $B$ could be the first, second, or third person chosen; that is we are computing $aab+aba+baa$.

Now, technically we choose without replacement, but with the numbers involved the probability will be almost exactly the same.

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  • $\begingroup$ So is it correct to say this: 100,000,000,000*.001%^2*99.9%Statistically speaking you would need 100 billion people in 1 room for the odds of 1 person from rest group to be in the same room as 2 people from the top group? Is this correct? This isn't true if you put these 7 billion in a room. But I'm trying to look at it differently. I other words you would need to see 100 billion people and statistically only 2 of them will be in the top group. (So seeing the same person again counts as a see) $\endgroup$ – nineyrold Apr 24 '14 at 19:16

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