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Topic: Evidence based on intersection of two sets of rare cases
Replies: 7   Last Post: Nov 18, 2012 1:23 PM

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divergent.tseries@gmail.com

Posts: 40
Registered: 7/29/12
Re: Evidence based on intersection of two sets of rare cases
Posted: Nov 12, 2012 11:05 AM
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I looked at this in a somewhat primitive way, since I do not have real data to work with.

I asked, what is the probability of being in Termin's group and winning a Nobel in a science over the period involved given that the person chosen was highly intelligent.

I used your standard of 4sd for intelligence, but Termin had to use a lower value. I guesstimated a world population of 2 billion at the time for the average population and sampled the science winners of the Nobel finding about five people won per year over the time period involved. Teams were not common then.

I solved it as:
P(Nobel and Termin|Intelligent)=P(Intelligent|Nobel AND Termin)P(Nobel and Termin)/P(Intelligent)

I treated the probability of a person being intelligent given they both won the Nobel in a science and were in Termin's group as 1.

This simply raises the question of "is winning a nobel and being in Termin's group independent or not."

Logically, the probability of winning a Nobel being dependent upon being in Termin's group is zero. It isn't a consideration.

The probability, unconditioned upon intelligence, of being a Nobel prize winner over the period was 8.75*10^-8 and the probability of being in Termin's group worldwide and unconditioned upon intelligence was 5*10^-7. Being in the 4 sd or above is 3.167*10^-5.

So the probability of being in Termin's group and winning a Nobel, given that you are an intelligent person of the day, was only 1.38*10^-9

Given a sample size of only 1000, it is nearly certain that the expected number of winners is zero.




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