How Can There Still Be a Sex Difference, Even When There Is No Sex Difference?

Re:How Can There Still Be a Sex Difference, Even When There ... by ksrm on Tuesday July 07, 2015 @07:27PM (#9848825)

I don't understand why you are taking the advantage between the two options to be the difference between their probabilities, rather than the ratio. If I am 2% successful at X and you are only 1% successful, I am 0.02/0.01 = twice as successful as you, not 0.02-0.01 = 1% more successful.

As dlss has explained, it doesn't matter what constant you multiply by, the ratio is the same.

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Re:How Can There Still Be a Sex Difference, Even When There ... by logfromblammo on Wednesday July 08, 2015 @10:46AM (#9851841)

You are taking the ratio of P(B|~A) over P(B|A):

  P(A) = P(~A) = 0.5
  P(B|~A) / P(B|A) = ( P(~A|B) * P(B) / P(~A)) / ( P(A|B) * P(B) / P(A))
  = P(~A|B) / P(A|B)
  = ( 1 - P(A|B)) / P(A|B)

I don't understand why you consider this to be a meaningful number. It isn't even a probability any more; it's just a dimensionless ratio.

In your example, your result is 2. That alone tells me it cannot be a probability, as those numbers are restricted to the domain from zero to one.

If you partition the population by sex, the probability of randomly selecting a good engineer from the all-male group is only 0.8% higher than picking from the all-female group. That is the advantage you realize by selecting that criterion for your partition. If you are taking a ratio of conditional probabilities, you are no longer partitioning the probability space by that condition.

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