Let's assume male heights are distributed according to a pdf p(x;μ,σ) for a given mean and standard deviation, and assume the female heights are identically distributed with a mean lower by two standard deviations p(x;μ-2σ,σ) (which seems to be reasonably accurate). For a given female height y, the odds a random man is smaller than her are given by the integral of p(x;μ,σ) where x goes from negative infinity to y. You then have to integrate this quantity where y ranges from negative infinity to positive infinity, with the measure p(y;μ-2σ,σ) dy.
It's easy to show this quantity is independent of either the mean or standard deviation, so I just numerically integrated it in Mathematica for a standard normal distribution, which again seems to be reasonably accurate for human heights.
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How did you calculate that? (7.8%)
Let's assume male heights are distributed according to a pdf p(x;μ,σ) for a given mean and standard deviation, and assume the female heights are identically distributed with a mean lower by two standard deviations p(x;μ-2σ,σ) (which seems to be reasonably accurate). For a given female height y, the odds a random man is smaller than her are given by the integral of p(x;μ,σ) where x goes from negative infinity to y. You then have to integrate this quantity where y ranges from negative infinity to positive infinity, with the measure p(y;μ-2σ,σ) dy.
It's easy to show this quantity is independent of either the mean or standard deviation, so I just numerically integrated it in Mathematica for a standard normal distribution, which again seems to be reasonably accurate for human heights.