There’s also the point that in terms of evaluation, the two were likely to get the benefit of every potential break. Any decision regarding them that is anywhere near 50/50 is going to be adjudicated in their favor.

Jehu, that’s a good point, but we can also plausibly argue the other way: things weren’t set up for women, so the first two bear the brunt, they are conscious of their pioneeringness, may not have been as well prepared as the men because no one was expecting them to do it, nor would they know exactly how to prepare, etc. Hence my arguing just from the data and not bringing in normal curves or anything.

(If you did do a normal distribution in the Bayesian analysis, obviously the results will be even more extreme in estimating the different means.)

It is probably worse than that. The two volunteers are probably substantially stronger than the mean woman Marine officer — considerably moreso than the average male Marine candidate is. The two that stepped up to the plate likely assumed they had a much better chance of making the cut than most of their counterparts.

If you consider most Marine infantry officers likely in the second sigma of physical ability among males (typically from 85th to 98th percentile), it’s likely these two women were in the third or fourth sigma among women. From this you can infer you’re talking about at least a three-sigma or so difference between population means.

To play around with some statistics:

0/2 is 100% failure while 26/107 is 24% failure; if we use Laplace’s Rule of Succession to try to estimate the chance the next woman will fail, from 2/2 we get (2+1)/(2+2)=3/4=75%.

In Bayesian terms, we can guess that the likelihood ratio for P(women more likely to fail | data) / P(equally likely to fail | data) is ~10.3 (integrating on a uniform prior 0-1).

From the frequentist perspective, if we assume the null hypothesis that women fail at exactly 24%, then the odds of two simultaneous failures would be 0.24*0.24=0.0576 or 5.76%. This is two categorical variables, so one wonders what a chi-squared test says:

> chisq.test(matrix(c(2, 0, 26, 107), ncol = 2))
Pearson’s Chi-squared test with Yates’ continuity correction
data: matrix(c(2, 0, 26, 107), ncol = 2)
Warning message: In chisq.test(matrix(c(2, 0, 26, 107), ncol = 2)): Chi-squared approximation may be incorrect

X-squared = 3.6357, df = 1, _p_-value = 0.05655

(Given the definition of a chi-squared test, the almost identical answer is not surprising.)