I’m quite sure now, I came to the same conclusion independently of GPT after getting a hint from it, which itself I had already almost guessed.

A woman having the top 10% of any characteristic is almost the same as rolling a 10 sided die and coming up with a 1 (this was the actual problem I presented GPT with, and when it answered it did so in what looked like a hybrid of code and text so I’m quite sure it is computing this somehow).

What was clearly wrong with the first math was that if I roll just three die, there would already be1*10^3 or1/1000 chance of getting all 1’s. And if I roll five die, there would be a much higher, not lower chance that I get at least three 1’s.

When rolling five die, there are 10 different possible combinations of those five die that have exactly three out of five 1’s, and it’s a little bit more complicated than this, but almost all of the probability mass comes from rolling three 1’s, since rolling four or five 1’s is far less likely. So you get very close (much closer than needed for a Fermi estimate) to the answer by simply multiplying the 10 possible combinations by 1/1000 chance that each of those combinations will be all 1’s, for a total of about ^{1}⁄_{100} or ~1%. Pretty basic once you see it, I would be surprised if this is incorrect.

Yes, that’s the main place I’m still uncertain, the ten combinations of three 1’s have to be statistically independent which I’m having trouble visualizing; if you rolled six die, the chance that either three pre-selected specific die would be 1’s or the other three die would all be 1’s could just be added together.

But since you have five die, and you are asking whether three of them will be 1’s, or another overlapping set will be 1’s, you have to somehow get these to be statistically independent. Part of that is actually what I left out (that GPT told me, so not sure but sounds sensible), you take the chance that the other two leftover die will both not be 1’s; there’s a

^{9}⁄_{10}chance that each will not be a 1, so .81 chance that both will not be ones, and you actually have to multiply this .81 by the 1/1000 for each set of three 1’s. So that slightly lowes that part of the estimate to (1/100010).81=.81%So you have excluded the extra 1’s from the sets of three 1’s but then you have to do the same calculation for the sets of four 1’s and the one set of five 1’s. The set a five 1’s is actually very easy, there’s a

^{1}⁄_{10}chance that each will land on one, so all of them together is 10^5=1/100,000, adding only .001% to the final calculation, and the four 1’s are also about a factor of 5 less likely then three 1’s because you have to roll another 1 to get four 1’s. So you have to roll four 1’s and one not-1, or (1/10,000).95=.045%.81+.001+.045=.856%

Still not 100% sure because I suck at combinatorials but this seems pretty likely to be correct. Mainly going off that

^{1}⁄_{1,000}intuition for any three sets of 1’s and that being repeated ~10 times because there are five die, and the rest sounds sensible