This is counterintuitive in an interesting way.
You’d think that since P(Q1|~∀xQx) = 1⁄2 and P(Q1|∀xQx) = 1, observing Q1 is evidence in favor of ∀xQx.
And it is, but the hidden catch is that this depends on the implication that ∀xQx->Q1, and that implication is exactly the same amount of evidence against ∀xQx.
It’s also an amusing answer to the end of part 1 exercise.
I don’t understand this yet, which isn’t too surprising since I haven’t read the background posts yet. However, all the “roughly speaking” summaries of the more exact stuff are enough to show me that this article is talking about something I’m curious about, so I’ll be reading in more detail later probably.