You need some _known_ distribution though, and it's shocking because the distribution is ommitted from the question, and the presence of the same distribution is snuck into the answer.
Not to be too dismissive, but the title text of this xkcd [0] seems relevant. No matter how nice the explanation is, the fact that the conclusion is wrong suggests that the reasoning has a flaw. I took a stab at what I think that flaw is when I responded to your rebuttal [1].
I dimissed it at first (you can see I made some embarrassing comments above), but the strategy is correct. I'd argue it's even intuitively correct once the problem in question is clarified.
It's not saying that after the player see the number in the first envelope, the strategy guarantees a >50% outcome.
It's saying that give any distribution, over all possible outcomes, >50% times the strategy will end up pick the larger number. You can say this >50% is the expected winning chance before the player see the number in the first envelope.
I'd say this is "intuitve" because, if your strategy can guarantee "when the player see a large number in the first envelope, he's less likely to switch than if he saw a small number", it would be better than blindly switching by coin toss. So intuitively such a strategy exists.
The only "trick" here is that since the player doesn't know the initial distribution, they can't tell "how large counts as large?" therefore they needs something that preserves some property over the whole real number line. That's why the strategy involves sampling from a another distribution whose PDF is non-zero everywhere.
