https://www.alexirpan.com/2015/09/09/the-other-two-envelope-...

If your solution is the same as this article's, it's plain wrong. Even the natural number case is plain strong.

It's very easy to demostrate as well: consider a trivia case where the distribution is just {P(1)=1/3, P(2)=1/3, P(3)=1/3} and you see 2 in the first envelope. There is no strategy to get a better chance than 50%. Therefore, any strategy that gives a better chance than 50% must implicitly make an assumption over the initial distribution (and therefore excludes a distribution like {P(1)=1/3, P(2)=1/3, P(3)=1/3})

Actually the article is even "wronger" than this, because "started A" and "switched" aren't independent and one can't simply use the product of their probability. The above example is a quick way to demonstrate it's not a general strategy without assumption to get >50% winning chance. Similarily, one can just use {P(1)=1/3, P(2)=1/3, P(3)=1/3} (this is a valid distribution over real numbers!) to demonstrate the real number strategy isn't general.

Again, for both natural number and real number case, the discussion over strategies is only meaningful is we know something about the distribution.

Interestingly, this article is wrong more or less in the same way as believing switching does give you more expected value in the original "twice money in another envelope" variation.

Edit: For people who are interested in the switching strategy, check Randomized Switching in the Two-Envelope Problem (2009). Spoiler: full of discussion over the initial distribution.

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.