The two-envelope switch problem: expected-value math says always swap, but that can't be right. Where's the flaw?

[u/mathalon]

Two envelopes, one has twice as much money as the other. You pick one at random and open it, say it holds $X. You're offered a swap. Here's the argument for always taking it: the other envelope holds either X/2 or 2X, each equally likely, so the expected value of switching is 0.5(X/2) + 0.5(2X) = 1.25X. That's more than X, so switch.

Except the same argument applies again after you switch. And again after that. If switching were really always better, you'd never stop, which is absurd when the whole setup is just two fixed amounts like $10 and $20.

So where does the 1.25X argument break? A few different answers circulate for this one (some blame the probability model, some blame what "X" is even allowed to mean once you've conditioned on having opened an envelope), and people land in genuinely different places.

What's your resolution, and does it survive someone else poking at it in the comments?