The paradox
Two envelopes; one holds twice what the other does. You pick one, open it, and find X. Should you switch?
The tempting argument: the other envelope holds either 2X or X/2, each with probability 1/2. So its expectation is
0.5(2X) + 0.5(X/2) = 1.25X
Switch! But by the same logic you should switch back. And again, forever. Something is wrong.
Where it breaks
The symbol X is doing two different jobs.
Let the true pair be (A, 2A). If you hold the smaller, switching gains you A. If you hold the larger, switching costs you A. Each is equally likely, so the expected gain is zero. Correct, and unsurprising.
The flawed version writes the gain as "2X or X/2" where X is your observed amount, which differs between the two cases. In the first case X = A; in the second X = 2A. Treating them as one quantity smuggles in an inconsistency.
The deeper version
If you open the envelope and see a specific number, the analysis genuinely depends on your prior beliefs about the amounts.
For switching to be favourable at every observed value, you would need a prior where, for all x, seeing x makes the pair (x, 2x) more likely than (x/2, x). No proper probability distribution does this - it would require unbounded expected value spread over infinitely many scales.
With any realistic prior, switching helps for small observed amounts and hurts for large ones, which is exactly what intuition says: find £10 and switching is tempting; find £10,000 and it is not.
What the interviewer wants
Not "you should not switch" - they want you to locate the error precisely. Say: the argument uses X to mean two different amounts across the two branches, and no valid prior makes the conclusion hold everywhere.
Related: expected value and the St Petersburg paradox, which is the other classic case of an expectation calculation misleading you.