The Two-Envelope Paradox

You're shown two envelopes, told one contains twice as much money as the other, and asked to pick one. You pick envelope A, containing some amount $x$, and are then offered the chance to switch to envelope B. A tempting argument says: B contains either $2x$ (if A was the smaller) or $x/2$ (if A was the larger), each with probability one half, so switching has expected value $0.5(2x) + 0.5(x/2) = 1.25x$ — strictly better than sticking with $x$. But this argument applies symmetrically to either envelope, meaning you should always want to switch no matter which one you're holding, which is absurd, since the two envelopes are identical before you look inside either one.

The flaw is in treating $x$, the amount actually in your hand, as a fixed known number while also averaging over what the other envelope might contain as though $x$ were still the smaller amount half the time and the larger amount half the time. Once you've actually opened the envelope and observed a specific number, the probability that this specific number is the smaller versus the larger share depends on the (unstated) distribution the amounts were originally drawn from — it is generally not a flat 50/50 once you condition on having seen a particular value, especially for distributions where very large amounts are rare. The "always switch" argument silently smuggles in an assumption — that any observed value is equally likely to be the smaller or larger member of the pair, for every possible value — that no proper probability distribution can actually satisfy.