Two Envelopes Paradox Calculator: Expected Value and the Switching Fallacy

No. The correct expected-value analysis shows that both envelopes have the same expected value, regardless of what you see. Switching does not improve your expected payoff. The naive argument that says you should switch is mathematically flawed: it treats the same symbol X as if it simultaneously represents both the smaller and the larger of the two unknown amounts, which is an equivocation.

The naive formula averages two scenarios: other envelope = 2X (if yours is the smaller) and other envelope = X/2 (if yours is the larger), each with probability 0.5. That gives 0.5 * 2X + 0.5 * (X/2) = 1.25X. The problem is that X in the two branches refers to different base amounts - in one branch it is the smaller amount m, in the other it is the larger amount 2m - so averaging them as if they are the same quantity is invalid.

No, they are different paradoxes with different resolutions. In Monty Hall, the host deliberately opens a losing door after your choice, adding new information that genuinely changes the probability distribution and makes switching beneficial (2/3 vs. 1/3 chance of winning). In the two-envelopes problem, no new information is added that breaks the symmetry between your envelope and the other, so switching offers no advantage.

Yes, provably. Before opening your envelope, pick a random number T from any continuous distribution over positive numbers. After seeing your amount X, switch if X < T, stay if X >= T. Because T is independent of the envelope amounts and is drawn from a continuous distribution, there is always a positive probability that T falls strictly between the smaller and larger amounts. When that happens, the strategy correctly identifies which envelope is larger. The overall probability of picking the larger envelope exceeds 0.5, though you can never know exactly by how much without knowing the envelope amounts.

For the naive conditional expectation E(other | you see X) > X to hold for every possible value of X, the prior distribution on the envelope amounts would need to be improper - meaning it cannot integrate to 1 over the positive reals. In practice, this means the expected value of the envelopes would be infinite. With any finite, proper prior distribution, the expected values of both envelopes are equal, and the switching argument collapses.

In the abstract setting where you have no prior knowledge of the envelope amounts, no: the symmetry between the two envelopes means that whatever you see, you cannot tell whether you are holding the smaller or the larger. However, if you have some prior belief about the likely range of amounts (for example, if you know the amounts are between $1 and $100), then seeing $90 gives you reason to believe you are probably holding the larger envelope, and you might rationally decide to stay.

The two-envelopes fallacy is essentially an improper Bayesian prior in disguise. For the naive argument to hold universally - for every possible amount X you might see - the distribution over the base amount m must assign equal probability density to every positive number, which is impossible for a normalized probability distribution. This is an improper prior, and reasoning with it produces nonsensical results such as the infinite chain of 'always switch' conclusions. The lesson is that Bayesian reasoning requires proper, normalizable priors to produce valid expected values.