Two Envelopes Paradox Calculator

The two-envelopes paradox is a probability puzzle about a choice between two sealed envelopes, one holding an amount X and the other twice that amount. It is famous less for being a good puzzle than for the fact that its usual solution is wrong in an instructive way.

The setup hides a genuine asymmetry. A right switch doubles your money, a wrong one halves it, so even a strategy that wins less than half the time can be profitable. Most statements of the paradox ignore this and reason purely about which event is more likely.

You open an envelope showing y. Either it holds X, so X = y, or it holds 2X, so X = y/2. Both readings are consistent with what you see, so the question is which is more likely.

A prior turns that into arithmetic. Each reading gets a weight from the prior density, and the doubled reading carries an extra factor of one half because the map x to 2x stretches the range and flattens the density. Dividing gives P(larger | y).

Switching gains y when you hold the smaller envelope and loses y/2 when you hold the larger one, so the expected gain is P(smaller)·y − P(larger)·y/2 and the break-even probability is two thirds.

The classic argument sidesteps all of this by assuming a uniform distribution over the positive reals, whose density 1/x has an infinite integral. The resulting "probability" grows past 1 as the cutoff climbs, which is the contradiction and also the diagnosis.

The probability that you hold the larger envelope is the number that organises everything else. Below the prior minimum times two it is zero, in the interior it is a fixed fraction, and above the prior maximum it is one.

The expected gain is measured against keeping the envelope, which is the zero baseline. A negative number means the swap costs you money in expectation, whatever the odds say.

The two-thirds threshold is the difference between the optimal strategy and the intuitive one. The intuitive rule swaps only when you certainly hold the smaller envelope, which leaves profitable swaps on the table.

A small positive expected gain is a statement about the bounds you supplied, not about the envelopes. Widen the range and the evidence gets rarer, so the edge shrinks.

• The range of the prior. It decides how often an amount you see can rule one envelope out, which is the only source of information in the problem.
• Which prior you choose. A log-uniform prior makes the interior completely uninformative, while a uniform one gives a fixed 1/3 chance of holding the larger envelope wherever both readings survive.
• Where the amount falls in the range. Near the ends the answer is certain; in the middle it is not.
• The factor between the envelopes. A factor of two makes the two branches look alike to a scale-invariant prior; a larger factor would not.

• Understanding the standard resolutions of the paradox, including the improper-integral diagnosis.
• Deciding what a prior-free version of the question can and cannot tell you.
• Teaching why some apparently reasonable probability arguments are not.
• Any practical version of the puzzle where the range of the hidden amount really is known.

• The result depends entirely on the prior you supply. There is no prior-free answer to this question, which is the whole content of the paradox.
• The puzzle assumes the two envelopes are equally likely to be the one you receive and that the amounts are exactly related by the stated factor.
• It says nothing about who chose X or why. If X is a prize designed to be interesting, that is information the prior has not captured.
• The simulation reports averages over the prior, which is a different quantity from the average a single player experiences, and the two can differ sharply.