Two Envelopes Paradox Calculator

Two envelopes hold two amounts, one exactly double the other. You pick one, see $A inside, and a voice of pure logic whispers: the other holds 2A or A/2 with equal chance — an expected value of 1.25A — so switch, always, whatever A is. That last clause is the fraud. Give the amounts an honest prior distribution, as this calculator does, and the truth is subtler: sometimes switching really is favoured, sometimes it is a guaranteed loss, and averaged over everything you might see, the edge cancels to the cent.

The Fallacy, Dissected at the Joint

The seductive argument’s flaw is not arithmetic — (0.5 × 2A) + (0.5 × A/2) really is 1.25A — it is the smuggled premise: that after seeing A, the other envelope is still a fair coin between double and half. Probabilities after evidence are not chosen; they are computed, from a prior, by Bayes’ rule — and the computation depends on A itself. Under this page’s prior (smaller amount uniform up to M), seeing a value A below the cap makes it more likely, not equally likely, that you hold the smaller envelope: the density of values arriving as “the larger amount” is spread over twice the range, so any specific A is half as dense arriving that way — posterior odds 2:1 in favour of holding the smaller, P = 2/3. The famous 1.25 is therefore wrong in both directions at once: too pessimistic here (the honest figure is 1.5A), and catastrophically optimistic when A exceeds the cap, where holding the larger is certain and the coin-flip would have you trade a fortune for its half. What cannot survive, under any proper prior, is the claim that switching gains for every A — the version that also “proves” you should switch back, forever, printing money from notation.

Worked Example: The Hundred-Dollar Envelope

The defaults run the canonical scene with the prior stated out loud: you see $100; the smaller amount was drawn uniformly up to $1,000. The naive whisper quotes $125 and urges the switch. The honest ledger: P(hold smaller) = 2/3, so EV(switch) = (2/3 × 200) + (1/3 × 50) = 133.33 + 16.67 = $150.00, against $100 for staying — under this prior, at this A, switching genuinely is favoured, and the calculator’s candour about that is what makes its refutation of the paradox credible. Now push the scene past the cap: see $1,200 with the same M and the regime flips — your envelope certainly holds the larger amount (the smaller cannot exceed $1,000), the other holds $600 surely, and staying wins by $600. One prior, three answers — switch, stay, or (before opening, averaged over all scenes) indifference — and the paradox’s single shouted answer is exposed as the average of the three, mistaken for each.

The Conservation Law Underneath

Why must the regimes cancel overall? Because envelopes are filled before they are chosen, and no ritual of swapping can change the total in the room: the two strategies’ unconditional expectations are the same number — for this prior, 0.75 × M, the mean of the amount you hold (the smaller averages M/2, the larger M, and you hold each half the time). A policy of always switching merely re-deals the same expectation; a policy of switching conditionally — at small A, under a known prior — harvests information, and its gains are paid for exactly by the scenes (large A) where the same information orders you to stay. Information, unlike notation, is a real resource with a conservation law. The paradox endures because the naive argument pretends to the gains of information while supplying none — it uses the sight of A in its formula (2A, A/2) while refusing A’s evidential weight in its probabilities (the fixed 50/50). Keep both or neither, and the magic stops.

The Symmetry That Still Holds

One symmetry survives every prior, and naming it keeps the resolution honest: before the envelope is opened, you and a friend holding the other envelope face mirror-image positions, and no strategy available to both of you can favour both. Whatever edge a conditional strategy earns at small A is earned against the scenes where the other envelope’s holder, seeing the doubled amount, is ordered by the same logic to stand pat — the gains are transfers across possible worlds, never creations within one. This is why casinos cannot be beaten by envelope rituals and why the paradox’s perpetual-motion flavour is the tell of its error: expected value is conserved across the whole ensemble of scenes, and any argument that finds a surplus in every scene has miscounted the ensemble. The calculator’s three regimes — switch, stay, shrug — partition that ensemble without remainder.

A Pocket Protocol for Paradoxes

The envelopes generalise into a method worth carrying to every probability puzzle that feels haunted: write down the sample space, name the distribution on it, compute expectations from those two artefacts and nothing else, and treat any conclusion reached without them as a draft, not a result. The paradox’s 1.25A fails at the second step — no distribution is ever named, which is exactly the gap the “50/50” assertion rushes to fill. Conditioning puzzles in general (the boy-or-girl family on this hub is the other exhibit) dissolve under the same protocol: the sample space, written out, is usually the whole exorcism. Keep the protocol; the puzzles will keep coming.

Related Tools

More from the Statistics Calculators hub: the Boy or Girl Paradox Calculator — the other famous case of conditioning changing everything — and the Monty Hall Problem Calculator, where switching really does pay unconditionally, for reasons that survive the same scrutiny.

Frequently Asked Questions

So is switching good or bad? Give me a straight answer.

Under a stated prior it depends on A, and the calculator gives the straight answer for the prior you state: with the uniform prior, switch below the cap (expected gain, up to 1.5A at the posterior 2/3), never above it (a certain halving waits). Without any prior, the question “should I switch?” has no answer at all — which is the paradox’s real lesson. What is never right, under any proper prior, is the original claim: a guaranteed 25% edge for switching at every A, sight unseen, forever. That version is not a strategy but a notation error with confidence.

Why is P(holding the smaller) 2/3 and not 1/2?

Bayes, with densities. Seeing A can happen two ways: A is the smaller amount (its draw density is 1/M at A, and you picked that envelope with probability 1/2 — contribution 1/2M), or A is the larger amount, whose values are the smaller ones stretched by a factor of two across twice the range, halving their density (contribution 1/4M). Posterior odds 2:1, hence 2/3. Intuition’s 1/2 ignores that the larger envelope’s probability is spread thinner — the same stretching that creates the 2A prize dilutes the chance that any particular value is it. (In a discrete world of whole dollars the lattice changes the bookkeeping a shade — the posterior there is even — but the qualitative lesson, that the probability is computed from the prior rather than asserted as a coin flip, is untouched.)

Before opening, can any strategy beat indifference?

No — and yes, in the famous refinement. No strategy based on the envelopes alone can beat 50% at holding the larger amount, by symmetry. But add one ingredient from outside — a random threshold of your own, drawn from any spread-out distribution — and a jewel appears: open your envelope, switch if its amount is below your threshold, stay if above. This beats 50% strictly whenever the two amounts can straddle the threshold, because your decision now correlates with the amounts through the threshold’s coin. It wins probability of the larger prize, not expected value, and it is a theorem, not a paradox: information smuggled in honestly, through a random device, does honest work.

Why can’t I use a uniform prior over all amounts, with no cap?

Because it is not a probability distribution: spread evenly over an infinite range, the total probability is infinite (or zero everywhere), and expectations computed under it can be made to say anything — including 1.25A for both envelopes at once, the contradiction that exposes the trick. Every proper prior (total mass 1) restores sanity, which is why the calculator demands a cap — not as a limitation of the tool but as the price of admission to probability itself. Different caps give different local verdicts and the same global cancellation, which is robustness of exactly the right kind.

How does this differ from the Monty Hall problem?

In Monty Hall, an informed host acts after your pick, and his action leaks real information — the probabilities genuinely move to 2/3 for switching, unconditionally, under the game’s stated rules. In the envelopes, no one acts; the only evidence is the amount, and its weight depends on a prior the puzzle refuses to state. The pair are the two bookends of conditional probability: one puzzle where intuition denies a real asymmetry the rules create, one where it invents an asymmetry no rules create. Mastering both is most of what “understanding conditioning” means.

Who first posed the two envelopes problem?

The exchange argument traces to Maurice Kraitchik’s recreational writings of the 1940s — with two neckties, each husband certain his was the pricier gift — and the doubling-envelope form was sharpened by Martin Gardner and the probability journals of the 1980s–90s, where the improper-prior subtleties became a small research industry (the “necktie paradox” literature). It still generates papers, which is the surest sign its hidden assumption — a probability asserted without a distribution — is one the human mind is built to smuggle past itself.