Advanced

Two Envelopes Paradox Calculator

You pick one of two envelopes — one holds twice as much as the other. You peek inside and see $X. Should you switch? Enter the amount and your belief about whether your envelope is the smaller one to compute the expected value of switching versus keeping.

$

The amount visible when you open your chosen envelope

%

50% = no prior information; >33% favours switching
EV of switching
125$

vs keeping: $100 — net gain from switching: $25

EV if you switch
$125
EV if you keep
$100
Decision
Switch envelopes
Break-even P(smaller)
33.3% — switch if above this
If you switch & smaller
$200 (double)
If you switch & larger
$50 (half)
50%
50%
P(you hold smaller) → get $200
P(you hold larger) → get $50
Probability-weighted outcomes of switching (p = 50%)
Step by step
  1. 1

    Probability (p)

    50% ÷ 100 = 0.5
    Your estimated probability that you are holding the smaller envelope.
  2. 2

    EV if you switch (smaller case)

    0.5 × 2 × 100 = 100
  3. 3

    EV if you switch (larger case)

    (1 − 0.5) × 100 ÷ 2 = 25
  4. 4

    EV of switching

    100 + 25 = 125
Results are estimates for general information only and are not professional advice — always verify important results independently before relying on them. Read the full disclaimer.
Quick answer

How does this calculator work?

EV(switch) = p × 2X + (1−p) × X/2; EV(keep) = X. Switch only when p > 1/3. The famous "1.25X paradox" (p = 50%) uses an improper uniform prior — with any proper prior the true EV of switching equals the EV of keeping. The calculator shows the conditional decision given your belief p about which envelope you hold.

Formula
EV(switch) = p × 2X + (1−p) × X/2 • EV(keep) = X • Switch if p > 1/3
How this is calculated

The two-envelopes problem is a classic puzzle in probability and decision theory. You are given two sealed envelopes, one containing twice as much money as the other. You open your envelope and see $X. A naïve argument says: "the other envelope has either 2X or X/2 with equal probability, so its expected value is (2X + X/2)/2 = 1.25X > X — I should always switch." But the same logic applies after switching, implying you should switch back forever — clearly paradoxical.

The resolution lies in the prior distribution of amounts. The naïve calculation implicitly assumes a uniform prior over all positive reals, which is an improper (non-normalising) distribution. With any proper prior, a careful Bayesian analysis shows that seeing $X gives you information about whether your envelope is the smaller or larger one. Let p = P(your envelope is the smaller). Then EV(switch) = p × 2X + (1−p) × X/2, and EV(keep) = X. Setting EV(switch) > EV(keep) gives p > 1/3. So you should switch only if — given the amount you see and your prior beliefs — there is more than a 33.3% chance your envelope is the smaller.

This calculator lets you explore that conditional decision. At p = 50% (no information about which envelope you have) the naïve paradox emerges: EV(switch) = 1.25X. As p approaches 0% (you are confident you have the larger envelope), EV(switch) falls to X/2 — clearly keep. The paradox is not about a computational error; it highlights how improper priors silently corrupt expected-value reasoning.

Frequently asked questions

With no information about the distribution of amounts, you cannot rationally prefer either envelope — the expected value of switching equals the expected value of keeping when calculated correctly with a proper prior. The naïve "1.25X" argument uses an invalid infinite uniform prior. In practice, real amounts have bounded distributions, and seeing $X gives you Bayesian evidence about where X sits in that distribution, which can rationally favour switching or staying.

Switching is rational when your posterior probability that you hold the smaller envelope exceeds 1/3. This occurs, for example, if you know amounts are drawn uniformly from a bounded range and the amount you see is near the bottom of that range — it is more likely the smaller envelope. Conversely, if $X is near the maximum plausible amount, p is low and you should keep.

Both involve conditional probability and counterintuitive switching decisions, but they are distinct. In the Monty Hall problem a new piece of information (opening a goat door) shifts the probabilities and switching is provably correct. In the two-envelopes problem without additional information no action is provably better, and the apparent advantage of switching is an artefact of an improper prior.

Also known as

should i switch envelopes probability
envelope switching expected value
two envelope problem solution
decision theory envelope paradox
bayesian two envelopes calculator
envelope swap ev calculator

APA

TG we-Calculate Editorial Team. (2026). Two Envelopes Paradox Calculator [Online calculator]. TG we-Calculate. https://we-calculate.com/calculator/two-envelopes-paradox-calculator

Chicago

TG we-Calculate Editorial Team. "Two Envelopes Paradox Calculator." TG we-Calculate. 2026. https://we-calculate.com/calculator/two-envelopes-paradox-calculator.

IEEE

TG we-Calculate Editorial Team, "Two Envelopes Paradox Calculator," TG we-Calculate, 2026. [Online]. Available: https://we-calculate.com/calculator/two-envelopes-paradox-calculator

BibTeX

@misc{wecalculate_two_envelopes_paradox_calculator, title = {Two Envelopes Paradox Calculator}, author = {{TG we-Calculate Editorial Team}}, howpublished = {\url{https://we-calculate.com/calculator/two-envelopes-paradox-calculator}}, year = {2026}, note = {TG we-Calculate} }

Did this calculator help you?