Two Envelopes Paradox Calculator
Analyze the famous two envelopes paradox. Calculate expected values and understand why the naive 'always switch' strategy is mathematically flawed despite seeming logical.
The amount of money you found in your envelope
What is the two envelopes paradox?
You have two envelopes with money, one has double the other. You pick one, open it, and see amount X. Should you switch? Naive reasoning: other envelope has X/2 or 2X (equally likely), so E(switch) = (X/2 + 2X)/2 = 1.25X > X. This suggests always switch - but both envelopes are symmetric, creating a paradox!
What is wrong with the "always switch" reasoning?
The flaw: After seeing X, you CANNOT assume P(other=X/2) = P(other=2X) = 50%. These probabilities depend on the initial amount distribution. If the setup only allows amounts {10, 20} and you see 20, the other MUST be 10 (not 40). The symmetry assumption breaks down.
Should I switch envelopes or not?
Without additional information about how the amounts were chosen, switching offers NO advantage. Both envelopes are equally good in expectation. The paradox arises from incorrectly applying probability theory. In practice: flip a coin, it doesn't matter mathematically!
How is this paradox resolved mathematically?
Resolution requires Bayesian analysis with a prior distribution on initial amounts. For ANY proper prior, E(keep) = E(switch). The naive calculation fails because it assumes an improper "uniform distribution over all positive numbers" which doesn't exist. You need bounded, realistic amount distributions.
Are there real-world applications of this paradox?
Yes! Similar reasoning errors appear in: (1) Stock trading (when to sell/hold), (2) Game shows (Monty Hall-type problems), (3) Decision theory (opportunity cost fallacies), (4) Sequential search problems. The lesson: Consider the full probability structure, not just observed outcomes.