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Two Envelopes Paradox Calculator

Two Envelopes Paradox calculator. Bayesian resolution, Monte Carlo simulation. See why E(switch) = E

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How: Enter inputs and compute results.

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STATISTICSProbability Paradoxes

Two Envelopes Paradox — Why Naive Switching Fails

Bayesian resolution and simulation. Naive: E(switch)=1.25X. Reality: E(switch)=E(stay). Run the simulation to see it.

Parrondo's Paradox →Monty Hall →

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Simulation Parameters

For educational and informational purposes only. Verify with a qualified professional.

Key Takeaways

  • The Paradox: Two envelopes contain money. One has twice the other. You pick one and see X. Naive argument: "E(switch) = 0.5(X/2) + 0.5(2X) = 1.25X. Always switch!" But that leads to infinite switching.
  • Resolution: The naive argument double-counts. With a proper Bayesian prior, E(switch) = E(stay). No advantage to switching.
  • Monte Carlo: Run this simulation — average gain from switch stays near 0. The paradox vanishes when the prior is properly specified.
  • Connection: Related to St. Petersburg paradox — both involve infinite expectation fallacies.

Did You Know?

📬The paradox appears in many forms: exchange paradox, envelope paradox, two-envelope problemSource: Wikipedia
🧮The flaw: assuming X/2 and 2X are equally likely when you see X. They depend on the prior.Source: Falk (2008)
📊Simulation proves: with proper setup, switching and staying have identical expected value.Source: Monte Carlo
📈St. Petersburg paradox: infinite expected value from a fair game — similar resolution.Source: Stanford Encyclopedia
🔬Falk (2008) and others showed the resolution requires specifying the prior distribution.Source: Academic literature
📚Stanford Encyclopedia of Philosophy covers the philosophical implications.Source: SEP

Expert Tips

Run multiple trials

100,000 trials — switch advantage stays ~0

Try different distributions

Fixed, uniform, exponential — all show no advantage

Connect to St. Petersburg

Both involve infinite expectation fallacies

Bayesian prior

The resolution requires specifying the prior

How It Works

1. The Setup

Two envelopes: one has A, the other has 2A. You pick one and see amount X. The other has either X/2 or 2X.

2. The Naive Argument (Flawed)

"E(other) = 0.5(X/2) + 0.5(2X) = 1.25X. Always switch!" But then you'd want to switch back endlessly — a paradox.

3. Bayesian Resolution

The problem: X/2 and 2X are not equally likely. With a proper prior (e.g., uniform on [1,∞)), E(switch) = E(stay). No advantage.

4. Monte Carlo Simulation

This calculator runs trials: randomly assign smaller/larger, pick one, compare stay vs switch. Average gain from switch stays near 0.

Why the Naive Argument Fails

The naive argument assumes: when you see X, the other envelope has X/2 or 2X with probability 50% each. But this is wrong:

  • If X is the smaller amount, the other has 2X.
  • If X is the larger amount, the other has X/2.
  • P(X is smaller) vs P(X is larger) depends on the prior distribution of amounts.
  • With proper prior, these probabilities balance so E(switch) = E(stay).

Connection to St. Petersburg Paradox

The St. Petersburg paradox: A fair game pays 2^n if the first heads appears on toss n. The expected payoff is infinite, yet people would pay only a finite amount to play. Similarly, the Two Envelopes paradox involves improper use of expectation: the naive argument leads to "always switch" which implies infinite expected gain. Both paradoxes are resolved by properly specifying the probability space and avoiding improper priors or infinite expectations.

Frequently Asked Questions

Why does the naive argument say always switch?

It assumes X/2 and 2X are equally likely when you see X. That assumption is wrong — the probabilities depend on the prior distribution.

What is the Bayesian resolution?

With a proper prior (e.g., uniform on amounts), the expected value of switching equals the expected value of staying. No advantage.

How is this related to St. Petersburg paradox?

Both involve infinite or improper expectation assumptions. St. Petersburg: infinite expected payoff from a fair game. Two envelopes: improper prior leads to infinite switching.

Does the simulation prove the resolution?

Yes. Run 100,000 trials — the average gain from switching stays near 0. The paradox vanishes when the setup is properly specified.

What is the double-counting fallacy?

The naive argument treats X as both the smaller and larger amount in the same calculation. You cannot condition on X without specifying the prior.

By the Numbers

1.25X
Naive E(switch)
0
Switch advantage
2/1
Ratio of amounts
50%
Pick either envelope

Disclaimer: This calculator demonstrates the Two Envelopes Paradox for educational purposes. The Bayesian resolution (E(switch) = E(stay)) is mathematically rigorous. Simulations converge with more trials.

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