The Two Envelopes Paradox
You're offered two envelopes. One contains twice as much as the other. Should you always switch? The math says yes... but that can't be right.
Play the Envelope Game
The Setup
You're presented with two identical envelopes. You're told:
- One envelope contains some amount of money, call it X
- The other envelope contains exactly 2X
- You don't know which is which
You pick one envelope and find it contains $100. Should you switch?
💡 The Tempting Argument
"If I have $100, the other envelope has either $50 (half) or $200 (double). There's a 50% chance of each. So the expected value of switching is:"
"Switching gives expected value $125 versus keeping $100. I should switch!"
The Problem
The argument above seems to prove you should always switch. But consider:
⚠ The Infinite Loop
If switching is always better, then after switching you should... switch again? And again? Forever?
Before even opening your envelope, the argument says switching is better. But both envelopes are symmetric! How can A be better than B, when you could just as easily have picked B first?
Something is deeply wrong with the expected value calculation. But what?
Finding the Flaw
The fallacious argument makes a subtle probability error. Let's denote:
- S = the smaller amount (unknown)
- 2S = the larger amount
- Y = what you see in your envelope
The error: The argument treats "my envelope has the smaller amount" and "my envelope has the larger amount" as equally likely given any specific value Y.
But these conditional probabilities depend on the prior distribution of S! If you see $100:
- If S = $50 is more likely than S = $100, you probably have the larger envelope
- If S = $100 is more likely than S = $50, you probably have the smaller envelope
📊 The Correct Calculation
Let p(S) be the probability distribution over possible values of S (the smaller amount).
The probability you have the smaller envelope, given you see Y, depends on:
These are generally NOT equal! The 50-50 assumption is unjustified.
Proposed Resolutions
Philosophers and mathematicians have offered several ways to understand this paradox:
1. No Prior Distribution
The problem doesn't specify how S was chosen. Without a prior, you can't compute the expected value of switching. The calculation is undefined, not 1.25X.
2. Mixed Reference Points
The fallacy mixes two scenarios: when your envelope has S and when it has 2S. Using "my amount" as a fixed reference point conflates different probability spaces.
3. Improper Priors
For the 50-50 argument to work for ALL values of Y, you'd need an "improper" probability distribution that assigns infinite total probability. Such distributions don't represent real situations.
4. Symmetry Resolution
By symmetry, E[other] = E[yours]. Whatever gain you might get by switching, you could equally lose. The expected value of switching is exactly what you have.
The Correct Expected Value
Let's work out the math properly. Let S be the smaller amount.
Before opening either envelope, switching provides no advantage. You have the same expected value either way.
The fallacy arises when you condition on seeing a specific amount Y and incorrectly assume 50-50 odds of it being small or large. The odds depend on the unknown distribution of S.
Interesting Variations
The Open Envelope Variant: What if you're allowed to open your envelope before deciding? Surprisingly, there exist strategies that do slightly better than random! If you use a threshold (switch if below $T, keep if above), you can gain an edge—but only because you're implicitly assuming a probability distribution.
Known Bounds: If you know the amounts are between $1 and $1000, seeing $1 means you must have the smaller envelope. Seeing $1000 means you have the larger. Intermediate values give partial information.
Infinite Case: If any amount is possible with equal probability, you get an "improper" distribution—like the uniform distribution over all positive numbers. This doesn't actually exist! The paradox relies on this impossibility.
Lessons from the Paradox
1. Expected value requires a distribution. You can't compute expectations without knowing (or assuming) probabilities. "I don't know" is not a license to assume 50-50.
2. Conditional probabilities aren't always equal. Seeing evidence changes what you should believe. The probability of holding the smaller envelope depends on what you see inside.
3. Symmetry is a powerful check. Before opening, the envelopes are symmetric. Any argument that favors one must be flawed—there's no asymmetric information yet!
4. Reference frames matter. Whether you measure expected gains in terms of S (the smaller amount), Y (what you see), or something else affects the calculation. Mixing frames causes errors.