Explore the classic probability puzzle where switching between two envelopes appears to give infinite expected value — revealing hidden assumptions in conditional probability reasoning.
Last updated: March 2026
One envelope contains x, the other contains 2x
Number of times to repeat the experiment
Stay Strategy
$149.97
Wins: 4997
Switch Strategy
$150.03
Wins: 5003
Both strategies converge to equal expected value. The paradox arises from conditioning on seeing a value A without knowing if it's the smaller or larger amount — this creates implicit probability assumptions that break the symmetry argument.
The Two Envelopes Paradox (also called the Exchange Paradox) is a famous probability puzzle. You're given two envelopes: one contains $x and the other contains $2x. You're shown the contents of one envelope, then asked whether you should stick with it or switch to the other.
The Paradoxical Argument:
If your envelope contains amount A, switching gives you:
E[switch | see A] = ½(2A) + ½(A/2) = 1.25A
This seems to suggest you should always switch — but both envelopes are chosen symmetrically, so neither should have an advantage! This apparent contradiction is the paradox.
Resolution:
The issue is that you don't know whether A is the smaller or larger amount. Conditioning on seeing A $\Rightarrow$ the prior probabilities of "A is smaller" and "A is larger" are not both ½. The implicit distribution of x matters: infinite expected value appears only under infinite or unbounded priors.
One envelope has amount x, the other has 2x. You observe the contents of one envelope (call it A). The paradox asks: is switching always better?
When you see amount A, you must ask: what is P(A is the smaller amount | I see A)? This depends on the prior distribution of x, not just symmetry.
The 'always switch' argument assumes P(small | see A) = P(large | see A) = ½, but this is only true under specific prior distributions — not universally.
Run many trials where x is fixed (e.g., $100) and envelopes are assigned randomly. You'll see both strategies average out equally, confirming no advantage.
If you use a truly uninformative (improper) prior on x, the expected value calculations break down, revealing why the paradox arises mathematically.
Scenario: A game show offers two prize envelopes. You're told one has twice the money of the other. The host shows you an envelope with $1,000. Should you switch?
Naive Paradox Reasoning:
If I switch, I get $500 or $2,000 with equal probability × expect $1,250. Switching seems better!
Expected value if I stay: $1,000
Expected value if I switch: 0.5 × $500 + 0.5 × $2,000 = $1,250
The Resolution: The above calculation is flawed. Before seeing $1,000, both envelopes had equal value by symmetry. Seeing $1,000 doesn't change that symmetry — the envelope you have and the other envelope are still interchangeable from your perspective. The apparent advantage comes from treating the conditional distribution as if it's uniform over the set {smaller, larger} — but you don't know the prior on x, so this assumption breaks down.
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