Two Envelopes Paradox Calculator

Enter the smaller envelope amount to see both envelope values, the expected value in your hand, the flawed 1.25X switching argument, and why switching gives no real gain.

Quick Facts

Method
Two envelopes hold X and 2X; each is equally likely to be yours
Expected value of either envelope is 1.5X, so switching gains nothing on average.

Your Results

Calculated
The two envelopes
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Smaller X and larger 2X
Expected value per envelope
-
1.5X, same for both
Flawed argument's claim
-
1.25 × amount held
Real gain from switching
-
Expected difference

Ready

Enter the smaller envelope amount and calculate.

What the Two Envelopes Paradox is

You are shown two sealed envelopes. You are told one contains exactly twice as much money as the other, but not which is which. You pick one at random. Before opening it, you are offered the chance to swap for the other envelope. Should you switch? Intuition says it cannot matter — the envelopes were interchangeable when you chose. Yet a plausible-looking calculation seems to prove you should always switch, and that is the paradox.

The setup in symbols

Let the smaller envelope hold an amount X. Then the larger holds 2X. Because you chose blindly, you are equally likely to be holding the smaller or the larger envelope:

  • With probability 1/2 you hold X (the smaller); switching gives you 2X.
  • With probability 1/2 you hold 2X (the larger); switching gives you X.

The expected value of the envelope in your hand is 0.5 × X + 0.5 × 2X = 1.5X. By identical reasoning, the expected value of the other envelope is also 1.5X. The two are equal, so on average switching changes nothing.

The flawed switching argument (and where it breaks)

The paradox comes from a tempting but wrong calculation. Call the amount in your hand A. The argument claims the other envelope is 2A with probability 1/2 (you hold the smaller) or A/2 with probability 1/2 (you hold the larger), so its expected value is:

0.5 × 2A + 0.5 × A/2 = 1.25A

Since 1.25A > A for any positive A, this appears to say you should always switch — and after switching the same logic says switch back, forever. The flaw is that A does not represent a single fixed number. In the "2A" branch, A is the smaller amount X; in the "A/2" branch, A is the larger amount 2X. These are two different dollar values, so you cannot treat A as one constant and average across the branches. Substitute the true value in each branch and the phantom 25% gain disappears: 0.5 × 2X + 0.5 × X = 1.5X, exactly the expected value of the envelope you already hold.

Reference points

  • If the smaller envelope holds $100, the pair is $100 and $200; every envelope has expected value $150; switching gains $0 on average.
  • If the smaller envelope holds $50, the pair is $50 and $100; expected value $75; switching gains $0.
  • The flawed argument applied to a held amount of $100 claims the other envelope is "worth" $125 — but that $100 is only the smaller value in one of the two equally likely worlds, so the figure is meaningless as a decision rule.

Does opening the envelope change anything?

If you open your envelope and see a specific amount, whether switching helps now depends on the prior distribution of the smaller amount X. Only if you have a genuine probability model for how the amounts were chosen — and that model makes the larger amount more likely than the smaller given what you observed — can switching have positive expected value. For any proper prior with finite mean, the advantage cannot hold for every possible observed value; it averages out to zero overall. With no prior information (the standard textbook version), seeing the amount gives you nothing actionable, and switching remains a coin flip.

Frequently Asked Questions

Should I switch envelopes or not?
In the standard symmetric problem, no. Both envelopes have the same expected value (1.5 times the smaller amount), so switching neither helps nor hurts on average. You may as well keep the one you have.
Why is the 1.25X calculation wrong?
It labels the amount in your hand "A" and then averages "2A" and "A/2" as if A were one fixed value. But the A in the 2A case is the smaller amount, while the A in the A/2 case is the larger amount — two different sums of money. Using the correct value in each branch gives 1.5X, not 1.25A, and the supposed gain vanishes.
Can opening the envelope ever justify switching?
Only if you have a real prior distribution for how the amounts were set and the value you observe makes the other envelope more likely to be the larger one. For any well-defined prior with a finite average, no rule can give an advantage for every possible observed amount — the gains and losses cancel to zero overall.
Is this a real mathematical paradox or just a mistake?
It is a veridical-style puzzle built on a subtle error rather than a genuine contradiction in mathematics. The probabilities and expected values are perfectly consistent once you stop conflating the two meanings of the "amount in hand." The lesson is about defining random variables carefully before averaging over them.

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