How this instrument works
The setup: there are two envelopes, one contains twice as much money as the other, and you pick one at random and open it. You see an amount — call it x. The tempting 'switching argument' goes like this: your envelope is equally likely to be the smaller one or the larger one of the pair. If yours is the smaller one, the other envelope holds 2x. If yours is the larger one, the other holds x/2. Averaging those two possibilities with equal 1/2 weights gives an expected value for the other envelope of 0.5(2x) + 0.5(x/2) = 1.25x — seemingly proving you should always switch, no matter what x turns out to be.
That conclusion is absurd once you notice the symmetry: the exact same argument would apply if you'd opened the OTHER envelope instead, telling you to switch back, and so on forever — you can't have a rule that says 'always switch' be simultaneously correct for both envelopes in the same pair. This is exactly why it's called a paradox rather than just a puzzle: it's a seemingly valid argument that leads to a self-contradictory conclusion, and identifying precisely where the reasoning breaks down is the actual mathematical content of the problem.
The flaw lives in treating the two cases as equally-likely 1/2-and-1/2 possibilities given only the observed amount x, without accounting for the actual prior probability distribution the envelope amounts were drawn from. Once you specify a genuine probability distribution for how the smaller amount was originally chosen, the paradox resolves — but there's no single universal fix, and different well-defined setups resolve it in different ways, which is part of why the two envelopes problem remains a popular topic in the philosophy and mathematics of probability.
- Enter the amount you see in your opened envelope into Amount in your envelope.
- Read Other envelope, if yours is the smaller one (2x) — what the other envelope would hold under that assumption.
- Read Other envelope, if yours is the larger one (x/2) — what the other envelope would hold under the opposite assumption.
- Read Naive 'expected value' of switching (the fallacy) — the flawed 1.25x calculation that the switching argument produces, shown here to illustrate the paradox rather than as genuine advice to switch.
Worked example — you open your envelope and find $100
Enter 100 into Amount in your envelope. The instrument computes both possibilities: if your $100 is the smaller amount, the other envelope holds 2×100 = $200 (Other envelope, if yours is the smaller one); if your $100 is the larger amount, the other envelope holds 100/2 = $50 (Other envelope, if yours is the larger one).
Naive 'expected value' of switching (the fallacy) reads $125.0000, computed as 0.5×200 + 0.5×50 = 100 + 25. This 1.25x figure is exactly the flawed conclusion the switching argument produces — and it would produce the exact same '25% higher, so switch' recommendation no matter what amount you'd actually opened, which is the tell that something in the reasoning has gone wrong, not genuine evidence that switching pays off.
Questions
Should I actually switch envelopes based on this calculation?
No — that's the entire point of the paradox. The 1.25x figure comes from an argument with a hidden flaw (assuming a flat 50/50 chance between 'my envelope is the smaller one' and 'my envelope is the larger one' regardless of the amount x you actually saw), and that same flawed argument would tell you to switch back again after switching, forever, which is self-contradictory. This calculator computes the naive figure specifically to illustrate the fallacy, not to recommend switching.
Where exactly does the switching argument go wrong?
It implicitly assumes that, no matter what amount x you observe, there's still an equal 50/50 chance your envelope is the smaller or larger one of the pair. That assumption silently requires a probability distribution over possible envelope amounts that treats every possible x as equally likely regardless of size — and no valid probability distribution can actually do that over all positive numbers. Once you specify a real, proper distribution for how the amounts were chosen, the 50/50 assumption breaks down for at least some values of x, and the paradox resolves.
Is this the same as the Monty Hall problem?
No, they're different classic probability puzzles that just happen to both be famously counter-intuitive. Monty Hall involves genuinely updating your probabilities based on new information revealed by a host who knows where the prize is. The two envelopes paradox involves a flawed argument that looks like a legitimate expected-value calculation but rests on a hidden, impossible assumption about the underlying probability distribution — the 'trick' is entirely different in nature.
Does the naive expected value change depending on the amount I enter?
Yes, proportionally — the naive figure is always exactly 1.25 times whatever amount you enter, since the flawed formula is a fixed multiple of x. That every single possible x produces the same '25% more, on average' conclusion, with no exceptions, is itself a strong clue that the argument can't be valid — a genuinely sound strategy wouldn't recommend switching unconditionally for every conceivable amount you might see.