The 25% Gain From Swapping Needs a Distribution That Does Not Exist

Requiring the fifty-fifty at every amount you could open forces the weights to satisfy f(x) = f(x/2)/2, whose only solutions are proportional to 1/x, and that integrates to infinity at both ends. Conditional on the pair, the swap gains the smaller amount or loses it with equal chance, which is zero and needs no assumption at all. The article carries a proper spread where the conditional answer is genuinely x/2, and the infinite-mean spread where swapping really is right at every observable amount.

Two envelopes, one holding exactly twice the other. You open yours and find 100. The other holds either 200 or 50, so it averages 125, a quarter more than you are holding. Swap. Then run the argument again from the other side and swap back, forever.

Something has to give, and it is not the arithmetic. Half of 200 plus half of 50 really is 125, and 125 really is five quarters of 100. The step that fails is the fifty-fifty, and it fails in a specific way: no probability distribution exists that would justify it.

What the loop actually assumes

Write $X$ for the amount you are holding. The computation says:

\mathbb{E}[\text{other} \mid X = x] = \tfrac12(2x) + \tfrac12\!\left(\tfrac{x}{2}\right) = \tfrac54 x

(1)

For the loop to be endless, (1) has to hold at every amount you might have opened, and 100 is only one of them. It has to hold at 50, at 800, at 12.5, at every rung of the doubling ladder in both directions. That is the assumption doing the damage, and stating it plainly is most of the resolution.

The weights that assumption forces

Where the pair comes from

The pair is $\{m, 2m\}$ for some positive $m$ , drawn from some distribution with density $f$ on the smaller amount, and you hold one of the two with probability a half each. Everything below is about which $f$ could make (1) true for all $x$ .

If the smaller amount has density $f$ , then the amount in your hand has density $g(x) = f(x)/2 + f(x/2)/4$ : the first term is the case where you hold the smaller, the second where you hold the larger of the pair $\{x/2, x\}$ . The chance you hold the smaller, given what you see, is therefore:

\Pr\bigl(\text{holding the smaller} \mid X = x\bigr) = \frac{f(x)/2}{f(x)/2 + f(x/2)/4}

(2)

Set that equal to a half for every $x$ , which is what (1) demands, and the equation collapses to a functional equation on $f$ alone:

f(x) = \tfrac12 f\!\left(\tfrac{x}{2}\right) \quad\Longrightarrow\quad f(x) = \frac{c}{x}

(3)

Check it: $\tfrac12 f(x/2) = \tfrac12 \cdot \tfrac{2c}{x} = \tfrac{c}{x}$ . The reciprocal is the unique shape that keeps the posterior flat, which makes sense on reflection: the fifty-fifty has to be invariant under doubling, so the weights have to be scale invariant, and $1/x$ is what scale invariance looks like on the positive line.

The reason no such distribution exists

A density has to integrate to one. This one integrates to infinity, and it does so at both ends:

\int_1^{\infty} \frac{dx}{x} = \infty, \qquad \int_0^{1} \frac{dx}{x} = \infty

(4)

No constant $c$ rescues that. There is no probability distribution on the positive reals for which the fifty-fifty holds at every amount, so the premise of the endless loop was never available. The 125 in equation (1) is a calculation performed inside a world that does not exist.

Fig. 1 — The weights the fifty-fifty demands. Going up they halve and the sum converges; going down they double and it does not. Either end is enough to kill it.

The divergence at zero is the half people miss, and it is worth checking on the discrete ladder rather than the integral. Requiring the fifty-fifty at every level pins the weights at $w_{t} = w_{t-1}/2$ along a doubling ladder, and the total over the rungs from $-N$ to $N$ grows like $2^N$ : 63.97 at $N = 5$ , 2048 at $N = 10$ , and about $2.2 \times 10^{12}$ at $N = 40$ . The mass all piles up at the small end.

The statement that survives with no assumptions at all

Once the pair is fixed, the whole thing becomes trivial and that is the point. Conditional on the pair being $\{m, 2m\}$ , you are holding one of the two with probability a half each. Swapping gains $m$ or loses $m$ :

\mathbb{E}\bigl[\text{gain} \mid \{m, 2m\}\bigr] = \tfrac12 m + \tfrac12(-m) = 0

(5)

For the pair on screen, $\{100, 200\}$ , swapping gains 100 or loses 100. This needs no distribution, no integrability, nothing. It is the symmetry of a coin flip, and it is the claim worth making, because it cannot be attacked.

A real spread, where the answer is not flat

Conditioning on the pair is not the same as conditioning on what you saw, and the difference is the interesting part rather than a technicality. Take a concrete spread: let the smaller amount be uniform on $(0,1)$ . Then for $x < 1$ you might be holding the smaller of $\{x, 2x\}$ or the larger of $\{x/2, x\}$ , with density $1/2$ and $1/4$ respectively. Those normalise to $2/3$ and $1/3$ , so:

\mathbb{E}[\text{gain} \mid X = x] = \tfrac{2}{3}\,x + \tfrac13\!\left(-\tfrac{x}{2}\right) = \tfrac{x}{2} \quad (x < 1)

(6)

Above 1 you must be the larger holder, and the gain from swapping is $-x/2$ . So under this spread swapping is genuinely right at small amounts and genuinely wrong at large ones. Averaging over the density of what you hold, which is $3/4$ on $(0,1)$ and $1/4$ on $(1,2)$ :

\int_0^1 \tfrac34 \cdot \tfrac{x}{2}\,dx + \int_1^2 \tfrac14\left(-\tfrac{x}{2}\right)dx = \tfrac{3}{16} - \tfrac{3}{16} = 0

(7)

Fig. 2 — Under an honest spread the answer depends on what you opened, and the two halves cancel exactly once weighted by how often each amount occurs.

The caveat that makes fixing the pair load-bearing

It is tempting to promote (5) to the unconditional slogan that swapping is worth zero on average. Do not, because it is false in general. Put weight $\tfrac13 \left(\tfrac23\right)^{k}$ on the pair $\{2^k, 2^{k+1}\}$ for $k = 0, 1, 2, \dots$ . That is a genuine probability distribution: the weights are positive and sum to one. Its mean is infinite, since the pair values grow like $2^k$ while the weights only decay like $(2/3)^k$ .

Under that spread, condition on seeing $X = 2^n$ with $n \ge 1$ . The posterior on holding the smaller is $2/5$ rather than a half, and:

\mathbb{E}\bigl[\text{gain} \mid X = 2^{n}\bigr] = \tfrac25 \cdot 2^{n} - \tfrac35 \cdot 2^{n-1} = \frac{2^{n}}{10} = \frac{X}{10} > 0

(8)

Strictly positive at every amount you could possibly observe, and at $X = 1$ the gain is $+1$ because you must be holding the smaller. So there really are honest distributions under which you should always swap, and the reason this is not a money pump is that the unconditional expectation does not exist: you cannot average (8) against a distribution whose mean is infinite and get a number. This is the same lesson the paradox has been teaching all along. When an answer is an average, ask which average.

None of the above says a single swap is a mistake. After opening a real envelope, with real beliefs about what the person filling it would plausibly put in, swapping can be exactly right. What is refuted is the version that never terminates, and the version that claims to know the answer without knowing anything about where the money came from.

Sources and further reading

The problem and its history — Two envelopes problem
What equation (3) produces, and why it is not a distribution — Improper priors
The object equations (5) to (8) compute — Conditional expectation
The other classic where an infinite mean does the work — St. Petersburg paradox

The functional equation in (3) was derived here from (2) rather than quoted, and its divergence was checked on the discrete ladder as well as in the integral. Equation (6) was confirmed by four hundred thousand draws under the uniform spread, which returned a mean gain of -0.000363 and matched the conditional profile to a worst bucket error of 0.0011. The counterexample spread in (8) was tabulated exactly: its weights sum to 1.000000000, its partial mean over sixty terms had already reached $3.1 \times 10^{7}$ , and the conditional gain is positive at every observable amount.