It started with neckties
The earliest known version comes from Maurice Kraitchik, author of a book on mathematical recreations. In it two men compare their neckties, and also their wallets.
Two envelopes hold money, and one holds twice as much as the other. You take one and see an amount A. The other holds 2A or A/2, so 1.25A on average, and switching pays. But you picked at random, so switching cannot help. There is no single number to find here: the task is to find the step where the reasoning breaks.
A host puts an amount x in one envelope and 2x in the other. You don't know which is which. The puzzle has been around since the mid-20th century, and papers on the best way to explain the flaw are still being written.
Two identical envelopes contain money. One holds twice as much as the other.
You take one envelope at random. You may open it and see the amount A, or leave it closed.
"The other envelope holds 2A or A/2, 50/50. So on average it holds ½·2A + ½·A/2 = 1.25A. Switching pays."
The same argument works for a closed envelope, and for the other envelope after you switch. So you should switch forever. Which step of the argument is wrong?
"I can win A or lose only A/2. I should switch"
With a closed envelope, switching gains nothing: you picked at random, and the two envelopes are interchangeable. The flaw hides in what the letter A stands for. This puzzle is about a line of reasoning, not about a single number.
If you see the amount, the answer depends on the host's rule. Rounds 4 and 5 cover this.
You have taken an envelope and not opened it. Lock in your answer before reading on.
A breakdown of your answer will appear here.
The host chooses the amounts by a hidden rule. You open an envelope and decide: keep or switch. On the right the game counts what you would have got by always keeping or always switching. If the 1.25A argument is right, the second total should be a quarter bigger.
Each round, "always keep" gets your envelope and "always switch" gets the other one. Same envelopes, only the decision differs. The host's rule is revealed in round 4.
The computer plays many rounds with three strategies. The chart shows how many times bigger the "always switch" total is than the "always keep" total. The naive argument promises 1.25. Theory says 1. The third strategy, Thomas Cover's trick, switches only when the amount is below a random threshold.
Cover's threshold: before each round we draw a random number T (exponential with mean 100). If the amount in the envelope is below T, switch, otherwise keep. For any rule the host uses, the chance of ending with the larger envelope is above 50%, because sometimes T lands between x and 2x. How far above depends on whether T's scale matches the amounts. For amounts up to a million and the same T, the edge would be only a few thousandths of a percent.
The 1.25A argument has two weak links. First, the letter A stands for different amounts in different cases. Second, 50/50 odds cannot hold for every amount.
Say the envelopes hold x and 2x. Each case has chance ½, so the average gain from switching is zero. The naive formula uses A = x in the first case and A = 2x in the second, then treats them as one amount. That is why the gain looks like A and the loss like A/2, when both are really x.
Apply the same argument to the second envelope and call its amount B. It says the first envelope holds 1.25B on average. So each envelope is on average a quarter bigger than the other, which is impossible. When an argument proves a claim and its opposite, it contains an error, even before we can see where.
The chance that the other envelope is bigger depends on how the host chooses amounts. For it to be ½ for every A, the amount A/2 would have to be as likely as A, A as likely as 2A, and so on forever in both directions. Infinitely many equally likely amounts add up to a total probability of infinity, not 1. No such rule exists.
If you don't open the envelope, you have no information at all. "Pick at random and switch" is the same as "pick the other one at random". Both give the larger envelope with chance ½. That is why the simulation shows equal totals for switching and keeping, whatever the host's rule.
In the game the host picked the smaller amount x from 5, 10, 20, 40, 80, 160, 320 with equal chances. For most amounts you can see, the odds really are 50/50 and switching averages 1.25A. But if you see 640, the other envelope surely holds 320. That one big loss balances all the gains, and the total for switching blind comes to zero.
| You see A | Chance to see it | Other is bigger | Other on average | Contribution to switching gain |
|---|
So if you know the host's rule and see the amount, the right answer depends on the amount: here you should always switch, except at 640.
Here is how the threshold works. If T is below both amounts, you always keep; if it is above both, you always switch: either way your chance is ½. But if T falls between x and 2x, you switch away from the smaller envelope and keep the larger, so you win for sure. When T has positive density on every number, that chance is always above zero.
What if you know the largest possible amount? What if the host picks amounts so that switching pays for every amount you can see? And where the puzzle came from.
The smaller amount x is uniform between 0 and 50. You see A and switch if A is at most the threshold. If you see more than 50, it must be the larger envelope.
The best threshold, 50, gives 75% and 25% more money on average. Here switching really pays, because you know the cap and see the amount.
In 1995 the philosopher John Broome described this rule: the smaller amount is 2ᵏ with probability ⅓·(⅔)ᵏ, k = 0, 1, 2, … If you see A = 1, the other holds 2 for sure. For any other A the other envelope is bigger with chance 2/5 and holds 1.1A on average. So whatever you see, switching pays. The table checks this on 100,000 rounds.
| A | Rounds | Other bigger | Other / A, simulated | Theory |
|---|
Under this rule the average amount is infinite. So "the average gain from switching" becomes ∞ − ∞ and has no value. Here is the average gain from switching blind in six separate runs of 10,000 rounds each. It doesn't settle on any number; it jumps up and down:
This is the case where the naive argument seems to hold for every amount. But the conclusion "so switching blind pays" is still wrong, because these gains cannot be added up.
An older relative of the puzzle comes from the Belgian mathematician Maurice Kraitchik. Two men argue over whose necktie cost more. They agree that the owner of the pricier tie gives it to the other. Each reasons: "If I lose, I lose my tie. If I win, I get a more expensive one. The bet is in my favour." A bet can't favour both sides.
The flaw is the same as with the envelopes: "my tie" is worth different amounts in the winning and the losing case. If the ties cost c and C, the winner gains C − c and the loser loses that same difference.
The earliest known version comes from Maurice Kraitchik, author of a book on mathematical recreations. In it two men compare their neckties, and also their wallets.
The mathematician John Littlewood gave a similar puzzle in his 1953 book and credited it to the physicist Erwin Schrödinger. In that version each card has a number on both sides. You see one side of a random card and decide whether to turn it over.
Martin Gardner first wrote about the necktie paradox in his Mathematical Games column in January 1958. Later, in 1982, he popularised it as a wallet game in the book Aha! Gotcha.
The economist Barry Nalebuff published two envelope problems, in 1988 and 1989. The second is the one most often seen today as "the two envelopes paradox".
The logician Raymond Smullyan showed that the puzzle breaks intuition even without probability. "If the envelopes hold x and 2x, I win or lose x." "If I hold A, I win A or lose A/2." Both sentences sound right but lead to different conclusions.
The random-threshold strategy is usually credited to the information theorist Thomas Cover. It wins more than half the time for any rule the host uses, even an unknown one. Dozens of papers have been written on the puzzle as a whole, and there is still no single explanation everyone agrees on.
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