Should You Always Switch? The Envelope Puzzle That Breaks Expected Value
A probability puzzle where flawless logic leads to an absurd conclusion (and the fix reveals something deep about how we reason with infinity)
This is The Curious Mind, by Álvaro Muñiz: a newsletter where you will learn about technical topics in an easy way, from decision-making to personal finance.
Today we’ll see a puzzle that had me going in circles for longer than I’d like to admit.
It looks simple: two envelopes, one contains twice as much money as the other, and you get to pick one. Before you open it, I offer you the chance to swap. Sounds trivial—until you realize there’s a perfectly reasonable mathematical argument that says you should always switch, and an equally reasonable one that says it doesn’t matter at all.
Both arguments can’t be right. So which one is?
The Two Envelope Problem
Imagine you are given two sealed envelopes, each containing some money.
All you know is the following: one contains twice as much money as the other.
You pick one of them randomly. Before opening it, you are offered the change to switch.
What do you do?
Why You Should Switch
I’m going to convince you that you should switch.
Imagine the envelope you chose, let’s call it A, has 100€ inside. How much money is there in envelope B?
If A is the envelope with the smaller quantity, then envelope B contains 200€ (recall one envelope contains twice as much money as the other one).
If A is the envelope with the larger quantity, then envelope B contains 50€.
Since you chose randomly, half of the time you chose the envelope with the smaller amount, and the other half the one with the higher amount. Therefore, your expected value of switching envelopes is:
Your current envelope has 100€. Switch, and on average you'll walk away with 125€. That's a 25% free lunch.
Therefore, you should switch.
Note that the argument works if the first envelope had 200€ or any other amount inside it: if we call X the amount inside the first envelope you picked, then switching gives you an average payoff of 1.25X (which is higher than X).
The Infinite Loop
Okay, so we concluded that you are better off switching. You go and pick the other envelope.
Now here’s the catch: we are back at the start.
You have now picked an envelope, and we can apply the same argument again: it has an amount X, and, by switching, you get 1.25X on average (by the exact same calculation as before).
So you should switch again!
You can keep going forever, with math telling you that the best option is to keep switching infinitely: like a dog chasing its tail.
Math telling you that the best option is to keep switching infinitely: like a dog chasing its tail.
Why You Should Not Switch
Now here is a different argument that says you should not switch (or, rather, that you should be indifferent).
We know that one envelope contains twice as much money as the other one. Let’s say the quantities are 100€ and 200€.1
If you picked the envelope with 100€ and you switch, you gain 100€ in profit.
If you picked the envelope with 200€ and you switch, you lose 100€.
Since you picked the envelope randomly, half of the time we are in the first case, and the other half in the second. Therefore, your expected gain if you switch is:
Switching gives you no gain (on average).
So…Which Argument is Right?
Given these two arguments, what would you do?
The answer: you should be indifferent. Switching gains you nothing
The second argument is correct. The first argument–the one saying you should switch–is wrong, but spotting the flaw is surprisingly subtle. It involves something called improper priors, and a full technical explanation would take us into the weeds. But here's an intuitive version.
Go back to the critical step:
Imagine the envelope you chose, let’s call it A, has 100€ inside. […]
Since you chose randomly, half of the time you chose the envelope with the smaller amount, and the other half the one with the higher amount.
This quietly assumes that (50€, 100€) and (100€, 200€) are equally likely pairs. Fair enough. But the argument works for any starting number—so (1M€, 2M€) should be equally likely to (2M€, 4M€). And (1 trillion, 2 trillion) equally likely to (2 trillion, 4 trillion). And so on, forever.
But think bout it: if you open your envelope and find 1 trillion euros inside, do you really think it’s 50/50 whether the other envelope contains 500 billion or 2 trillion? Of course not. Intuitively, the more money you see, the more likely it is you got the bigger envelope.
That intuition is mathematically correct. It turns out to be impossible for the pairs (X/2, X) and (X, 2X) to be equally likely for every possible X—the probabilities wouldn’t add up to 1. The first argument hides in an assumption about how the money was distributed that can’t actually be true. Once you account for realistic priors, the apparent 25% gain vanishes.
The dog stops chasing its tail.
Why This Matters Beyond Envelopes
The two envelope problem isn’t just a cute puzzle: it’s a warning. It shows how an expected-value calculation, done with apparent rigor, can give you an answer that is obviously absurd (switch forever!). The math feels right, but it’s quietly built on a flawed assumption about what’s equally likely.
This happens all the time outside of toy problems. Investment strategies, insurance bets, and even some published scientific results have fallen for similar traps: reasoning that looks clean on paper but rests on a prior that can’t actually hold up.
The lesson? When a calculation tells you there’s a free lunch, don’t reach for the fork. First, check the menu.
In Case You Missed It
These numbers don’t really matter and the argument works with X and 2X (instead of 100€ and 200€).