icon

The two envelopes problem solved

Share this page
Envelope

The two envelope problem is a famous paradox from probability theory (which we first presented on Plus back in September). Imagine you are given two envelopes, one of which contains twice as much money as the other. You're allowed to pick one envelope and keep the money inside. But just before you open your chosen envelope you are given the chance to change your mind. Should you stick with the envelope you picked first or switch?

To find out write $x$ for the amount that's in your chosen envelope. This means that the amount of money in the other envelope is either $2x$ or $x/2$. The probability that it's $2x$ is $1/2$ and so is the probability that it's $x/2$. So the expected amount you'll get for switching is

  \[ \frac{1}{2}\left(2x+\frac{x}{2}\right) =x+\frac{x}{4} = \frac{5x}{4}. \]    

Since that’s bigger than $x$, you should swap envelopes. But what if you are given another chance to swap envelopes after you have changed your mind once? By the same reasoning as above you should swap back again. And then, by the same argument again, you should swap a third time, and so on, forever. You end up in an infinite loop of swapping and never get any money at all. What’s wrong with this reasoning?

A resolution

Let’s write $A$ for the envelope you picked at first and $B$ for the other one. We write $x$ for the amount of money in $A$. Now since we haven’t opened envelope $A$, $x$ isn’t a fixed amount: it’s a random variable. It can take one of two values: the smaller amount of money that’s hidden in the two envelopes or the larger amount of money. Let’s write $y$ for the smaller amount and $2y$ for the larger amount (recall that one envelope contains twice as much money as the other). Since you have picked $A$ randomly, there’s a 50:50 chance that $A$ contains either of the two amounts. This means that the expected amount of money $E(A)$ in envelope $A$ is

  \[ E(A) = \frac{1}{2}\left(y+2y\right) =\frac{3y}{2}. \]    

We said above that the expected amount in envelope $B$ is

  \begin{equation} E(B) = \frac{1}{2}\left(2x+\frac{x}{2}\right).\end{equation}   (1)

But recall that $x$ isn’t a fixed amount but can take one of two values. In the case that envelope $B$ contains $2x$, envelope $A$ contains the smaller amount of money, so $x=y.$ In the case that envelope $B$ contains $x/2$, envelope $A$ contains the larger amount of money, so $x=2y.$ So in formula (1) above, the first $x$ really stands for $y$ and the second $x$ stands for $2y$. The two $xs$ in the formula are actually different and shouldn’t have been added up to give $5x/4.$

Substituting the $y$ for the first appearance of $x$ in (1) and $2y$ for the second gives

  \[ E(B) = \frac{1}{2}\left(2y+y\right)=\frac{3y}{2}. \]    

Thus $E(A) = E(B)$ so there is no incentive to switch envelopes and hence no paradox.

What if you open envelope A?

What if we had already opened envelope $A$, to find $\pounds x$ inside, before being offered the chance to switch? Can we still produce the apparent paradox?

If you have opened envelope $A$ then $x$ is a fixed amount of money. There’s a 50:50 chance of finding $2x$ or $x/2$ in envelope $B$, so the expected amount in envelope $B$ is

  \[ E(B) = \frac{1}{2}\left(2x+\frac{x}{2}\right) =x+\frac{x}{4} = \frac{5x}{4}. \]    
Martin Hairer

We heard about the two envelopes problem in a talk by the Fields medallist Martin Hairer at the Heidelberg Laureate Forum 2017. Foto: Bernhard Kreutzer for HLF (c) Pressefoto Kreutzer.

The formula is now correct. It tells you that on average (if you repeated the same wager many times with the same amount $x$ in envelope $A$), you’d do better by switching. The paradox doesn’t arise. If after switching to envelope $B$ you are given the chance to switch back again, you won’t because you already know that the amount in $A$ is less than the expected amount in $B$. The paradox arose in the original version because both envelopes could be treated the same — the situation was symmetric. Once you have opened envelope $A$, however, the symmetry is broken.

Notice, however, that opening envelope $A$ and seeing the amount $x$ may change your mind about the probability that envelope $B$ contains $2x$ or $x/2.$ For example, if $x$ is a very large amount, then you might think it very unlikely that envelope $B$ contains the even larger amount $2x$. Writing $p$ for the probability that envelope $A$ contains the larger amount, the expectation $E(B)$ becomes

  \[ E(B) = 2x(1-p)+\frac{px}{2}. \]    

This is greater than $x$ if and only if $p<2/3.$ In other words, as long as you’re confident that $p$ is less than $2/3$ you should switch envelopes.

To us the above resolution of the supposed paradox appears satisfactory, but not everyone would agree. People have spent a lot of time thinking and writing about the two envelopes problem. Its Wikipedia page is a good start if you'd like to find out more.

Read more about...

Comments

Permalink

This problem causes us to confuse prior (before information is learned) and posterior (after information is learned) probabilities. The prior probability that A has the lower amount is 1/2. The same posterior probability is:
Pr(A=x & B=2x)/[Pr(A=x & B=2x) + Pr(A=x & B=x/2)].

The posterior probability that A has the higher amount is:
Pr(A=x & B=x/2)/[Pr(A=x & B=2x) + Pr(A=x & B=x/2)].

These are the expressions you must use in the expectation under "What if you open envelope A?". In general, they can't both be 1/2 - your benefactor would have to possess an infinite supply of money.

I'm inclined to agree and suggest going a bit deeper. The confusion between prior and posterior probabilities results from the fact that no decision is in fact made, however much we think we're seeing into what would happen were it made, so no new information is in fact ever learned. And that decision was never made because it was revoked before being implemented and nothing changes, including information. If a decision does change anything, then it can't be revoked or switched, at least not without cancelling the resulting information changes too.

In this crucial respect the two envelopes problem differs from Monty Hall (which some people compare it to), since in the latter the first decision does result in new information which the player can then use for their next. But switching doors doesn't mean they're revoking their first decision. They can't.

(When I finally hit the SAVE button, I can't change anything either. Oh well, here goes)

... but what it changes is unknown. This is why the issue is indeed the distribution.

An example might help. Say your monk (from the link) put a $10 bill in one envelope. He then put nine $5 bills and one $20 bill into a bucket. With his eyes closed, he picked one bill from the bucket and put it in the second envelope.

The statement "one envelope contains either half, or twice, what is the in other" is still a true statement. The chances that you pick the higher, or lower, envelope are indeed both 50%. But if you open your envelope and see $10, the chances are 90% that you have the larger envelope, not 50%.

The formula Exp(other) = (X/2)*Prob(higher) + (2X)*Prob(lower) is correct. What is wrong in the calculation is assuming Prob(higher) = Prob(lower) = 1/2 when you claim your envelope contains X. They are actually the relative probabilities that the pair of envelopes contained (X/2,X) and (X,2X), which you have no way of knowing.

Permalink

Note that the ONLY assertion being made in the original problem formulation is that the two envelopes contain x and x/2 with EQUAL probability. But because an absurdity follows from this - namely that it is profitable to continually keep switching envelopes - the assertion itself must therefore be false.

Indeed, there is no distribution in which the probabilities are both 1/2. Such a distribution does not exist so the probabilities are never 1/2. So the paradox does not arise at all.

If you change the game slightly to the following: You receive X amount of money and you can flip a coin. If you flip Tails they have to pay you 2X, if you flip Heads they have to pay you X/2, obviously you always should flip because in that game, the probabilities are 1/2.

Permalink

If you ask a child to choose a random number from 1 to a million, and he says something like 5, then it's likely he really didn't really randomize his choice.

If you are asked to choose a random positive number (with no other restrictions) then:
P(The number is less than 10) = 0.
P(The number is less than 100) = 0.

P(The number is less than any arbitrarily chosen number, N) = 0.

It is thus impossible to choose a random number with no constraints.

More technically:
P(A | B) where:
A: The number you have chosen is truly random
B: The number you have chosen is less than any arbitrary number, N

is given by:

P(A|B) = P(A Intersect B) / P(B)
The numerator is 0. The denominator is not.

Thus the problem is poorly formed.

If you restrict the maximum amount of money so some value, then the problem is trivial and whether or not you should switch (assuming you examine the contents of the chosen envelope before deciding whether or not to switch) is an easy calculation.

(I think)

Permalink

The reasoning is flawed for the scenario: 'What if you open envelope A?'

Per Marianne
"It tells you that on average (if you repeated the same wager many times with the same amount $x$ in envelope $A$), you’d do better by switching."

If you open envelope A no new relevant information is gained (assuming no practical limitations of likely $ ranges). Suppose person Y opens A and person Z opens B. Would they not conclude using Marianne's logic that each is better off switching? There is no advantage to switching regardless of the amount revealed in the envelope (again assuming normal $ range estimates arising from practical considerations are excluded).

True, if you had a chance at a 50/50 wager of either losing 0.5x or winning 1.0x, then it would be to your benefit to take the wager. But in the example you open envelope A, you still don't know if the fair value of the game. The fair value of the game is 1.5x (where x is the smaller amount). After opening you don't know if you have the smaller or larger amount by definition. Choosing to play the game you in essence wager 0.5x against the 1.5x fair value. Half the time you lose and are left with 1x, half the time you win and have 2x. Switching after opening is no different.

Permalink

By all means calculate the expected amount in envelope B as 5x/4 if you like. But since you're switching, you're giving up the amount in envelope A, which is x. So the expected gain on the switch must be 5x/4 - x = x/4

But what if it's A that contains 2x or x/2 and you switch it for B containing x? Then the expected gain is the result of reversing the subtraction in the above EV equation, namely x - 5x/4 = -x/4. Symmetry.

If you can't know which envelope is which, then you can't know either the advantage or disadvantage, the gain or loss, to switching. The expected gains, positive and negative, cancel. Symmetry.

If you know that A contains x, then sure, you switch. If you know A contains x/2 or 2x, then sure, you don't switch. Symmetry.

  • Want facts and want them fast? Our Maths in a minute series explores key mathematical concepts in just a few words.