This post is about a classic probability puzzle. It goes something like this: I place two envelopes on the table in front of you. One of them contains a Prize, which is an amount of money in pounds, but you don’t know how much it is. The other one contains a Special Bonus Prize, which is worth exactly twice as much money as the Prize. It’s your lucky day — but you can only choose one envelope. Which do you choose?
“Well,” you say to yourself, “it doesn’t matter, they’re both the same,” so you pick one at random. Let’s say it’s the one on the left. But now I ask you if you want to change your mind.
“Well,” you might say to yourself, “let x be the amount of money in the envelope I’m holding. This envelope has a 50% chance of being the Prize, in which case the other envelope contains 2x. On the other hand, there’s a 50% chance that this is the Special Bonus Prize, in which case the other envelope contains 0.5x. But still, the expected value of the other envelope is 0.5*2x + 0.5*0.5x = 1.25x. So on the balance of probabilities I should definitely switch.” But then I offer to let you switch again, and again, and again, and every time you go through the same reasoning, never managing to settle on a particular envelope because each one seems like it should contain more money than the other. Clearly something is wrong with this reasoning, but what is it?
In this post, I’ll solve this problem in what I consider to be the proper Bayesian way, pinpointing exactly where the problem is. You might want to think about the question for a bit and come up with your own idea of its solution before reading on.
One thing to note before we start is that the problem goes away if I tell you how much money the Prize is worth. For example, if the Prize is £10 and the Bonus Prize is £20, the reasoning goes like this: If the envelope in my hand is the Prize then the other envelope is worth £20, otherwise the other envelope is worth £10, so its expectation is £15. That’s the same as the expected value of the envelope in your hand, so there’s no problem. So whatever the difficulty is, it has something to do with the idea that the value of the Prize is unknown.
How can we model this problem using probability theory? I decided to use three jointly distributed random variables: A, which represents the value of the envelope on the left (call it envelope A); B, which represents the amount of money in the other envelope; and E, which can take on two values, a or b. The variable E represents whether envelope A or envelope B contains the Special Bonus Prize.
We have to come up with a prior distribution for these three variables. In order to make them represent the things they’re supposed to represent, this distribution must have the following properties:
- The marginal distributions must be the same for the amount of money in the two envelopes. That is, , for every x.
- Each envelope should have a 50/50 chance of containing the Bonus Prize: .
- The Bonus envelope should contain twice as much money as the Prize envelope: and .
- The joint distribution shouldn’t change if you swap the two envelopes: .
These aren’t necessarily independent, since the last one implies the first one and possibly the first three imply the last one (I’m not sure), but they’ll all be used below.
Now, with this notation, let’s go through the paradoxical argument again. I’ve marked where the problem is.
- Envelope A has a 50% probability of being the Special Bonus envelope. , or .
- (The incorrect step.) Let the contents of A be x. Then, from (1), envelope B contains 2x with probability 0.5, and 0.5x with probability 0.5. That is, and , for any given x.
- Therefore the expected value of B is equal to , where x is the value of envelope A.
- Therefore the expected value of B is greater than the expected value of A, and I should switch
- By symmetry I should also switch if I choose B, so by induction I should keep switching forever and become infinitely rich.
Step 2 is the problem. It’s true that , but step 2 (and hence the whole argument) hinges on claiming that this is still the case when conditioned on the actual value of A. That is,
for every x. But in fact is given by
whereas is given by
Thus can only be equal to if , for every x. That is, step 2 only works if A is conditionally independent of E.
In other words, it only works if knowing whether A is the bonus prize tells you nothing about how much money is in it. This should raise a red flag already — if I told you that the envelope you’d selected was, in fact, the bonus prize, it would be quite strange if you didn’t then expect it to contain more money.
But an argument one sometimes hears in regards to this thought experiment is that if you know nothing — literally nothing — about the value of A, then this will in fact be true. Let’s try to codify this idea mathematically and see where it leads us.
Now, it’s quite difficult to say, a priori, what it really means to “know nothing” about something in probability theory. There’s a whole theory of so-called ignorance priors in Bayesian probability theory, but they’re quite fiddly and subtle things, so I’m not going to start out by trying to construct one. Instead I’ll just accept the claim in the previous paragraph (that knowing nothing means A is conditionally independent of E) and see where it leads.
Now, with this assumption of conditional independence, we have that
But . By similar reasoning to (i), this is also equal to , and by exchange of A and B this is equal to .
So we have that this particular notion of “literally not knowing anything” implies that the marginal prior for A has the property that
for every x. You can construct various fancy priors that have this property, such as , but the one that looks most like an ignorance prior is the uniform prior. Uniform priors for unbounded quantities are a bit odd and have a few formal subtleties, but you can deal with them. They’re improper priors, meaning that you can’t normalise them, but essentially, this ignorance prior assigns the same infinitesimal probability density to every value of x.
Using such a uniform prior for an amount of money is a bit weird. It means, for example, that the probability that the envelope contains £10 is the same as the probability that it contains £, which is the same as the probability that it contains £. But more than that, it means that the expected amount of money in the envelope is infinite. (This is also true of the “fancy” priors mentioned above.)
So now, finally, we can fully explain the paradox. If you really really “didn’t know anything” about the amount of money in the envelopes, you might be justified in assigning a uniform marginal prior to the value of each envelope’s contents. Then when you select envelope A, you can ask yourself “what is the expected amount of money in this envelope?” The answer is infinity. Should you switch? Well, the expected amount of money in envelope B is infinity, which is equal to infinity*5/4. So there’s no paradox. It doesn’t matter if you switch or not, because you’ll expect to become infinitely rich in any case.
But of course, if this was a real situation then the uniform prior would be a rather silly one. The contents of my envelopes cannot be less than 1p, and they can’t be more than the total amount of money in existence, which I guess is in the trillions of pounds. (Of course, you’re free to pick a smaller upper bound if you want.) Any prior you can come up with that fulfils these constraints will have finite expectations, and won’t allow you to conclude step 2 from step 1 in the argument above.
So to conclude, the two envelopes paradox is not a paradox at all, but just an intuitively reasonable argument that has a hard-to-spot error. (Confusing a conditional distribution with an unconditional one.) If you work through the problem in a proper Bayesian fashion, you realise that you can’t avoid considering your prior knowledge of the envelopes’ contents. As long as you choose a sensible prior, the problem evaporates.