Two envelope paradox

This is a deceptively simple paradox that shows how easy it is to make false assumptions about probabilities.

The two envelope problem

Here is the scenario. You have won a game show, and you get to choose between two envelopes that each contain a different prize. The two envelopes are identical, but one contains £10 and the other contains £20. There is no way to determine which envelope is which. You are invited to choose one of the envelopes, and you get to keep the money inside.

Since the envelopes are identical, there is no way to know which contains the larger amount of money. So you choose an envelope at random.

However, before you open the envelope, you are offered the chance to swap envelopes. Should you swap? It seems pointless, because both envelopes are equally likely to contain the £20, so what possible benefit could there be in swapping?

But the host points out that if your envelope contains X amount of money, then the other envelope must either contain X/2 or 2X, with equal probability.

What the host says is certainly true. So if you swap, you will either lose half your money or double your money.

The expectation value is the average amount you would expect to win if you repeated the game lots of times. Half the time you would get X/2 and half the time you would get 2X. So on average, you would get:

That amount is greater than X (the amount in your current envelope) so it is worth swapping. You might end up with less money, or you might end up with more money, but on average you would be better off!

Or so it seems.

The paradox

But how can that be? You picked the original envelope at random, so how could it suddenly be better to swap envelopes? After all, if by chance you had selected the other envelope, then the same reasoning above would also tell you to swap.

There is further problem with this. Let's call the envelope you chose A, and the other envelope B. Having chosen envelope A, the logic above tells you it is better to swap envelopes. Envelope B has an expectation value that is 1.25 times X, the amount in envelope A.

So you swap envelopes. You are now holding envelope B.

But now the host offers you the chance to swap back. This time, of course, envelope A either has half the amount in envelope B, or twice the amount. So by the same logic we used above, it is better to swap. If we previously decided it was better to choose envelope B, how can it now be better to choose envelope A?

What went wrong?

There is clearly something wrong here. Both envelopes can't be better than each other. So what is the problem?

We previously said that, if we have envelope A, then envelope B must either contain half the amount in A, or twice the amount in A.

That statement is true, but it doesn't tell the whole story. In reality:

• Envelope B might contain half the amount in A, but only if A contains £20.
• Alternatively, envelope B might contain twice the amount in A, but only if A contains £10.

So when we say that if A contains X then B might contain X/2 then that is only true if X is £20. And when we say that if A contains X then B might contain 2X then that is only true if X is £10. But those two statements cannot both be true for the same value of X which means that the expectation value calculation from before is not valid.

Instead, let's use the value Y to represent the amount in envelope A, and the value Z to represent the amount in envelope B. Now Z depends on Y, but it isn't a formula, instead it is a simple rule:

• If Y = £10 then Z = £20 or
• If Y = £20 then Z = £10

Of course, we don't know the value of Y, but we do know that it must be either £10 or £20, with equal probability. We can therefore calculate the expectation value assuming we don't swap the values. It is:

So the expectation value if you just open envelope A is £15. This is exactly as you would expect, it is the average of the two equally likely possible values.

What if you do swap the envelopes? Envelope B contains the amount Z, which is either £20 or £10. After you swap, you are holding envelope B. So the expectation value if we swap is:

So if you swap the envelopes, the expectation value is £15, the same as if you don't swap.

Now of course we don't know how much is in envelope A. But what this shows is that the expectation value in both cases is £15. Given that you don't know what is in your envelope, then swapping envelopes will definitely either make you £10 better off or £10 worse off, with equal probability. So statistically there is no advantage or disadvantage in swapping.

The source the confusion?

The key point in this apparent paradox is that the total amount of money in both envelopes is fixed. But the paradox is phrased in such a way as to lead you into treating it as if the amount in envelope A is fixed, and the amount in envelope B is either half of A or twice A. As explained above, that is not true (B can only be half A if A is £20, and can only be twice A if A is £10).

Let's consider a different case where the assumption above is true. Say you have been given an envelope C that you know contains £10. There is an envelope D that you are told contains either £5 or £20, with equal probability.

In this new case, it is statistically better to swap, because you stand to either lose £5 or gain £10 by swapping.

See also

• Monty Hall problem
• Pigeonhole principle
• Drunk man and keys problem
• Raven paradox
• Painter's paradox
• Potato paradox
• Sleeping Beauty problem
• Russell's paradox
• Surprise exam paradox
• Coin rotation paradox