Two envelopes hold an amount and its double. You open one and find 8, reason that the other holds 4 or 16 with equal chance, and conclude that switching is worth 10. The same reasoning works whatever you find, so it says to switch without looking. The lab computes the worth of switching from a stated prior and finds the step that fails: the chance of “double” cannot be one half at every amount.
Core ideas
Three ideas locate the fault. The first is the argument in its strongest form, the second is the quantity it quietly assumes, and the third is what a finite prior does to that quantity.
From the premise that the other envelope holds half your amount or double it, each with probability one half, the expected value of switching is five quarters of your amount, which is more than keeping it. With the amount in hand the arithmetic has no slip, and if there is a fault it is in a premise. Run without opening, the same words hide a slip, and the lesson says where.
Before you open anything, your envelope is the smaller one with probability one half. After you see 8, the right probability is the chance that 8 is the smaller amount of a pair, given how the pairs were chosen. That is a posterior, and it depends on a prior over the pairs.
If the pairs come from a finite list, there is a largest amount. Seeing it tells you your envelope is the larger one, so the chance that the other is double is zero there. The premise cannot hold at every amount, and the argument needed it to.
The material
The argument first, then the quantity it assumes, then the lab.
Two sealed envelopes hold money. One holds some amount and the other holds exactly double. You are given one at random and may keep it or swap. You open yours and find 8. The other, you reason, holds 4 if yours is the larger and 16 if yours is the smaller, and the two cases are equally likely because you were handed yours by chance.
Ten is more than eight, so switch. Nothing in this depends on the number 8. If you had found x, the other would be worth (1/2)·(x/2) + (1/2)·2x, which is 5x/4, more than x. Since you would switch whatever you found, you may as well switch without opening. And having switched, the same argument about the new envelope says to switch back. A rule that recommends swapping forever, between envelopes that were symmetric to begin with, is what makes this a paradox.
The two envelopes paradox is the argument that you should switch whatever amount you find, from the premise that the other envelope holds half or double with equal probability, and the unacceptable conclusion that two symmetric envelopes are each better than the other.
Which exit is best depends on how the argument is run. Run without opening the envelope, it equivocates. The letter x stands for the amount in your envelope in both cases, but if the pair is some amount and its double, your envelope holds the smaller amount in one case and the larger in the other, so the x in x/2 and the x in 2x are not the same number, and averaging them as if they were is a step that only looks valid. That is exit two, and it is the standard diagnosis of the closed version: written with the pair as y and 2y, switching is worth (1/2)·2y + (1/2)·y whichever envelope you hold, and so is keeping. Run after opening, with the 8 in hand, the step is sound: 4 and 16 are two definite amounts and their average at equal weights is 10. Accepting the conclusion would mean swapping forever, which no one defends. So in the open version look at the premises. One is that the envelopes hold an amount and its double, which is the setup. Another is that you should pick the act of greater expected value, which Decision and Rationality defended. The third is that the chance of “the other is double” is one half, and that one is not part of the setup.
Why was it thought to be? Before you open anything, the chance that yours is the smaller envelope is one half, because it was handed over by chance. The argument carries that one half across the act of opening. But what you learn on opening is the amount, and the amount is evidence about which kind of envelope you hold. A small amount suggests the smaller one, and a large one the larger. The chance after the opening is a posterior, as in “Updating on Evidence”, and a posterior depends on a prior over the pairs.
Now suppose the prior is bounded, so that the pairs come from a finite list. Then there is a largest amount that can appear, and when you see it you know your envelope is the larger. The other holds half, with probability one, and switching is worth half of what you have. The same holds in mirror image at the smallest amount, where you know the other is double. So the chance of “double” is zero at one end of the list and one at the other, and in between it is whatever the prior makes it. It cannot be one half at every amount.
The lab takes the chance as the prior over two hypotheses, “mine is the larger, so the other holds 4” and “mine is the smaller, so the other holds 16”, and takes the payoffs of keeping and switching under each. In the first preset the prior is one half each, which is the premise of the argument for this amount. The lab reports that switching is worth 10 and keeping 8, as the argument said. In the second preset the prior is 1 on the first hypothesis: 8 is the largest amount there is. Keeping is worth 8 and switching only 4. In the third preset the prior is 1 on the second, and switching is worth 16.
At 8, with one half on each hypothesis, the lab gives switching 10 against 8. If the largest amount in the list is 8, the prior on the second hypothesis there is 0, and the lab gives switching 4 against 8. The same arithmetic, fed a different prior, reverses the advice. No single act is best at every amount, and that is why the always-switch rule cannot be right.
The reply has a price, and it should be stated. Giving up the premise means giving up a very natural thought, that when you know nothing about the amounts the two cases are equally likely. A bounded prior does not say that thought is always wrong. It says the thought cannot be right at every amount, and it leaves the reader free to hold it at the amounts in the middle of the list, as the first preset does.
The lab's limit is the limit of the reply. Every prior here is over finitely many hypotheses, and the payoffs are the amounts, chosen by you. Priors over unboundedly many amounts, some of which give the switching argument an infinite expected value, belong to a harder form of the puzzle that is named in the course home and not built. What the lab computes is that a prior that can be written down, with a largest amount, cannot make the argument work at every amount.
Interactive
prior times likelihood, normalised; the Bayes factor is the weight of the data
Method
Five steps, for this puzzle and for any argument that a symmetric choice is better on one side.
For the amount you hold, say what the other envelope would contain in each case, and what keeping and switching pay in each.
Write down the chance it gives to each case. The always-switch argument gives one half to each, for every amount.
Before opening, one half is right. After opening, the chance is a posterior, so ask which prior over the pairs would give it.
Take the largest and the smallest amount the prior allows. There the chance of “double” is 0 and 1, and the advice reverses.
The premise that the chance is one half at every amount is the one to deny. The price is a natural thought, that ignorance of the amounts gives equal chances, which holds only in the interior.
Take the envelope you hold to contain 8, and compare the two priors on what the other contains. The first is the argument's own, the second puts 8 at the top of the list.
At 8 in the interior, switching is worth 10 against 8. At the largest amount, the prior on “the other is 16” is 0 and switching is worth 4. The argument used one half at both.
Given the amount in your envelope and a prior over the two cases, you should be able to compute the expected value of keeping and of switching, state the chance the always-switch argument assumes, show with a bounded prior where that chance fails, and name the exit and what it costs.
Something to change about this lesson? Write it down, tick it off when it is handled, and export the list when you want to act on it. Kept in this browser only: nothing is sent anywhere, and nobody is notified.
No recommendations yet for this lesson.