Skip to main content.
December 23rd, 2003

Two Envelopes

Despite this blog’s title, there really aren’t enough rants here. Let’s make up for that.

I was just reading another less than interesting paper on the two-envelope paradox when I started thinking, “Why is anyone still writing about the two-envelope paradox? Surely everything that needs to be said about it has been said.” And I was right. But perhaps everything that needs to be said about it has not been said in the one place. So I’ll say it all here. Note that nothing that I’ll say here is even close to being original – the real message of this rant is that this is a puzzle that’s well past its use-by date. (And I use lots of italics when I’m ranting.)

The argument for the paradoxical conclusion, that you’re better off switching no matter which envelope you get, relies crucially on an inference like the following.

The amount of money X in your envelope is from the set {x1, x2, …, xn, …}. (Note this is a countably infinite set. There are probably versions of the paradox where the set is uncountable. The same things can be said about that version of the paradox.) Call this set S.

For all x in S, the conditional expected utility of swapping given X=x is positive.

Therefore, it is in your interest to swap.

Call the inference here (CC). (CC) is a kind of conglomerability principle – it says if something is good according to every member of a particular partition, then it is good simpliciter. Given some standard Bayesian assumptions, (CC) is equivalent to the following principle.

Let Y and Z be bets. For any proposition p, and bet W, let W & p be the bet that pays what W pays if p, and nothing otherwise. (I assume bets can have negative ‘payouts’, so all choices are bets.) Let (p1, p2, …, pn, …} be a countable partition of possibility space. Then if for all i, Y & pi is preferable to Z & pi, then Y is preferable to Z.

It’s really important to keep in mind here that (CC), or something very much like it, is just essential to the paradoxical reasoning. There’s simply no argument that you should swap that doesn’t use as a premise the principle that you should swap whatever is in your envelope. And this premise doesn’t get you to the conclusion without (CC), or something stronger than it.

Now (CC) in either its intuitive or formal versions is a very plausible principle. To prove this, just note how many people have tacitly appealed to it in setting up the two-envelope paradox. But unfortunately it is inconsistent. Vann McGee showed this “An Airtight Dutch Book” Analysis 1999. The only agents that can satisfy (CC) are those that have either (a) bounded utility curves or (b) ‘opinionated’ belief states – more precisely, there are only finitely many propositions about which their credence is neither 0 nor 1. And, as has been known since at least John Broome’s 1995 Analysis paper, the two-envelope paradox only gets going if you assume the agent in question satisfies neither condition.

Now as far as I can tell, that’s all one needs to say about the paradox. The paradoxical conclusion is only reached by taking an inconsistent principle of reasoning, and applying it in just the case where we know on independent grounds that it cannot safely be applied. But is that all people say? Well, no.

If you want to find out what they do say, Google is your friend. Note that some of the links you’ll find, such as the first one, do say reasonable things – i.e. something similar to what I say. But not many I’m afraid.

To be sure, there still is something odd here. Principles like (CC) are very intuitive. It’s hard to know what to do when faced with a situation where you know your preferences will change as soon as you find out something, and you know you’re about to find it out. As McGee says, it looks like you face an Airtight Dutch Book in that situation. But that’s all the counterintuitiveness – there’s simply nothing special about the two-envelopes here, because there’s simply no argument from consistent premises that says you should switch.

Happy holidays!

Posted by Brian Weatherson in Uncategorized

3 Comments »

This entry was posted on Tuesday, December 23rd, 2003 at 12:56 am and is filed under Uncategorized. You can follow any responses to this entry through the comments RSS 2.0 feed. Both comments and pings are currently closed.

3 Responses to “Two Envelopes”

  1. Joel Velasco says:

    Brian,

    While I agree that any writer who doesn’t even mention the failure of the CC principle is being irresponsible, the problem still seems very interesting to me. Imagine you are in a Broome-like situation and the chances of the smaller value of two envelopes containing $2^n is 2^n/3^n+1. Presumably you think that if offered a chance to switch before opening your envelope, you have no reason to, but if you have opened your envelope, you should switch no matter what is inside. I find this very hard to swallow even though I am aware that this kind of dominace reasoning is invalid.

    What if you do open the envelope? Have you gained any relevant information? You knew already that you would want to switch no matter what you saw. If you haven’t gained any relevant information, is there a good argument for switching after opening?

    I don’t know your tastes, but there is a technical paper by Clark and Shakel (2000) who end up arguing that even if you see the amount in your envelope you have no reason to switch.

    Joel Velasco

  2. nick shackel says:

    I’ve only just come across this. I agree that the pattern of inference you criticise is not generally correct and if used in the two-envelope paradox it is mistaken. However, that is only the first step in what is going wrong, and is eliminated as soon as people realize that evaluating the expected gain on swapping by E(B |A)–A doesn’t even begin to apply the apparatus of probability theory correctly (for a start, E(B |A)–A is a random variable, not an expectation). Instead it is necessary to calculate E(E(B |A)–A).

    Doing so eliminates the dependence on CC inferences, since E(E(B |A)–A) = E(B-A) (standard theorem) and so now you rest your decision not on what is in each envelope but simply on the unconditional expected difference. Doing so also eliminates a whole class of paradoxical cases: those for which E(E(B |A)–A) =0, and this fact proves that this method does not depend on CC inferences, since these cases are still ones in which E(B |A)–A are positive for each value of A. (Provably, all cases for which E(A) is finite belong to this eliminated class.)

    So it eliminates one class of paradoxical cases — but not all! It leaves in place and untouched two classes of paradoxical cases, those for which E(E(B |A)–A) is infinite (examples of which had been discussed in the literature prior to our original Mind 2000 paper and which we called ‘unbounded paradoxical’) and those for which E(E(B |A)–A) is finite (which had not been discussed as such in the literature prior to our 2000 paper and which we called ‘best paradoxical’).

    So these kinds of cases, unbounded and best paradoxical, are not addressed by criticising the inference from ‘doing X rather than Y is better in each member of a partition of the event space’ to ‘doing X rather than Y is better’. We claim they are addressed by our paper (the quick summary of which is in our later paper Mind 2003).

  3. Jeff Johnson says:

    Nick,

    I agree with you that not everything is explained by the failure of (CC). Suppose that the amount x in your envelope is selected from S, and the amount in the other envelope is either .5x or 2x, based on a coin flip. In this case, it seems plausible that you should blindly switch, so if the (CC) principle is bad, then there are apparently other principles of reasoning at work.

    I was a little confused, however, by the second paragraph of your comment. When you said that E(E(B|A)-A) = E(B-A), were you only talking about the cases in which E(E(B|A)-A) = 0? As you indicate on p. 425 of your Mind 2000 paper, they would not be equivalent in other cases. Also, it doesn’t seem right that in these cases, E(B|A)-A are positive for each value of A. In the example you gave on p. 421 of the same paper, the partial sums of the infinite series E(E(B|A)-A) are all positive, but the expected gains themselves were negative for some values of A.