January 25, 2004

The Reversed Two-Envelope Problem

The traditional two-envelope problem can be described as follows:

You have an envelope with x in it. The angel has an envelope with either x/2 or 2x in it; each is equally likely. If you want to increase the expected value of the number in your envelope, should you switch envelopes with it (the angel)?

[UPDATE 1/27: I don't think this is an accurate description anymore. See here.]

I'm going to try to show that the answer is "Depends on how things got that way."

In the traditional two-envelope problem, God randomly picks a number, S/He writes that number and its double in two envelopes, and then the angel randomly gives you one of the two envelopes. If you open your envelope and see that it has x in it, it's meant to be equally likely that it's the smaller envelope (and hence the angel has 2x) and that it's the larger envelope (and hence the angel has x/2).

(Brian pointed out to me in D.C. that it's actually impossible to define a probability distribution such that each rational number is equally likely, and that makes it hard to say that it really is equally likely, given that you have x, that it's the smaller or the larger number. But let me bracket that--which, as Jerry Fodor says, means "try not to think about it"--it shouldn't matter for the larger point.)

Consider this:

(Reversed Two-Envelope Problem) God randomly picks a rational number x. He writes down x in a blue envelope. The angel then flips a fair coin: heads, it writes down 2x in a red envelope; tails, it writes down x/2 in the red envelope. The angel hands you the blue envelope, and then offers to switch envelopes with you. Should you switch to the red envelope?

The answer, I think is yes. The case is no different than if you had opened the envelope, and the angel had offered a fair coin flip that would double or halve your stake. The EU of taking the coin flip is 5x/4, and that's better than the x you have.

Note also that this argument doesn't work in the other direction (and a good thing, too). If you already have the red envelope, your EU is 5x/4, and you don't want to exchange that for the x in the blue envelope.

What does my Chronological Ordering Principle say? Well, the infinite process Q consists of God's putting x in the blue envelope. The finite process R consists of the angel's flipping the coin. So the partition PQ is determined by the value x. Each cell of PQ consists of two equally likely outcomes, one with x is in the blue envelope and x/2 in the red, one with x in the blue and 2x in the red. In each cell, the EU of the blue envelope is x, and the EU of the red envelope is 5x/4. So in every cell you're better off taking the red envelope. According to the Chronological Ordering Principle, you should take the red envelope.

Note that the description at the top of this entry applies equally well to the Reversed Two-Envelope as to the traditional two-envelope problem. In each case, one envelope contains twice the other, and it's equally likely that your x is the smaller number as it is the bigger number. But it matters whether they started by giving you x or by determining what the two numbers in the envelopes were going to be.

I think that this is related to the notorious Monty Hall problem, but I'm sticking my chin out when I say that. More on this later, maybe.

Posted by Matt Weiner at January 25, 2004 03:56 PM

Your alternative is interesting (and I agree with your analysis). I think the point your buddy Brian made is actually the larger one, though. I think it's telling that God, rather than some mundane entity, was introduced to the puzzle as a dispenser of envelopes. In the case of a mortal dispenser with regular limitations you can immediately start making arguments about the infeasibility of obtaining truly random distribution across an infinite range, whereas once God is introduced you need a more rigorous argument to show that it is inherently impossible.

Posted by: bbartlog at January 27, 2004 07:18 AM