## August 26, 2004

### Another Two-Envelope Problem

Brian Weatherson blogs a paper by Josh Dever and Eric Schwitzgebel on the two-envelope problem, and thinks their solution can't handle this case:

Consider the following example. God partitions the reals in [0, 1] into two unmeasurable sets, S1 and S2. He picks a real at random from [0, 1]. If its in S1, He puts \$10 into a red envelope, if its in S2 He puts \$20 into that red envelope. He then rolls two fair and independent dice. If they land double-six, he puts an amount into a blue envelope equal to the amount in the red envelope plus \$5. Otherwise, he puts an amount into that blue envelope equal to \$5 less than the amount in the red envelope. Got it? (Its easier with tables, but tables are hard in blogs.)

You are not told which number He picked, or how the dice landed, but you are told all of the above. You are then given a choice of the red or blue envelopes. How should you choose?

I take it that its obvious you should pick the red envelope. After all, whatever is in it, you have a 35/36 chance of getting \$5 less with blue, and only a 1/36 chance of getting more. So I say, pick red.

I asked myself--how do the various solutions to the two-envelope problem that I proposed handle this? It turns out they do pretty well.

Suppose that probability space is partitioned by performing process Q, which has an infinite number of outcomes, and then process R, which has a finite number of outcomes. Let PQ be the partition induced by the outcomes of Q alone--that is, two ultimate outcomes are in the same cell of PQ iff they result from the same outcome of Q. Suppose that, in every cell C of PQ, strategy S yields a higher EU than strategy T given that you're in C. Then strategy S is preferred over strategy T

should take care of it, with an easy modification--for "finite number of outcomes" substitute "finite number of outcomes with well-defined probabilities," and for "infinite number of outcomes" substitute "not [what I just said]." Then God's initial pick can be treated as process Q, the throw of the dice can be treated as process R, and you should pick the red envelope because no matter what the outcome of Q picking the red envelope has a higher EU.

(In fact, Brian's case is basically the Extra Coin Flip, with a process involving undefined probabilities substituted for the initial St. Petersburg.)

So I rule OK. I don't think that my solution runs afoul of Brian's stricture against attempts to resolve the two-envelope paradox by EU reasoning, because I only invoke EU reasoning for finite processes with well-defined probabilities; and anyway Brian's point is that those attempts can't handle this case, whereas my solution does. Perhaps he'll disagree.

Also, I have a quicker answer than Brian does to Eric's response (in Brian's comments) that the reasoner should use subjective decision theory--I do have a problem with subjective decision theory. But since many won't find that convincing, I'm glad that Brian has a more substantive answer.

Posted by Matt Weiner at August 26, 2004 03:06 PM