Two Envelope Sophistry

I find the Two Envelope Problem fascinating. The problem is simple: there are two opaque envelopes that contain some money. You don't know how much is in each envelope, but you know that one envelope has twice as much as the other. You pick one, but then you get the option to switch to the other one. Should you switch?

Here is an argument for switching: your envelope has $x. Since we know that one of the envelopes contains twice as much as the other one, the envelope you didn't pick should either have $2x or $.5x in it, each with equal likelihood. So, the expected value in the other envelope is the average of $2x and $.5x, or $1.25x, greater than the expected value of $x in your envelope.

But, of course this is absurd: you picked an envelope at random, and you could have just as easily picked the other one. If you switch, then the exact same argument applies to the new envelope, and you should switch back to your original one! And so on forever. This paradox doesn't have an obvious solution, though Wikipedia tries very hard to explain it.

Lots of smart people have proposed resolutions to the Two Envelope Problem, and I won't rehash them here. The point I want to make is an extension of a point made by Raymond Smullyan in his book Satan, Cantor, and infinity and other mind-boggling puzzles.

My idea is that we only have two data points (the money in each envelope), so describing the quantities as doubles/halves of each other is only one of a literally infinite number of functional forms that could fit the two points. So just like we can make the argument above to always switch, we could just as easily make arguments to be indifferent or to never switch by assuming different functional forms.

Below are some example arguments for taking different attitudes toward switching based on the functional form that fits the two data points. For each of these, I will use the example that one envelope contains $4, and the other contains $8, though of course other numbers could work just as well.

Argument to be indifferent (adapted from Raymond Smullyan): One envelope has $x, and the other one has $x+$4. So if the envelope you picked has $y in it, the envelope you didn't pick either has $y+$4, or $y-$4, each with equal probability. So, the expected value in the other envelope is the average of $y+$4 and $y-$4, which is $y, so you should be indifferent to switching.

The argument to switch assumes that the function 2x described the relationship between the $4 and the $8, and this leads to an expected value of $1.25x in the "other" envelope. Smullyan's argument rests on the fact that instead of using 2x to describe the relationship between the $4 and the $8, we can use the function x+4 and get the rational result. Both functions fit the two data points equally well.

The simple idea I have is to look at other functional forms of the relationship between the two quantities, and see what conclusions they lead to. I'll start with an argument to never switch.

Argument to never switch: The relationship between the money in the envelopes is log(exp(x)+2926.36). So, one envelope has $4, and the other envelope has log(exp(4)+2926.36)=$8. So if the envelope you picked has $y, the envelope you didn't pick either has log(exp(y)+2926.36), or log(exp(y)-2926.36), but since this function is concave, the expected value in the other envelope is less than $y! So you should never switch.

We could imagine many other possible functional forms that fit these two data points. For example, we could say that the function exp(x+0.6931472) describes the relationship between 4 and 8. Then, the expected value of switching is even higher than $1.25x - we want to switch even more than the original paradox described! All of the infinite possible functional forms will lead to different preferences for switching.

Applying different functional forms to the data leads to infinite possibilities for Two Envelope sophistry - would could convince ourselves (or someone else) to switch or not switch with many varying degrees of expected gain or loss. The fact that this sophistry is possible may point to the solution to the paradox. The comparison between x and 2x is a red herring. Really, you are deciding between twice a small amount, or half of a large amount, so the losses are (of course) symmetric. The argument for indifference, since it's symmetric, is the only argument that we can take at face value and believe.

One thought on “Two Envelope Sophistry

  1. I guess another sophist e.g. could be ...?

    "Argument to never switch. The envelope you didn't pick contains x. The one you DID pick contains either $2x or $0.5x with equal probability, so stick with it."

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