# Non-Normalizable Probability Measures for Fun and Profit

By Sean Carroll | May 17, 2010 9:43 am

Here’s a fun logic puzzle (see also here; originally found here). There’s a family resemblance to the Monty Hall problem, but the basic ideas are pretty distinct.

An eccentric benefactor holds two envelopes, and explains to you that they each contain money; one has two times as much cash as the other one. You are encouraged to open one, and you find \$4,000 inside. Now your benefactor — who is a bit eccentric, remember — offers you a deal: you can either keep the \$4,000, or you can trade for the other envelope. Which do you choose?

If you’re a tiny bit mathematically inclined, but don’t think too hard about it, it’s easy to jump to the conclusion that you should definitely switch. After all, there seems to be a 50% chance that the other envelope contains \$2,000, and a 50% chance that it contains \$8,000. So your expected value from switching is the average of what you will gain — (\$2,000 + \$8,000)/2 = \$5,000 — minus the \$4,000 you lose, for a net gain of \$1,000. Pretty easy choice, right?

A moment’s reflection reveals a puzzle. The logic that convinces you to switch would have worked perfectly well no matter what had been in the first envelope you opened. But that original choice was complete arbitrary — you had an equal chance to choose either of the envelopes. So how could it always be right to switch after the choice was made, even though there is no Monty Hall figure who has given you new inside information?

Here’s where the non-normalizable measure comes in, as explained here and here. Think of it this way: imagine that we tweaked the setup by positing that one envelope had 100,000 times as much money as the other one. Then, upon opening the first one, you found \$100,000 inside. Would you be tempted to switch?

I’m guessing you wouldn’t, for a simple reason: the two alternatives are that the other envelope contains \$1 or \$10,000,000,000, and they don’t seem equally likely. Eccentric or not, your benefactor is more likely to be risking one dollar as part of a crazy logic game than to be risking ten billion dollars. This seems like something of a extra-logical cop-out, but in fact it’s exactly the opposite; it takes the parameters of the problem very seriously.

The issue in this problem is that there couldn’t be a uniform distribution of probabilities for the amounts of money in the envelopes that stretches from zero to infinity. The total probability has to be normalized to one, which means that there can’t be an equal probability (no matter how small) for all possible initial values. Like it or not, you have to pick some initial probability distribution for how much money was in the envelopes — and if that distribution is finite (“normalizable”), you can extract yourself from the original puzzle.

We can make it more concrete. In the initial formulation of the problem, where one envelope has twice as much money as the other one, imagine that your assumed probability distribution is the following: it’s equally probable that the envelope with less money has any possible amount between \$1 and \$10,000. You see immediately that this changes the problem: namely, if you open the first envelope and find some amount between \$10,001 and \$20,000, you should absolutely not switch! Whereas, if you find \$10,000 or less, there is a good argument for switching. But now it’s clear that you have indeed obtained new information by opening the first envelope; you can compare what was in that envelope to the assumed probability distribution. That particular probability distribution makes the point especially clear, but any well-defined choice will lead to a clear answer to the problem.

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CATEGORIZED UNDER: Mathematics

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