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Fri, 16 Jun 2006

The envelope paradox
This is on my mind because someone asked about it in IRC yesterday and I was surprised at how coherently I was able to explain it on the spur of the moment. There are several versions of this paradox. My favorite version goes like this: you're going to play a game with an adversary. The adversary writes two different numbers on slips of paper and puts them in an envelope. The numbers are completely arbitrary; they could be absolutely any numbers whatsoever: zero, or π, or -1428573901823.00013, or anything else.

You pick one slip at random from the envelope and examine the number written on it. You then make a prediction about whether the other number is larger or smaller. If your prediction is correct, you win a dollar; if it is incorrect, you lose a dollar.

Clearly, you can break even in the long run simply by making your prediction at random. And it seems just just as clear that there is no strategy you can use that does better than breaking even. But this is the paradox: there is a strategy you can use that does better than breaking even. (This is what W.V.O. Quine calls a "veridical paradox": it's something that seems impossible, but is nevertheless true.)

Spoilers follow, so you might want to stop reading here for twenty-four hours and try to figure out a winning strategy yourself.


Let's call the number you get from the envelope A and the number still in the envelope B. You can see A, and you are trying to predict whether B is larger or smaller than A.

Here's your winning strategy. Before you see A, choose a random number R. If A < R, then conclude that A is "small", and predict that B is larger. If A > R, make the opposite prediction.

There are three possibilities. Either (1) A and B are both less than R, or (2) they are both greater than R, or else (3) one is less than R and one is greater.

In case 1, you predict that B > A, and you have a 50% probability of being correct. In case 2, you predict that B < A, and you have a 50% probability of being correct.

But in case 3, you win every time! If A < R < B then you see A, conclude that A is "small", and predict that B > A, which is correct; if B < R < A then you see A, conclude that A is "large", and predict that B < A, which is correct.

Since you're breaking even in cases 1 and 2, and you have a guaranteed win in case 3, you have a better-than-even chance of winning overall. There's some positive probability p (which depends on the method you use to choose R) that you have case 3, and if so, then your expected positive return on the game is p dollars per game.

The paradoxical part is that it initially seems as though you can't get any idea, just from looking at A, of whether it's larger or smaller than the unknown number B. But you can get such an idea, because you can tell from looking at A how big it is, and big numbers are more likely to be larger than B than small numbers are.

What you've done with R is to invent a definition of "large" and "small" numbers: numbers larger than R are "large" and those smaller than R are "small". It's an arbitrary definition, and it doesn't always succeed in distinguishing large from small numbers—it thinks that R+1 and 1000000R+1000000 are both "large"—but it can distinguish some large numbers from some small numbers, and it never gets confused and concludes that x is large and y is small when x is actually smaller than y. So it may be arbitrary, and extremely coarse, but it is never actually wrong.

In the cases where this very coarse method of deciding "large" from "small" fails to distinguish A from B, you get no new information, but that's okay, because you can still break even. But if you get lucky and the adversary has chosen numbers that you can distinguish, then you win.

Another way to look at the paradox is like this: suppose the adversary is required to choose his two numbers at random. Then you have a simple winning strategy: if A is positive, predict that B is smaller, and if A is negative, predict that B is larger. Even when both numbers are positive or both are negative, you win half the time; if one is positive and one is negative, you are guaranteed a win.

If the adversary knows that this is what you are doing, he can cut you back to merely breaking even, by limiting himself to always choosing positive numbers. But you can foil this strategy of his by choosing your "positive" and "negative" classes to be divided somewhere other than at 0: instead of "positive" being "> 0" and "negative" being "< 0", you make them mean "> R" and "< R". The adversary still wants to choose two numbers that are always positive, but since he doesn't know how big R is, hw doesn't know how large he has to make his own numbers to get them both to be "positive".

Still, this suggests the best strategy for the adversary: choose two very very large numbers that are close together. By doing this, he can make your expected win close to zero.

The envelope paradox is often presented in a different form: you are given two envelopes. One contains a bunch of money, say x dollars. The other contains twice as much. You open one envelope at random and examine its contents. Then you choose one envelope to keep.

A naïve analysis goes like this: I open the first envelope and see x. I can keep this envelope and collect amount x. If I switch, I have a 50% chance of ending up with 2x and a 50% chance of ending up with x/2, for an expected outcome of 5x/4. Since 5x/4 > x, I should always switch.

This is what Quine calls a "falsidical paradox": the reasoning seems good, but leads to an impossible conclusion. The strategy of always switching can't possibly be correct, because you could apply it with without even seeing what is in the envelope. You could keep switching back and forth all day, never opening either envelope, and increasing your expected winnings to infinity.

The tricky part, again, is that having seen x in the envelope, you cannot conclude that there is exactly a 50% chance of x being the larger of the two amounts. You get some information from the size of x, and if x is a large amount of money, then the probability that x is the larger of the two amounts is thereby greater than 0.5.

To do a full analysis, one has to ask the question of how the original amounts were selected. Say that the two amounts are b and 2b; let's call b the "base amount". How did the adversary select b? Let's say that the probability of the base amount being any particular amount x is P(x). It is impossible that b has an equal probability of being every number, because $$\int_{-\infty}^\infty P(x) dx$$ is required to be 1, and if P(b) is the same for every possible base amount b, then it is a constant function, and constant functions do not have the required property.

When you see x in the envelope, you know that one of two situations occurred. Either x is the base amount, and so is smaller, which occurs with probability P(x), or x/2 is the base amount, and x itself is the larger, which occurs with probability P(x/2).

Since these are the only possibilities, the a posteriori likelihood that x is the smaller number is P(x)/(P(x/2) + P(x)). This is equal to 1/2 only if P(x) = P(x/2). Although this can occur for particular values of x, it can't be true for every x. As x increases, P(x) approaches zero, so for sufficiently large x, we must have P(x/2) > P(x), so P(x/2) + P(x) > 2P(x), and P(x)/(P(x/2) + P(x)) < P(x)/2P(x) = 1/2.


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