### drcabana.orgComputational divertissements

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Here is how Wikipedia describes the two envelopes paradox:

You are given two indistinguishable envelopes, each of which contains a positive sum of money. One envelope contains twice as much as the other. You may pick one envelope and keep whatever amount it contains. You pick one envelope at random but before you open it you are given the chance to take the other envelope instead.

Should you switch? You don't know how much is in either envelope, so symmetry suggests there is no point to switching. But there would not be much of a paradox absent some reason to think that you should switch. The argument for switching is as follows.

Assume the envelope you are holding is the worth X > 0. Then the other is worth either 2X or X/2, with equal likelihood. So the expected value of the other envelope is (1/2 * 2X) + (1/2 * X/2) = X + X/4. This exceeds X, so you should switch. The paradox is that if you do switch, the same expected value argument still holds. So you should switch again, ad infinitum.

A little online searching will find all manner of complicated responses to this paradox. I think the matter is easily resolved. The first step is to realize the expected value computation above is nonsense.

Expected value is defined as the sum of p * probability(p) over all points p in the sample space. In the computation above we are invited to accept 2X and X/2 as the points in the space, each having probability 1/2. But there is a problem with that. There are exactly two points in our sample space, the unknown values in the envelopes. We can call those values X and 2X. We can call them X and X/2 if you prefer. Either choice means the same thing, that one value is twice the other. But the two points cannot be 2X and X/2 because 2X and X/2 differ by a factor of 4. The premise of the paradox is that one envelope contains twice as much as the other. The proposed expected value computation is performed over an inadmissable sample space.

There is another problem with the proposed EV calculation. Not only are we invited to consider 2X and X/2 as points in the sample space, we are asked to do so while holding an envelope containing X. So there are three distinct points in the sample space. 2X and X/2 occur each occur with probability 1/2 per the EV calculation. The sum of probabilities is bounded above by 1, so the X that we are assumed to be holding must have probability zero. Even though we are holding it.

There are a couple of legitimate ways to analyze the 'paradox' via expected values. One is to compare the expected value of one envelope to the expected value of the other. Let's say the envelopes contain X and 2X. We don't know which envelope contains which amount, nor does it matter. Both envelopes have the same expected value, (X + 2X)/2 = 3X/2. Since both envelopes have the same expected value, the EV computation offers no reason to switch.

The above does not depend on the choice of notation that the envelopes contain X and 2X. We could write the unknown values as Y and Y/2. The expected values of both envelopes are still equal, now both are (Y + Y/2)/2 = 3Y/4. The outcome looks a bit different, but the change of variables makes no difference. What matters is that both envelopes have the same EV.

Here is another way to think about it. One envelope contains X, the other contains 2X. You don't know which you are holding, either is equally likely. If you are holding X and then switch you gain X. If you are holding 2X and then switch you lose X. You are as likely to lose X as to gain X by switching, so the expected gain is zero. There is no reason to switch, just as there is no reason not to switch. But there is nothing left of the alleged paradox.