In a paradox like this one, our intuition fails to tell us the correct answer. In this case, our intuition says that the always switching strategy is ridiculous, why not stick with the original envelope? Why should it matter if we switch every time? I framed the question using the leprachaun with the extra envelope partially to convince myself that in fact intuition is wrong here: you should always switch envelopes in the problem. A direct calculation of expected return conditioned on the value in the first envelope leads to the same conclusion.

But the leprachaun point of view of the problem illustrates why our intuition fails: you can view the two envelope problem set up as being inherently unphysical. You cannot perform the experiment in the real world, and so you should not be surprised that the intuition that we get from performing experiments in the real world does not apply.

To summarize my three points: the correct answer to your problem is that you should always switch. The intuitive answer to the problem is switching should not matter. The intuitive answer is wrong because this experiment cannot be conducted in the real world, and so real world rules do not apply.

>The intuitive
>answer to the problem is switching should not matter.
Since (b): Exchanging the envelopes before opening does not affect the probability distribution for the amounts of money in the envelopes.

>The
>intuitive answer is wrong because this experiment cannot be
>conducted in the real world, and so real world rules do not
>apply.
More specifically, we have (c): The expected amount of money in either envelope is infinite.

First note that if money is put at random into two envelopes in such a way that both (a) and (b) hold then so must (c).

For any particular value for the amount of money in X, the expected value of the amount in Y is greater than the expected value of the amount in X (by (a)). Averaging over all possible values for the amount of money in X, E(Y) > E(X) (if both are finite). But from (b) we know that E(X) = E(Y). Hence E(X) > E(X), which is a contradiction if E(X) is finite. So E(X) is infinite, which is (c).

Now (c), on the face of it, does not seem to be a paradoxical condition (however, there is a closely linked paradox). But it can be seen to easily produce both (a) and (b). The following thought experiment constructs a situation in which both (a) and (b) hold, and in which the way that both arise from (c) is far more obvious:

Let two envelopes X and Y be independently filled with money with the same probability distribution as one another, and in such a way that the expected amount in either envelope is infinite.

When envelope X is opened, the amount of money it contains is found to be finite. But the expected value of the amount in envelope Y is infinite. So, given the chance, you should switch. But by symmetry the same would be true if envelope Y were to be opened first.

Thankyou

sfwc
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Looking at the problem "locally" in terms of conditional probabilities, the initial symmetry of the game is broken as soon as you choose a specific envelope, because the rules exclude the possibility that the other envelope contains the same amount of money as the one you chose, so if you would have chosen the other envelope first, you would have been presented with a DIFFERENT set of possible outcomes. If you think about the set of possible events conditioned upon the value of the envelope, you would be looking at a different space of probability events.

--Stuart Anderson

Now the argument for switching envelope says the expected value would
increase if you switch. But here our intuition plays a trick with us. We are used to expected values that converge. In these 'normal' situations, when you repeat an experiment a number of times, the average gain you get will approach the expected value.

When the expected value is infinite on the other hand, it means that we cannot expect the average gain to converge to anything.

Even if the expected value does not converge, it is possible to carry out a long series of the experiment in reality, so we have to understand what it means that the average gain does not converge.

If you were to try to run a series of experiments with a stochastic variable with an expected value that is infinite (easily simulated in a spreadsheet), you would see that the average gain would keep jumping around. This happens because the distribution is such that the largest gain in the series will be comparable to the sum of all the other gains. As a result it doesn't matter what you have done in all the experiments that didn't have the largest gain. Just one experiment decides what you have gained after the long series of
experiments. As just one experiment decides the outcome, the argument for switching envelopes based on expected values disappears. There is a 50% chance that you picked the envelope with the largest amount of money in the dominating experiment in the first place in which case keeping the envelope will give you the largest gain for the series. This is then the reason why arguments based on expected values fail when the expected value does not converge.

Yours truly

Ib Jørgensen