User:Martin Hogbin/Two envelopes problem

The first error in the formulation is in step 1. You cannot denote an unknown amount chosen by chance by a variable as you would a generally unknown amount.

The most common way to explain the paradox is to observe that A is used inconsistently in the expected value calculation, step 7 above. In the first term A is the smaller amount while in the second term A is the larger amount. To mix different instances of a variable or parameter in the same formula like this isn't legitimate, so step 7 is thus the proposed cause of the paradox.[2]

For example, if the lower of the two amounts is denoted by C, the expected value calculation may be written as

   \frac{1}{2} C + \frac{1}{2} 2C = \frac{3}{2} C.

Here C is a constant throughout the calculation—as it should be—and we learn that 1.5C is the average expected value in either of the envelopes. So according to this new calculation there is no contradiction between the decisions to keep or to swap, and hence no need to swap indefinitely.

Although this may be the simplest solution, it is not the best one because it assumes that step 6 is correct. Step 6 is false because no distribution of unbounded values can be uniform. If every number in the distribution (i.e. every possible amount of money in the envelope) had the same non-zero probability, b, then the sum of the probabilities would be infinity times b, instead of 1. Thus it is not possible that A/2 and 2A are equally probable, ie the probabilities of large values must fall off.[3]

Another way to see the problem that keeps A constant is as follows. Let A be the amount of the first envelope you pick. Then it follows that the expected value of the money in the other envelope is

   \frac{1}{2} \frac{A}{2} + \frac{1}{2} 2A = \frac{5}{4} A.

You switch because the expected return is higher than what you hold in your hand (\frac{5}{4}A > A). If you switch back, you will only have A. Therefore, you should stick with only one switch. The paradox then becomes a question of why should you choose one and then switch, when you could have just as easily chosen the other one first.

The disconnect between intuition and the math stems from the fact that the formula in step 7 is a valid Bayesian formulation of a problem, but it is of the wrong problem. The formula inadvertently introduces a third possible outcome into the scenario. The three envelope values are A, 2A, and \frac{A}{2}. This conflicts with the problem statement that there are two envelopes.

The situation described by the formula in step 7 is one in which you are offered to play a game where you must choose between two envelopes, containing 2A and \frac{A}{2}, where A is the amount of money you forfeit in order to play. In this case, it is worth your while to play, since the expected return is \frac{5}{4}A.

A reformulation of step 7 would be this: \frac{1}{2}A + \frac{1}{2}2A = \frac{3}{2}A. This means that you have a fifty-fifty chance of choosing A over \neg A, which is what is expected. There is no reason to switch.