Two variants of the necktie/envelopes problem

Fro wikipedia: (Try it without googling :)

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Variant 1

The setup: The player is given two indistinguishable envelopes, each of which contains a positive sum of money. One envelope contains twice as much as the other. The player may select one envelope and keep whatever amount it contains, but upon selection, is offered the possibility to take the other envelope instead.

The switching argument:

1. Denote by A the amount in the selected envelope.
2. The probability that A is the smaller amount is 1/2, and that it's the larger also 1/2
3. The other envelope may contain either 2A or A/2
4. If A is the smaller amount, the other envelope contains 2A
5. If A is the larger amount, the other envelope contains A/2
6. Thus, the other envelope contains 2A with probability 1/2 and A/2 with probability 1/2
7. So the expected value of the money in the other envelope is:
8. This is greater than A, so swapping is favored
9. After the switch, reason in exactly the same manner as above, but denote the second envelope's contents as B
10. It follows that the most rational thing to do is to swap back again
11. This line of reasoning dictates that envelopes be swapped indefinitely
12. As it seems more rational to open just any envelope than to swap indefinitely, the player is left with a paradox.

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Variant 2:

The solution above does not rule out the possibility that there is some non-uniform prior distribution of sums in the envelopes to give the paradox force.

Suppose that the envelopes contain the integer sums {2ⁿ, 2ⁿ⁺¹} with probability 2ⁿ/3ⁿ⁺¹ where n = 0, 1, 2,… and the value of n is not known in advance.^[1]

A sensible strategy that guarantees a win is to swap only when the opened envelope contains 1, as the other must contain 2. Furthermore, suppose the envelope contains 2. Now there are only two possibilities; the envelope pair in front of us is either {1, 2} or {2, 4}. The conditional probability that it's the {1, 2} pair is

and consequently the probability it's the {2, 4} pair is 2/5 since all other envelope pairs have zero conditional probability.

It turns out that these proportions hold in general unless the first envelope contains 1. Thus, denote by x the amount found where x = 2ⁿ for some n ≥ 1, then the other envelope containsx/2 with probability 3/5 and 2x with probability 2/5. So the expected gain by switching is

which is more than x. This means that the player should switch in all cases.

But once again, the player may go through this reasoning before opening either envelope, and deduce that the other envelope should always be chosen. This conclusion is just as clearly wrong as it was in the first and second cases. But now the flaws noted above don't apply; the x in the expected value calculation is a known constant (in every single case) and the probabilities in the formula are obtained from a specified and proper prior distribution.