Direct simple solution of the envelopes paradox

[EnumaElish]

The problem

You have two indistinguishable envelopes in front of you. Both of them contain money, one of them twice as much as the other. You're allowed to choose one of the envelopes and keep whatever's in it. So you pick one at random, and then you hesitate and think "Maybe I should pick the other one". Let's call the amount in the envelope that you picked first A. Then the other envelope contains either 2A or A/2. Both amounts are equally likely, so the expectation value of the amount in the other envelope is

$$\frac{1}{2}\cdot 2A+\frac{1}{2}\cdot \frac{A}{2}=A+\frac{A}{4}=\frac{5}{4}A$$

Since this is more than the amount in the first envelope, we should definitely switch.

This conclusion is of course absurd. It can't possibly matter if we switch or not, since the envelopes are indistinguishable. Also, if the smaller amount is X, then both expectation values are equal to

$$\frac{1}{2}\cdot X+\frac{1}{2}\cdot 2X=\frac{3}{2}X$$

so it doesn't matter if we switch or not.

The "paradox" is that we have two calculations that look correct, but only one of them can be. The problem isn't to figure out which one is wrong, because it's obviously the first one. The problem is to figure out what's wrong with it.

The proposed (in)equality is A ^>=_< p_2A 2A + p_A/2 A/2. Since the two amounts are either "A" and "2A," or "A" and "A/2," p_2A > 0 implies p_A/2 = 0, and vice versa. Therefore p_2Ap_A/2 = 0. If p_A/2 > 0 then

p_A/2 A ^>=_< p_A/2p_2A 2A + p_A/2² A/2

p_A/2 A ^>=_< p_A/2² A/2

A ^>=_< p_A/2 A/2

Clearly A > p_A/2 A/2, and one would not switch.

Alternatively, if p_2A > 0 then

p_2A A ^>=_< p_2A² 2A + p_2Ap_A/2 A/2

p_2A A ^>=_< p_2A² 2A

A ^>=_< p_2A 2A

1/2 ^>=_< p_2A, and one would switch only if 1/2 < p_2A. Taking p_2A = 1/2 in the original problem as a given, one would not switch.

[Edit: this does not work in the general case where the values are kA and A/k with k > 2.]

 

[Valkarie]

ConclusionThe paradox is resolved by realizing that the proposed inequality is not true. The two expectation values are equal, so switching does not give us an advantage either way.

 

[tacman]

The solution to the envelopes paradox lies in the incorrect assumption that the expectation value of the amount in the other envelope is always greater than the amount in the first envelope. This is not always the case, as shown in the alternative calculations above.

The flaw in the first calculation is that it assumes that the probability of the smaller amount being A/2 is equal to the probability of the larger amount being 2A. This is not true, as the smaller amount can only be A/2 if the larger amount is 2A, but the larger amount can be 2A if the smaller amount is either A or A/2.

In other words, the probabilities are not independent of each other. This is known as the fallacy of the transposed conditional, where the probability of two events occurring is not the same as the probability of one event occurring given that the other event has occurred.

In the envelopes paradox, the probability of the smaller amount being A/2 is actually higher than the probability of the larger amount being 2A. This is because if the smaller amount is A/2, then the larger amount must be 2A, but if the smaller amount is A, the larger amount can be either 2A or A/2.

Therefore, the expectation value of the amount in the other envelope is not always greater than the amount in the first envelope, and switching envelopes is not always the best strategy. The paradox is resolved by understanding the fallacy in the first calculation and acknowledging the dependence of probabilities in this scenario.