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EnumaElish is offline
Dec19-06, 11:39 PM
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Quote Quote by Fredrik View Post
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

[tex]\frac{1}{2}\cdot 2A+\frac{1}{2}\cdot \frac{A}{2}=A+\frac{A}{4}=\frac{5}{4}A[/tex]

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

[tex]\frac{1}{2}\cdot X+\frac{1}{2}\cdot 2X=\frac{3}{2}X[/tex]

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 >=< p2A 2A + pA/2 A/2. Since the two amounts are either "A" and "2A," or "A" and "A/2," p2A > 0 implies pA/2 = 0, and vice versa. Therefore p2ApA/2 = 0. If pA/2 > 0 then

pA/2 A >=< pA/2p2A 2A + pA/22 A/2

pA/2 A >=< pA/22 A/2

A >=< pA/2 A/2

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

Alternatively, if p2A > 0 then

p2A A >=< p2A2 2A + p2ApA/2 A/2

p2A A >=< p2A2 2A

A >=< p2A 2A

1/2 >=< p2A, and one would switch only if 1/2 < p2A. Taking p2A = 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.]
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