You are told there are two envelopes. One contains twice as much money as the other one. You pick one but are allowed to change your mind after picking it. (You are equally likely to pick the one with less money as the one with more money.)
To figure out how much on average the other envelope should contain, one might average x/2 and 2x because one is equally likely to pick one as picking the other. Since this comes out to 5x/4, one might always change his or her mind. But wouldn't this end up with one never making up his or her mind?
Solution : The difficulty lies in not knowing what the amounts are.
So, you need to have some idea of what the prior distribution of money in the envelope is before you can do the calculation.
So, if I tell you that the 2 amounts are, for example, $10 and $20, then if you took the $10 envelope (i.e., if x = 10), there is no chance that the amount in the other envelope is $x/2.
It must be $2x = $20
Similarly, if you took the $20 envelope, the other envelope must be $10.
So conditioning on whether you took the $10 0r $20 envelope,
the expected value of the other envelope is actually
(1/2)*(20) + (1/2)*(10) = $15.
However, if you observe what's in your envelope, then you can condition on what you see,
the expected value of the other envelope is $20 if you see $10 in yours, or vice versa.
If you see $10, then you should switch, and if you see $20, you should not.
So there is no paradox in this case.
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