[PairTheBoard]

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Suppose you open the envelop and there is 20 dollars in it.
So in the other envelop there is either 10 dollars or 40 dollars.


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I'd say: as soon as you say this, realize it's a mistake, a contradiction, and "erase" your statement. Seeing one envelop tells you nothing about the other.


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No. That's not where the mistake is being made. If you accept the information that one envelope contains twice the other, seeing $20 does tell you something you didn't know before about the other envelope. You certainly now know that it doesn't contain $100. You didn't know this before seeing the $20. Logically, the conclusion is forced that your new, incomplete, state of knowledge about the other envelope is that it must have either $10 or $40. Another way to put this would be, seeing the $20 now eliminates all other possible amounts for the other envelope except for $10 and $40.

The mistake comes from taking these two remaining possiblities for the other envelope and assigning them both probabilities of 50%. As you pointed out before, that makes no sense. That would mean that when you open the other envelope, half the time you would see $10 and half the time you would see $40. That just doesn't happen. Every time these envelopes are offered and someone sees $20 there is only one possibility for the other envelope. That's where you're correct. As far as we know now, it could be $10 or $40. That's where TimWillTell's statement is correct. But whichever it is, that's what it will be every time. That's where the 50% probabililty assumption is wrong for purposes of computing EV with respect to switching the $20 envelope.

PairTheBoard