Re: Exchange Paradox
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Of course. The quantity x is itself a random variable.
To make it more clear, look at a concrete example, imagine there's $1 in one and $2 in the other. Then the original formulation is equivalent to stating that the EV is:
EV = 0.5*(2-1) + 0.5*(1/2 - 1) = 0.25
But this is incorrect since when you switch from the $2 to the $1 you are losing $1 and not $0.50.
The real random variable is ONLY your initial choice. So then you can formulate it as X being the initial choice. Then there's a 0.5 prob. of having started at $1 and thus the gain from switching is $1 and a 0.5 prob. of having started at $2 and thus the loss from switchin is $1. The EV is then clearly 0.
This isn't so much a paradox as much as just the wrong formula IMO.
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agreed.
You cannot substitute x for "N or 2N". The expected result of switching is always 1.5N (although achieving this result is not actually possible in the game)
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