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Jeff Kaufman's Writing on expected_valueen-us/p/expected-value-and-the-two-envelope-problemExpected Value and the Two Envelope Problem
https://www.jefftk.com/p/expected-value-and-the-two-envelope-problem
expected_value04 Sep 2011 08:00:00 EST<p><span>
Someone puts $X in one envelope and $2X in another. They hand you
one at random. Before you open it and keep the contents, they offer
to let you switch your envelope for the other one. You can't detect
any differences between the two envelopes. Should you accept their
offer to switch?
</span>
<p>
Common sense says no: the two are identical, you have even odds of
getting $X and $2X. Expected value, however, says yes. Whatever
envelope you have has some amount of money $Y in it. The other
envelope has, with even odds, $2Y and $0.5Y. The expected value of
switching, then, is the average of $2Y and $0.5Y, or $1.25Y, minus
the $Y in the envelope you have. That would be $0.25Y as the
expected value of switching, so you should switch. Unfortunately,
now that you have switched, you would repeat all this logic,
determine that the other envelope had more money, and switch again.
Forever. Does expected value trap us in eternal loops of envelope
cycling?
</p>
<p>
<a href="http://en.wikipedia.org/wiki/Two_envelopes_problem">This</a>
is often considered to be a demonstration of how expected value can
give unreasonable answers. But perhaps a solution to this is to
frame things differently. We know one of the envelopes has $X and
the other has $2X. In switching you are either going from $X to $2X
or the reverse; either gaining $X or losing $X. So the expected
value of switching is the average of $X and -$X, or zero.
</p>
<p>
The second framing gives a more reasonable answer, but is there any
other reason to prefer it to the first?
</p>
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