Expected Value and the Two Envelope Problem |
September 4th, 2011 |
expected_value [html] |
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?
This 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.
The second framing gives a more reasonable answer, but is there any other reason to prefer it to the first?
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