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Suppose we are given two boxes, with one of gift valued $n$ dollars and the other one valued twice as much. We can pick a box, and after open it we have the choice of switching to another box. Shall we switch?

I was asked about this during some graduate student seminar. My reasoning is since two choices are equally likely, it is equally likely I will get a box half of the original value and a box twice the original value. So I should switch as the net expectation is positive. However, this implies that my original choice is somehow inferior to the choice after switch, which violates symmetry principle. Can someone help?

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closed as off-topic by Steven Landsburg, Alexey Ustinov, Alex Degtyarev, RP_, Stefan Kohl Oct 26 '16 at 8:37

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question does not appear to be about research level mathematics within the scope defined in the help center." – Steven Landsburg, Alexey Ustinov, Alex Degtyarev, Stefan Kohl
If this question can be reworded to fit the rules in the help center, please edit the question.

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Did you see the value in the box before deciding whether to switch?

If you did not see it: You are calculating the expectation wrong. You said there are two events, the second box containing $2v$ or $v/2$. But this conditions on a particular $v$ being in the first box. You did not see what's in the first box, so this is inadmissible. The correct pair of events is that either the second box contains $2n$ and the first box contains $n$, or else the second box contains $n$ and the first contains $2n$, with probability $0.5$ for each event. (Of course you are therefore indifferent to switching.)

If you did see it: You are calculating the expectation wrong. Conditioned on opening the box and seeing $v$, there is a posterior probability that the other box contains $2v$ and a posterior probability that the other box contains $v/2$. These probabilities are determined via a Bayesian update from their prior probabilities, updating on the event that the first box contains $v$.

In general these probabilities will not be $0.5$, but we cannot compute them because you did not tell us a prior distribution on the values in the boxes. Without a prior, the posterior is not defined and you cannot calculate the expected value of switching.

For example, if you are told $n = 50$ in advance, i.e. guaranteed that one box contains $50$ and the other $100$, then of course you should switch if the first box contains $50$ but should not switch if the first box contains $100$.

If you are told that $n$ is drawn uniformly from $[0,N]$, then you should not switch if you see $v > N$ because you must already have the larger box. Etc.

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