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# System prepended metadata

title: To switch or not to switch

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To switch or not to switch
==

Suppose someone shows you two identical sealed envelopes each with a cheque in it. You are told that one of the cheques has twice the amount of money of the other. You are allowed to open one envelope to see what’s inside and based on what you see, either switch to the other envelope or accept the cheque in the first envelope. What are you going to do?

We can compute the expected value of switching to see if it is beneficial to switch. Write $X$ for the amount that you see on the cheque of the envelope that you opened. The other envelope has either $2X$ or $\frac{1}{2}X$ written on the cheque. So the expected payoff of switching is
\begin{equation}
\frac{1}{2}(2X) + \frac{1}{2}(\frac{1}{2}X) = 1.25X.
\end{equation}
The expected value of switching is larger than $X$?! Whatever you see on the first envelope, you have to switch to maximize expected value. But if you don't open the envelope and keep switching, you can get infinite money (in expectation)!

Something has gone terribly wrong. But what? 

Let's try to understand what is going on using an example. Suppose that the two envelopes contain $100 and $200 and you picked randomly the one with $100. In the argument above, I wrote that the other envelope contains either $200 or $50 each with probability $\frac{1}{2}$. It is now clear that this where the above calculation went wrong. There is no envelope with $50.

It looks like we are mixing up two possible situations. The situation where the two envelopes contain $50 and $100 or $100 and $200. Both situations are valid instantiations of the game that we can play, but they won't occur simultaneously.

The correct calculation of the expected gain of switching is done as follows. Suppose the two envelopes contain $M$ and $2M$ where $M$ can be anything, for example $100.  and you choose randomly one of them, then given that you chose $M$, if you switch you gain $M$. But if you chose $2M$ and you switch, you lose $M$. Making the total expected gain 0. The same calculation works if the two envelopes contained $M$ and $M/2$. This makes much more sense :).


