Houjun Liu

Double Envelope Problem

One envelope has 10 times the money in the other money.

WLOG let \(x\) be the envelope in Cary’s hand. The money in \(y\), then, \(y = \frac{1}{2}\qty(\frac{1}{10}x)+\frac{1}{2}\qty (10x) = 0.05x+5x = 5.05x\). Wat.

Basically; regardless if Cary took the envelope \(x\) or \(y\), the other envelope is expected to have \(5\times\) more money. What.


There’s a bug in this:

\begin{equation} y = \frac{1}{2}\qty(\frac{1}{10}x)+\frac{1}{2}\qty (10x) \end{equation}

is not true! There is a human PRIOR BELIEF!! Its very unlikely that mykel/chris put 10000 dollars into an envelope; so each individual amount in an envelope has an exogenous probability of it happening!