---
title: "Double Envelope Problem"
source: https://www.jemoka.com/posts/kbhdouble_envelope_problem/
---

One envelope has 10 times the money in the other money.
WLOG let \(x\) be the envelope in Cary&rsquo;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.
Why? There&rsquo;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!
