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}\left(\frac{1}{10}x\right)+\frac{1}{2}\left(10x\right) = 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’s a bug in this:
\begin{equation} y = \frac{1}{2}\left(\frac{1}{10}x\right)+\frac{1}{2}\left(10x\right) \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!