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@jemoka / Jemoka Knowledge Base / wiki/concepts/double_envelope_problem.md
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--- title: "Double Envelope Problem" type: concept source: https://www.jemoka.com/posts/kbhdouble_envelope_problem/ confidence: high status: active --- 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. Why? 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!