The two-envelopes paradox

HardProbability Conditional Probability & Expectation Expected Value~7m

Two envelopes contain money, one twice the other. You open yours and see $X$ . The "switch" argument says the other envelope holds $\tfrac12 X$ or $2X$ , each with probability $\tfrac12$ , for an expected $\tfrac54 X > X$ - so you should always switch. Where does this reasoning break down?

Show hints (2)+

Ask what prior over amounts would justify the $\tfrac12 / \tfrac12$ split at every $X$ .
No proper distribution is flat over an unbounded range.

Answer

Reveal answer →

The $\tfrac12 / \tfrac12$ split cannot hold for all $X$ under any proper prior, so the $\tfrac54$ edge is illusory.

Create a free account to see the full worked solution and track your progress.

Create a free account

Interview format: Market Maker, Systematic & Data-Driven Trading

The two-envelopes paradox

Answer

Related questions