A classic puzzle for Christmas.
You are presented with two sealed envelopes. Inside each envelope is some amount of money, and you know the following:
- One envelope contains exactly twice as much money as the other.
- You are allowed to choose one envelope and keep the money inside.
After you select an envelope, the game host offers you a deal: you can switch to the other envelope if you want.
You reason as follows:
Let the amount in the envelope you chose be
x.If you switch, the other envelope could either:
- Contain
2x(if our current envelope contains the smaller amount), or - Contain
\frac{x}{2}(if your current envelope contains the larger amount).
On average, switching seems to give you:
\text{Expected Value} = \frac{1}{2}(2x) + \frac{1}{2}\left(\frac{x}{2}\right) = \frac{5x}{4}This is greater than the
x you currently hold, so your reasoning seems to indicate you should always change!What's the mistake in this reasoning? As this is a famous problem, please don't look up the solution online and only answer if you haven't heard of this problem before~~
new information being gainedcondition is harder to assess...xvariable for the two cases, as they are different. See #823807.