Envelope Paradox 1

You are given two sealed envelopes, one of which contains twice as much money as the other. You select one at random and may either keep it or switch to the other. Should you switch?

Answer

Right Approach:

Without opening (works for any prior), let $$Y$$ be the smaller amount and $$2Y$$ the larger. You hold one at random, so

\begin{aligned} &\tfrac12\text{ chance you hold }Y\quad\Longrightarrow\quad\text{switching gains }=+Y,\\ &\tfrac12\text{ chance you hold }2Y\quad\Longrightarrow\quad\text{switching loses }=-Y. \end{aligned}

Therefore

\tfrac12(+Y) + \tfrac12(-Y) = 0,

and you are indifferent.

Wrong Approach:

People get it wrong by first peeking and seeing an amount $$x$$ , then (incorrectly) assuming the other envelope must contain either $\tfrac12x$ or $$2x$$ with equal probability, giving

\tfrac12(2x) + \tfrac12\bigl(\tfrac12x\bigr) = 1.25\,x,

and suggesting an endless switch.

The reason for this paradox is, because this reasoning implies a certain distribution for $$Y$$ which is not a probability distribution.

To see this, see that for any fixed $$x_0$$ we would have

\begin{aligned} \text{total mass} &\;\ge\; P(Y = x_0) \;+\; P\!\Bigl(Y = \tfrac{x_0}{2}\Bigr) \;+\; P\!\Bigl(Y = \tfrac{x_0}{4}\Bigr) \;+\;\cdots\\ &=\; P(Y = x_0)\,\bigl(1 + 1 + 1 + \cdots\bigr)\\ &=\; +\infty. \end{aligned}