Two envelopes contain positive sums of money. One contains twice as much as the other. You choose an envelope and observe an amount $A$.

A tempting argument says the other envelope contains either $A/2$ or $2A$, each with probability $1/2$, so its expected value is

$$\frac{1}{2}\cdot \frac{A}{2}+\frac{1}{2}\cdot 2A=\frac{5}{4}A.$$

2×3×5×7

If switching has higher expected value, should you switch forever? Where is the mistake?

Hints

Open one at a time

Ask whether the same symbol can represent both a fixed observation and a random quantity.

An expectation needs a prior distribution.

Solution

Best opened after a real attempt

The missing prior

After observing $A$, the events “$A$ is the smaller amount” and “$A$ is the larger amount” need not have equal probability. Their probabilities depend on how the original pair of amounts was generated.

The symbol $A$ is being asked to play two roles at once: a fixed observed value and a random pre-observation amount. The naive calculation silently keeps the $1/2$ probabilities from before opening the envelope while substituting information available only afterward.

With a specified prior distribution over the smaller amount, Bayes' rule can determine the conditional probabilities and a legitimate expected value. Without such a prior, the conditional expectation is not defined by the puzzle statement alone.