34  Two envelope paradox*

Example 34.1 (Two-envelope paradox) Imagine you are given two identical envelopes, each containing money. One contains twice as much as the other. You may pick one envelope and keep the money it contains. Having chosen an envelope at will, but before inspecting it, you are given the chance to switch envelopes. Should you switch?

The paradox arises when you try to solve the expectation. Let \(A\) denote the amount of money in the envelope you have chosen, and \(B\) denote the amount of money in the other envelope.

We know \(B\) is either twice as much as \(A\), or half as much as \(A\). Each with probability \(1/2\). So

\[E(B) = \frac{1}{2}(2A) + \frac{1}{2}(A/2) = \frac{5}{4}A\]

Since \(E(B)>A\), you should always switch! However, after you switch to \(B\), by the same argument, you should switch back to \(A\). You you switch back and forth indefinitely!

Where do things go wrong? The error in this calculation lies in a subtle misunderstanding: the two \(A\)s in the calculation actually represent different values, that are incorrectly equated. In particular, the \(2A\) represents the expected value in the other envelope given that it is the larger one, and the \(A/2\) represents the expected value in the other envelope given that it is the smaller one.

\[E(B) = E(B|B<A)P(B<A) + E(B|B>A)P(B>A)\]

Suppose the amount of money in the two envelopes are \(a\) and \(2a\) respectively. \(E(B|B<A)=a\) and \(E(B|B>A)=2a\). Therefore,

\[E(B)=\frac{1}{2}a + \frac{1}{2}2a = \frac{3}{2}a.\]

The same calculation applies to \(E(A)\). Thus, \(E(A)=E(B)\).