PS Pratyush Sharma

Probability and paradox

The flaw in the two envelopes paradox

Question

Imagine you are presented with two identical envelopes. One contains some amount of money and the other contains exactly twice that amount. You choose one at random, open it, and find an amount \( x \). You are then given the option to switch to the other envelope. Since the other envelope must contain either \( x/2 \) or \( 2x \) with equal probability, the expected value of switching is \( 0.5(x/2) + 0.5(2x) = 1.25x \), which is strictly greater than \( x \). You should therefore switch. But before opening the second envelope, you are presented with the same choice again.

If that were valid, switching would always increase your expected winnings. But that cannot be true, because the envelopes are identical and you could just as easily have picked the other one first. The game is entirely symmetric. Where is the flaw in the reasoning?

The mathematical flaw

Let the two envelopes contain amounts \( A \) and \( 2A \), where \( A \) is drawn from some prior probability distribution \( P(A) \). Let \( X \) be the amount in the envelope you picked, and \( Y \) be the amount in the other envelope.

The expected value of switching, conditioned on seeing \( X = x \), is \( E[Y \mid X = x] \). The other envelope contains either \( x/2 \) or \( 2x \). Therefore:

\[ E[Y \mid X=x] = 2x \cdot P(Y=2x \mid X=x) + \frac{x}{2} \cdot P\left(Y=\frac{x}{2} \mid X=x\right) \]

To understand why the paradox fails, we look at Bayes' theorem. Knowing that your envelope contains \( x \), the true smaller amount \( A \) must be either \( x \) (you picked the smaller envelope) or \( x/2 \) (you picked the larger envelope).

Let's calculate the exact probability that you are holding the smaller envelope, \( P(A=x \mid X=x) \). Since you pick one of the two envelopes at random, the conditional probability of pulling \( x \) is \( 0.5 \), regardless of the amounts inside. Applying Bayes' theorem:

\[ \begin{aligned} P(A=x \mid X=x) &= \frac{P(X=x \mid A=x) \cdot P(A=x)}{P(X=x \mid A=x) \cdot P(A=x) + P(X=x \mid A=x/2) \cdot P(A=x/2)} \\[10pt] &= \frac{0.5 \cdot P(A=x)}{0.5 \cdot P(A=x) + 0.5 \cdot P(A=x/2)} \\[10pt] &= \frac{P(A=x)}{P(A=x) + P(A=x/2)} \end{aligned} \]

The naive paradox unconditionally assumes that the other envelope is equally likely to contain \( 2x \) or \( x/2 \). In our formula, that assumption literally requires setting the probability \( P(A=x \mid X=x) \) to exactly \( 0.5 \). If we substitute \( 0.5 \) into the left side of our equation above and simplify, it forcefully demands the mathematical requirement that \( P(A = x) = P(A = x/2) \) for absolutely any real number \( x \) you might open.

Because money in the real world has financial limits, you simply cannot have a probability distribution where \( x \) and \( x/2 \) are always equally likely to be put in the envelopes everywhere across the spectrum. If you open an astronomically massive sum, it is highly probable you hold the larger envelope, which violently shifts the probabilities away from \( 0.5 \).