Bertrand's Box Paradox
Two boxes sit in front of you. Box 1 contains one $\$1$ bill and one $\$5$ bill. Box 2 contains two $\$1$ bills.
You pick a box uniformly at random, then draw a bill from it at random. The bill you drew is a $\$1$ bill.
What is the probability that the second bill in your chosen box is also a $\$1$ bill?
Hints
- The observation that you drew a $\$1$ bill is not equally likely under the two boxes. Which box makes this observation more probable?
- Apply Bayes' theorem: $P(B_2 \mid E) = \frac{P(E \mid B_2) P(B_2)}{P(E)}$. Compute $P(E)$ by conditioning on the box.
- There are three $\$1$ bills total across the two boxes. Two of them belong to Box 2. If you drew a $\$1$ bill, what fraction of the time did it come from Box 2?
Worked Solution
How to Think About It: This is a classic Bayes' theorem exercise disguised as a simple box problem. The trap is thinking "I picked one of two boxes, so it's 50-50." But the observation -- drawing a $\$1$ bill -- gives you information about which box you are in. Box 2 (two $\$1$ bills) is more likely to produce a $\$1$ bill on the first draw than Box 1 (which only has one). So seeing a $\$1$ bill should shift your belief toward Box 2. The answer should be greater than $1/2$.
Quick Estimate: There are three $\$1$ bills total across both boxes. Two of them live in Box 2. If you drew a $\$1$ bill, each of those three bills was equally likely to be the one you grabbed (since box choice is $1/2$ and bill choice within a box is $1/2$, each bill has probability $1/4$ -- but the $\$5$ bill is ruled out). So conditional on drawing a $\$1$ bill, the probability it came from Box 2 is $2/3$. If you are in Box 2, the remaining bill is certainly $\$1$. So the answer is $2/3$.
Approach: Formally apply Bayes' theorem, conditioning on which box was chosen.
Formal Solution:
Let $B_1$ = event you chose Box 1, $B_2$ = event you chose Box 2, and $E$ = event the first draw is a $\$1$ bill.
We want $P(B_2 \mid E)$, since if you are in Box 2 the second bill is guaranteed to be $\$1$, and if you are in Box 1 the second bill is the $\$5$ bill.
By Bayes' theorem:
$$P(B_2 \mid E) = \frac{P(E \mid B_2) \, P(B_2)}{P(E \mid B_1) \, P(B_1) + P(E \mid B_2) \, P(B_2)}$$
The pieces: - $P(B_1) = P(B_2) = \frac{1}{2}$ (uniform box selection) - $P(E \mid B_1) = \frac{1}{2}$ (Box 1 has one $\$1$ bill and one $\$5$ bill) - $P(E \mid B_2) = 1$ (Box 2 has two $\$1$ bills, so you always draw a $\$1$ bill)
Plugging in:
$$P(B_2 \mid E) = \frac{1 \cdot \frac{1}{2}}{\frac{1}{2} \cdot \frac{1}{2} + 1 \cdot \frac{1}{2}} = \frac{\frac{1}{2}}{\frac{1}{4} + \frac{1}{2}} = \frac{\frac{1}{2}}{\frac{3}{4}} = \frac{2}{3}$$
Answer: The probability the second bill is also a $\$1$ bill is $\dfrac{2}{3}$.
Intuition
This is the simplest version of Bertrand's Box Paradox, and it illustrates one of the most common errors in probability: ignoring the likelihood of the evidence. People see two boxes and instinctively say 50-50, but the act of observing a $1 bill is itself informative -- it is twice as likely if you are in the all-$1 box. Bayes' theorem is just the formal machinery for incorporating that information.
This exact pattern shows up constantly in quant work. A market maker sees a large buy order -- does it come from an informed trader or a noise trader? The order itself shifts the posterior, just like the drawn bill shifts your belief about which box you are in. Failing to update properly is the core of adverse selection, and it is exactly Bertrand's Box in disguise.