The Other Ball in the Box
There are three boxes, each containing two balls:
- One box has two blue balls (BB).
- One box has one blue and one red ball (BR).
- One box has two red balls (RR).
You pick a box at random and draw one ball from it -- it comes out blue. What is the probability that the other ball still in that box is also blue?
Hints
- Don't treat the two blue-containing boxes as equally likely -- one of them was twice as likely to produce the blue ball you drew.
- Think at the ball level, not the box level: you drew one of 3 blue balls uniformly, and 2 of those 3 live in the BB box.
- Apply Bayes' theorem with likelihoods $P(\text{blue} \mid \text{BB}) = 1$ and $P(\text{blue} \mid \text{BR}) = \frac{1}{2}$, then normalize.
Worked Solution
How to Think About It: Your first instinct might be: "I drew from one of two boxes that had any blue balls -- BB or BR -- so it's 50/50." That's wrong, and it's wrong in a way that comes up constantly in conditional probability puzzles. The issue is that BB was more likely to produce the blue ball you drew than BR was. You need to weight by how easy it was for each box to give you blue.
Quick Estimate: There are 6 balls in total across all three boxes: 3 blue and 3 red. Each individual ball was equally likely to be the one you drew. Given that you drew a blue ball, you drew one of the 3 blue balls uniformly. Two of those 3 blue balls live in the BB box -- so there's a 2/3 chance the box you picked is BB, meaning the other ball is also blue. The answer is $\frac{2}{3}$, not $\frac{1}{2}$.
Approach: Confirm with Bayes' theorem to make it airtight.
Formal Solution:
Label the boxes: BB (two blue), BR (one blue, one red), RR (two red). Each box is equally likely a priori: $P(\text{BB}) = P(\text{BR}) = P(\text{RR}) = \frac{1}{3}$.
The likelihoods of drawing a blue ball from each box are:
$$P(\text{blue} \mid \text{BB}) = 1, \quad P(\text{blue} \mid \text{BR}) = \frac{1}{2}, \quad P(\text{blue} \mid \text{RR}) = 0.$$
Marginal probability of drawing blue:
$$P(\text{blue}) = 1 \cdot \frac{1}{3} + \frac{1}{2} \cdot \frac{1}{3} + 0 \cdot \frac{1}{3} = \frac{1}{3} + \frac{1}{6} = \frac{1}{2}.$$
Posterior probability the box is BB given we drew blue:
$$P(\text{BB} \mid \text{blue}) = \frac{P(\text{blue} \mid \text{BB}) \cdot P(\text{BB})}{P(\text{blue})} = \frac{1 \cdot \frac{1}{3}}{\frac{1}{2}} = \frac{2}{3}.$$
If the box is BB, the remaining ball is blue with certainty. If the box is BR (posterior probability $\frac{1}{3}$), the remaining ball is red. So:
$$P(\text{other ball is blue} \mid \text{drew blue}) = \frac{2}{3}.$$
Answer: $\dfrac{2}{3}$.
Intuition
The mistake almost everyone makes is to say "there are two boxes with blue balls, I'm in one of them, so it's 50/50." The flaw is ignoring the selection mechanism. The BB box produces blue balls twice as often as the BR box, so observing a blue ball makes BB twice as probable as BR. This is Bayes' theorem in its most basic form: the data you observe shifts your beliefs in proportion to how likely each hypothesis was to produce that data.
This type of reasoning shows up constantly in quant work -- in adverse selection models (a counterparty who wants to trade with you is more likely to have seen bad news than good), in signal processing (a sensor that fires more often under hypothesis A gives you more evidence for A), and in Bayesian updating generally. The ball-counting shortcut -- 3 blue balls total, 2 are in BB, so the chance is 2/3 -- is a clean sanity check that works whenever the prior is uniform and the balls are exchangeable. Whenever you catch yourself ignoring selection probabilities, stop and weight by likelihood.