The setup
There are three identical boxes, each with two drawers, and each drawer holds one coin.
- One box holds two gold coins (call it GG).
- One box holds two silver coins (SS).
- One box holds one gold and one silver coin (GS).
You pick a box at random, open one drawer at random, and see a gold coin. What is the probability that the coin in the other drawer of the same box is also gold?
The tempting reasoning goes like this: the box cannot be SS, so it is either GG or GS. Those are two boxes, so the chance is one half. The correct answer is two thirds. The puzzle is credited to the French mathematician Joseph Bertrand, who published it in the nineteenth century.
Why one half is wrong
The flaw is treating the two remaining boxes as equally likely. They are not, because seeing a gold coin is more likely to happen from one of them.
Think about which box you might have been holding when you saw gold. The GG box shows gold whichever drawer you open. The GS box shows gold only if you happened to open the gold drawer, which is half the time. So given that you saw gold, the GG box is the more likely explanation.
Counting the coins
The cleanest way to see this is to label all six coins and count. The labels are only for bookkeeping:
| Box | Coin | Colour |
|---|---|---|
| GG | G1 | gold |
| GG | G2 | gold |
| GS | G3 | gold |
| GS | S1 | silver |
| SS | S2 | silver |
| SS | S3 | silver |
Picking a box at random and then a drawer at random makes each of the six coins equally likely to be the one you see, each with probability 1/3 × 1/2 = 1/6.
Three coins are gold: G1, G2 and G3. Given that you saw gold, each of the three is equally likely, with probability 1/3.
- If you saw G1, the other coin in the box is G2, which is gold.
- If you saw G2, the other coin is G1, which is gold.
- If you saw G3, the other coin is S1, which is silver.
So in two of the three equally likely cases the other coin is gold. The probability is 2/3, about 66.7 percent.
The same answer by formula
This is a direct use of conditional probability. Write the event “first coin is gold” as G and “other coin is gold” as B.
- The chance of seeing gold at all is (1/3 × 1) + (1/3 × 1/2) + (1/3 × 0) = 1/3 + 1/6 = 1/2.
- The chance of seeing gold and the box being GG is 1/3 × 1 = 1/3.
- Dividing, 1/3 ÷ 1/2 = 2/3.
The first line is an application of the law of total probability: add up the chance of the observation through each possible box. The division is the core step of Bayes theorem, which weighs how well each box explains what you saw. The same structure shows up in probability trees, where each branch is one box and one drawer.
Counting by natural frequencies
Another way to make the answer feel real is to use natural frequencies. Imagine repeating the experiment 600 times, with the boxes chosen evenly. About 200 trials use each box.
- In the GG trials, you see gold every time: about 200 trials.
- In the GS trials, you see gold about half the time: about 100 trials.
- In the SS trials, you never see gold.
That makes about 300 trials where you see gold. In about 200 of those the box is GG. So 200 out of 300, or two thirds, have gold behind the other drawer. The numbers are illustrative averages, not exact guaranteed counts, since real trials vary.
Why the paradox matters
The Bertrand box paradox is not mainly about coins. It shows a general habit: after an observation, people reduce the problem to “which options remain” and then count the options, instead of asking how likely each option was to produce the observation. The same trap sits underneath the Monty Hall problem, where the host’s behaviour carries information, and it is close in spirit to base rate neglect, where the starting proportions are forgotten.
Whenever a puzzle begins “given that you saw…” it is worth asking two questions. What did each possibility predict I would see? And were the possibilities equally likely to start with? Here, the GG box predicted a gold coin with certainty and the GS box only half the time, so the evidence favoured GG two to one.
A quick check by experiment
If the argument still feels slippery, you can test it. Put three pairs of objects in three envelopes, two of one kind, two of another, and one of each, then repeat the draw many times and record only the trials where the first object is the “gold” kind. The fraction with a matching partner should settle near two thirds as the number of trials grows, though any short run will wander. Sampling like this is the idea behind a Monte Carlo simulation, and it is a good habit for any puzzle where intuition and algebra disagree.