Bertrand’s Box Paradox Answer: 2 in 3 [Solved]

The Bertrand’s box paradox answer is 2 in 3. You pick a box at random, pull out one coin, and it is gold. The other coin sitting in that same box is gold with probability 2 in 3, not 1 in 2. Yet the 50-50 guess is the trap, and it catches almost every candidate who hears this question for the first time.

First posed by Joseph Bertrand in his 1889 book Calcul des probabilités, this puzzle still shows up in quant and trading interviews. It tests one skill: whether you weight evidence by how likely each suspect was to produce it. A clear modern walkthrough of Bertrand’s argument is on Allen Downey’s Probably Overthinking It.

The Question: Three Boxes, One Gold Coin

Three boxes sit on the table. They look identical from the outside. The first box holds two gold coins, the second holds two silver coins, and the third holds one gold coin and one silver coin. You pick a box at random, reach in, and pull out one coin at random. It is gold. Now the interviewer asks: what is the probability that the other coin in that same box is also gold?

Most candidates answer 1 in 2 without a pause. Their reasoning sounds airtight. The silver-silver box is out, because it could never have produced a gold coin. That leaves two boxes, and only one of them has a second gold coin. Two boxes, one favorable. Fifty-fifty. The reasoning is neat, confident, and wrong.

Why the Bertrand’s Box Paradox Answer Is 2 in 3

Forget the boxes for a moment and count the gold coins. There are exactly three gold coins in the whole setup. Two of them sit in Box 1, one sits in Box 3, and Box 2 holds none.

Next, ask a different question: which gold coin are you holding? You drew a gold coin, and every gold coin in the setup had the same chance of being the one you drew. Two of the three gold coins live in Box 1, where the other coin is gold. Only one of the three lives in Box 3, where the other coin is silver. So the coin in your hand came from Box 1 twice as often as it came from Box 3. The probability the other coin is gold is 2 in 3.

That is the whole trick. The two surviving boxes were never equally likely to hand you a gold coin: Box 1 produces gold every single time, while Box 3 produces gold only half the time. When you hold a gold coin, bet on the box that makes gold coins more often.

Run It 300 Times and Watch the Numbers

If the coin-counting argument still feels like a trick, run the game 300 times in your head. Each box gets picked about 100 times.

Box picked	Times picked	Gold coins drawn	Other coin is gold
Box 1: gold, gold	100	100	100
Box 2: silver, silver	100	0	0
Box 3: gold, silver	100	50	0
Total	300	150	100

Out of 150 gold coins drawn, 100 of them have a gold partner in the same box. That is 100 out of 150, which is 2 in 3. The other 50 came from Box 3, and their partners are all silver. No formulas, just counting.

The Bayes Table Behind the Bertrand’s Box Paradox Answer

The same result also falls out of Bayes theorem, and it is worth knowing both versions. The formal version impresses an interviewer who asks “can you prove it,” while the coin-counting version is the one you lead with. Here is the full table.

Box	Prior probability	Chance of drawing gold	Contribution	Probability it is this box, given gold
Box 1: gold, gold	1 in 3	1	1/3	2 in 3
Box 2: silver, silver	1 in 3	0	0	0
Box 3: gold, silver	1 in 3	1/2	1/6	1 in 3
Total	1	–	1/2	1

Read the table from left to right. Each box starts with a prior probability of 1 in 3. The chance of drawing gold is 1 for Box 1, 0 for Box 2, and 1 in 2 for Box 3. Multiply the prior by the chance of gold and you get each box’s contribution: 1/3, 0, and 1/6. Add them up and the total chance of drawing gold is 1 in 2, which makes sense, because half of all six coins are gold. Now divide each contribution by that total. Box 1 gives (1/3) divided by (1/2), which is 2 in 3, while Box 3 gives (1/6) divided by (1/2), which is 1 in 3.

Why 1 in 2 Feels Right and Is Wrong

The 50-50 argument makes one silent mistake: it counts cases instead of weighing them. “Two boxes left, one favorable” treats Box 1 and Box 3 as equals. They are not equals once a gold coin is on the table. Box 1 was always going to show gold, while Box 3 had to get lucky. Evidence does not just remove suspects, it favors the suspects who were most likely to leave that evidence behind. That is the point Bertrand was making in 1889: merely counting cases is not always proper.

The Silver Version Gives the Same Answer

Similarly, the puzzle works in reverse. Suppose the coin you pulled out had been silver instead of gold. What is the probability the other coin in the box is also silver? Count the silver coins: three in total, two in Box 2 and one in Box 3. The coin in your hand is twice as likely to come from Box 2, so the answer is 2 in 3 again. Symmetry is a good habit in probability interviews.

What the Interviewer Is Testing in the Bertrand’s Box Paradox Answer

This question looks like a party trick, but it measures four things a trading desk or quant team actually cares about.

• Conditional probability under pressure. Can you update a belief when new information arrives, or do you freeze on the first count that looks right? The 50-50 answer is what a candidate says when they stop thinking after the first step.
• Likelihood weighting. The core skill is noticing that evidence has a strength, not just a direction. A gold coin is stronger evidence for Box 1 than for Box 3, because Box 1 produces gold twice as often.
• Plain-words explanation. Anyone can write down Bayes theorem. The strong candidate explains it by counting three gold coins, in one sentence, without a formula. Interviewers remember the candidate who makes the answer obvious.
• Resistance to the obvious trap. The 50-50 answer is so tempting that defending 2 in 3 against a skeptical interviewer is part of the test. The 300-times table is your defense: it turns a debate about intuition into arithmetic.

A strong thirty-second answer sounds like this: “Two in three. There are three gold coins in total, two in the double-gold box and one in the mixed box, so the gold coin I am holding is twice as likely to come from the double-gold box.” Then walk through the 300-times table if the interviewer asks for proof.

How This Connects to the Monty Hall Problem

If Bertrand’s boxes felt familiar, there is a reason. The Monty Hall problem runs on the same error: treating the surviving options as equals after new information arrives. In Monty Hall, the host opens a goat door, and most people say the two remaining doors are 50-50. They are not, because the host was always going to open a goat door, while your first pick was right only 1 time in 3. In Bertrand’s boxes, the draw plays the host’s role: the gold coin was always going to appear from Box 1, but only half the time from Box 3.

Try Another Puzzle Next

If you liked the Bertrand’s box paradox, the next puzzle to try is the Monty Hall problem: three doors, one car, and a host who knows where it is. Should you switch? Then test yourself on the two envelopes paradox, where switching looks free and still gains nothing, and the birthday paradox, where 23 people are enough for a shared birthday.

Frequently Asked Questions

Is the Bertrand’s box paradox answer really not 1 in 2?

Yes, it really is 2 in 3. The 1 in 2 answer counts the two surviving boxes as equals. They are not equals: the double-gold box produces a gold coin every time, while the mixed box produces one only half the time. Count the three gold coins instead of the two boxes, and the answer is 2 in 3.

What if the first coin drawn is silver?

The answer is 2 in 3 that the other coin is silver too. Three silver coins exist, two in the double-silver box and one in the mixed box, so the silver coin you hold is twice as likely to come from the double-silver box.