ProbabilityVeridical paradoxResolved

Bertrand's Box Paradox

Also known as Bertrand's Box Problem, The Three Boxes Paradox

You reach into one of three boxes and pull out a gold coin. The chance the other coin in that box is gold is not one half.

When
1889
Difficulty
Beginner
Reading time
10 min
People
Joseph Bertrand

Veridical paradox. Quine's term. The result seems absurd but is demonstrably correct. Closely related to counterintuitive results, used where a formal demonstration exists.

Discover

Three wooden boxes sit on a table. Each has two little drawers, and each drawer holds one coin.

  • The first box holds two gold coins.
  • The second box holds two silver coins.
  • The third box holds one gold and one silver.

The boxes look identical from the outside. You cannot tell which is which.

You pick a box at random, pull open one of its two drawers, and out comes a gold coin.

Good news: you now know you are not holding the silver-silver box. It is either the gold-gold box or the mixed box. Two possibilities left.

So — what is the chance the other drawer of your box also holds gold?

Everybody says one half. Two boxes it could be, one of them all gold. Half.

It is two-thirds.

Understand

The mistake is subtle and it is the same mistake almost everybody makes. You counted the boxes when you should have counted the coins.

Forget boxes for a moment. Label the six coins:

  • Box A: gold G1, gold G2
  • Box B: silver S1, silver S2
  • Box C: gold G3, silver S3

You opened a drawer at random and found gold. Which gold coin are you looking at? There are exactly three gold coins in the whole setup, and — because you chose the box at random and then the drawer at random — you were equally likely to end up staring at any one of them.

So you are looking at G1, G2 or G3, each with probability one third.

Now ask what is behind the other drawer in each case:

  • If you are looking at G1, the other drawer holds G2. Gold.
  • If you are looking at G2, the other drawer holds G1. Gold.
  • If you are looking at G3, the other drawer holds S3. Silver.

Two of the three cases give gold. Two-thirds.

Why "one half" feels right and is wrong

The intuitive argument says: there are two boxes it could be, so it is 50/50. That would be correct only if the two surviving boxes were equally likely to have produced what you saw.

They are not. The gold-gold box has two ways of showing you a gold coin. The mixed box has only one. So seeing gold is twice as good evidence for the gold-gold box as it is for the mixed one, and the odds shift from 1:1 to 2:1.

Bertrand's own way of putting it: the question "is the other coin gold?" is really the question "did I pick the first box?" — and that question was settled before you opened anything.

Bertrand's proof that 1/2 is impossible

Bertrand did not just give the right answer; he showed the wrong one is self-defeating, and the argument takes four lines.

Suppose the answer really were 1/2. Then:

  • If you see gold, the chance the box is same-metal is 1/2.
  • If you see silver, by the identical argument, the chance is also 1/2.

But you must see one or the other. So before opening any drawer, the chance that your box is a same-metal box would be 1/2.

And that is plainly false. Two of the three boxes are same-metal, so the chance is 2/3.

Something you have not yet observed cannot depend on an observation you have not yet made. The assumption of 1/2 contradicts a fact you already knew, so it must go. Run the same argument with 2/3 and everything agrees.

The shape of it

opened 1/3 2/3 0
What the evidence was answering. Something gets opened, and the odds on everything else move. How far they move depends entirely on why that one was opened.

Examples

Bertrand's own scaling illustration. He asks you to imagine three hundred boxes instead of three: 100 gold-gold, 100 silver-silver, 100 mixed. Open one drawer of every box. You are now looking at 300 coins.

Boxes How many Coin you see
Gold-gold 100 gold, always
Silver-silver 100 silver, always
Mixed 100 gold about 50 times, silver about 50 times

Count the gold coins on display: 100 from the gold-gold boxes and about 50 from the mixed ones. 150 gold coins, of which 100 sit in a box whose other drawer is also gold.

$100/150 = 2/3$. No conditional probability notation required — just a tally.

The complete sample space. Every equally likely outcome of "pick a box, pick a drawer":

Box Drawer opened Coin seen Other coin
A (GG) 1 Gold Gold
A (GG) 2 Gold Gold
B (SS) 1 Silver Silver
B (SS) 2 Silver Silver
C (GS) 1 Gold Silver
C (GS) 2 Silver Gold

Six rows, each with probability 1/6. Cross out the three where you saw silver. Of the three remaining, two have gold behind the other drawer.

Do it with playing cards. Take three cards: one red on both sides, one black on both sides, one red on one side and black on the other. Shuffle them in a bag, draw one, and slap it on the table without looking at the underside. If the face up is red, bet even money that the underside is red too. Over 30 rounds you will win about 20 of them. This version is sometimes sold as the "three-card swindle", and it works because the mark counts cards instead of faces.

Explore

What Bertrand was actually arguing. The box paradox is not presented in Calcul des probabilités as a curiosity. It opens the…

What Bertrand was actually arguing. The box paradox is not presented in Calcul des probabilités as a curiosity. It opens the book, and it opens it as a methodological warning: a probability problem is not well posed until you have said how the evidence was generated. The same volume contains the far more famous "Bertrand paradox" about the length of a random chord in a circle, where three different and perfectly reasonable readings of "at random" give three different answers — 1/2, 1/3 and 1/4. The two problems make the same point from opposite directions. In the chord case the procedure is genuinely ambiguous and no answer is uniquely correct. In the box case the procedure is specified, and readers still get it wrong.

The procedure matters here too. The 2/3 answer depends on a detail that is easy to skim past: you opened a drawer at random, and reported what was in it. Change that and the answer changes.

  • Suppose instead someone who can see inside deliberately opens a drawer containing gold whenever one is available. Then seeing gold tells you only that your box is not the silver-silver box — the gold-gold and mixed boxes were both certain to produce this display — and the answer drops to 1/2.
  • Suppose you are told merely "this box contains at least one gold coin", with no account of how anyone came to say that. Then the answer is genuinely indeterminate, and you have walked into the boy or girl paradox.

A dissenting note. Some commentators have argued that although Bertrand reached the right answer, his reductio is right for the wrong reason — that it establishes the inconsistency of the 1/2 answer without isolating the actual error, which is the failure to weight the boxes by how many gold coins they contain. The criticism is fair as far as it goes, but Bertrand's argument has an advantage the modern one lacks: it needs no formal machinery at all.

Why this counts as veridical. Nothing here is contradictory, and nothing dissolves under analysis. The reasoning is elementary, a formal demonstration exists, and the surprising answer is simply the true one. That is Quine's definition of a veridical paradox, and Bertrand's box is about as clean an example as exists.

Historical context

1889. Joseph Bertrand published Calcul des probabilités with Gauthier-Villars in Paris. He was 67, permanent secretary of the…

1889. Joseph Bertrand published Calcul des probabilités with Gauthier-Villars in Paris. He was 67, permanent secretary of the Académie des Sciences, and had spent decades teaching at the École Polytechnique and the Collège de France. The box problem appears near the very start, in his original framing as three identical chests each with two drawers, each drawer holding one coin.

Bertrand posed it, gave the 2/3 answer, supplied the reductio, and added the 300-box illustration. There is no mystery about the origin of this one: it is signed, dated and solved in a single place.

1959. Martin Gardner published the Three Prisoners problem in Scientific American. Three prisoners, one to be pardoned; A asks the warden to name one of the other two who will be executed, and the warden names B. Does A's chance improve from 1/3 to 1/2? It does not — it stays at 1/3, while C's rises to 2/3. This is Bertrand's box with the coins replaced by pardons and a warden supplying the observation. Gardner's readers reacted much as Bertrand's had.

1975. Steve Selvin posed the same structure again in terms of a television game show, and it acquired the name it now carries. The Monty Hall problem has three objects, one of which is revealed by a rule-following agent, and probability 2/3 landing on the option you did not initially choose. Bertrand got there 86 years earlier.

The line of descent is real, not decorative. Boxes with two drawers, then prisoners and a warden, then doors and a host: the same conditional structure, restated three times, each time provoking the same public disbelief. That recurrence is the most interesting historical fact about the problem. Each generation encountered it as new.

Why it matters

Bertrand's box is the ancestor of the most argued-about puzzle in probability, and it is much the better teaching example —…

Bertrand's box is the ancestor of the most argued-about puzzle in probability, and it is much the better teaching example — because it strips away everything that makes Monty Hall contentious.

With Monty Hall, arguments stall on the host's motives. Does he always open a door? Does he know where the car is? Would he have offered the switch if you had picked differently? Those questions are legitimate and they are also a distraction, because they let a reader who has not grasped the conditioning believe the disagreement is about game-show trivia.

Bertrand's boxes have no host. There is nobody to have motives. The drawer opens at random. And the answer is still 2/3, still for the same reason. Whatever is going on, it is not about Monty.

The transferable lesson is a rule about evidence:

When evidence could have arisen in several ways, weight each hypothesis by how many ways it could have produced exactly what you saw — not by how many hypotheses survive.

Counting surviving hypotheses gives 1/2. Counting the ways they could have generated your observation gives 2/3. That difference is the entire content of the likelihood term in Bayes' theorem, and Bertrand demonstrated it decades before anyone was routinely teaching Bayes' theorem to undergraduates.

Modern relevance

Teaching Bayesian reasoning. The box problem is standard in introductory probability courses precisely because the sample space…

Teaching Bayesian reasoning. The box problem is standard in introductory probability courses precisely because the sample space is small enough to write out completely. A student who can enumerate six equally likely rows and cross out three of them has performed a Bayesian update without needing the formula, and can then be shown that the formula agrees.

Diagnostic testing. A positive result should be weighted by how likely each condition was to produce that particular result — not merely by which conditions remain possible. Clinicians shown "the test is 99% accurate" routinely give the box-paradox answer, and the corrective is the same: count the ways the observation could have arisen.

Forensic and genetic evidence. A DNA profile, a fibre match or a partial fingerprint is evidence whose strength depends on how many sources could have generated it. The prosecutor's fallacy — treating "this evidence is unlikely if innocent" as "innocence is unlikely given this evidence" — is the 1/2 answer in a courtroom.

Machine learning and class imbalance. A classifier's posterior depends on both the likelihood and the prior. Models trained on rebalanced data implicitly change the prior, and calibrating them back is the same arithmetic as re-weighting Bertrand's boxes by their coin counts.

Anywhere a system reports partial state. A monitoring system that says "at least one replica is unhealthy" tells you something different depending on whether it checked a random replica or searched for an unhealthy one. That is Bertrand's distinction between a random drawer and a deliberately chosen one, and it is why observability tooling documents its sampling policy.

Sources & further reading

Primary source

  • Joseph Bertrand(1889)Calcul des probabilitésGauthier-Villars, Paris.Chapter I, §2.

    The original statement, together with Bertrand's own proof by contradiction.

Book

  • Jason Rosenhouse(2009)The Monty Hall Problem: The Remarkable Story of Math's Most Contentious Brain TeaserOxford University Press.Chapter 2.

    Traces the line from Bertrand's boxes through the Three Prisoners problem to Monty Hall.

    ISBN 9780195367898

Academic reference

  • Charles M. Grinstead and J. Laurie Snell(1997)Introduction to ProbabilityAmerican Mathematical Society (2nd revised edition).Chapter 4.

Encyclopedia

Elsewhere on the web

How this connects

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Contrasts with
  • The Birthday ProblemTwo beginner-level probability shocks with opposite causes — Bertrand's trap is bad conditioning, the birthday trap is bad counting.

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