The Bertrand's box paradox calculator above solves the three-box coin puzzle, a conditional-probability problem whose correct answer is two thirds and whose intuitive answer is one half. It is not the same problem as Bertrand's paradox, which asks about the length of a random chord in a circle and famously has three legitimate answers; the two share a name because Joseph Bertrand published both in the same 1889 book, and confusing them is extremely common.
Arb Digital publishes this because the puzzle is the cleanest available demonstration of a specific reasoning failure: counting the containers rather than the outcomes. The tool generalises the problem to any mix of boxes so that the mechanism, rather than the memorised answer, is what you take away.
What This Bertrand's Box Paradox Calculator Does
The setup is fixed and simple. There are boxes, each holding two coins. Some hold two gold, some two silver, and some one of each. A box is chosen at random, and a coin is drawn at random from inside it. That coin turns out to be gold. The question is the probability that the other coin in the same box is also gold.
The calculator counts individual coins rather than boxes, which is the whole trick. It reports how many coins of the observed colour exist across all the boxes, how many of those sit in a matching pair, and the ratio between them. It also gives the prior probability of picking a matching-pair box before any coin is seen, so the shift caused by the observation is visible.
Because the box counts are adjustable, you can watch the answer move. With one box of each type it is two thirds. Add mixed boxes and it falls; remove the two-gold box entirely and it becomes zero, and the tool says so in words rather than printing a bare figure. Our conditional probability calculator handles the general P(A given B) form, and our Bayes theorem calculator works the same update from priors and likelihoods.
How to Use It
- Set how many boxes of each type there are. The classic problem uses one of each.
- Choose which colour of coin you drew. The problem is symmetric, so gold and silver behave identically when the box counts mirror each other.
- Read the headline probability that the second coin in the same box matches the first.
- Compare it against the prior in the first grid item, which is the answer before any coin was drawn.
- Switch to the fraction view to check the classic two-thirds result exactly.
The Reasoning and How It Is Calculated
With one box of each type there are six coins in total: two gold in the first box, two silver in the second, and one of each in the third. Every one of those six coins is equally likely to be the one drawn, because the box is chosen uniformly and then the coin within it is chosen uniformly.
Observing a gold coin eliminates three of the six: both coins in the two-silver box, and the silver coin in the mixed box. Three equally likely possibilities remain, and two of them — the two coins in the two-gold box — have a gold partner. The answer is therefore 2/3, not one half.
In general, if there are g boxes with two gold, s with two silver and m mixed, the number of gold coins is 2g + m, and the number of those whose partner is also gold is 2g. So P = 2g / (2g + m). With g = 1 and m = 1 that is 2/3, and the silver-coin version is the mirror image with s in place of g.
Note which quantity does not appear in that expression. The count of two-silver boxes cancels out entirely once a gold coin has been observed, because none of their coins could have been the one drawn. That is Bayes' theorem doing its work: the observation removes those branches rather than reweighting them. Wolfram MathWorld's page on conditional probability derives the theorem this argument depends on.
Why One Half Feels Right and Is Wrong
The intuitive argument runs: seeing a gold coin rules out the two-silver box, leaving two boxes, one of which has a second gold coin, so the answer is one half. Every step of that is true except the last, and the last fails because the two remaining boxes are not equally likely to be the box you are holding.
The two-gold box had two chances to produce the gold coin you saw. The mixed box had only one. So conditional on having seen gold, the two-gold box is twice as likely to be the box in front of you. Counting boxes treats those two possibilities as equal; counting coins does not. Bertrand's own stated purpose in constructing the example was to show that counting cases is not always proper — that you have to weight each case by how likely it was to produce what you observed.
The general lesson generalises well beyond coins. Any time evidence is more likely under one hypothesis than another, the hypotheses stop being equally probable even if they started that way. Medical test interpretation, spam filtering and fault diagnosis all turn on the same point, and all of them are misread in the same way when people count hypotheses instead of weighting them. Penn State's STAT 500 applied statistics course covers the conditional probability rules this argument rests on.
The Family of Puzzles This Belongs To
Three well-known puzzles share this structure, and recognising the shared skeleton is more useful than memorising three answers.
The Monty Hall problem is the same argument in a game show. Three doors, one prize; you pick one, the host opens a different door revealing no prize, and switching wins two thirds of the time. As in the box problem, the reason is that the host's action was more likely under one arrangement than another.
The boy or girl paradox is the same argument applied to siblings, and it is the one where the wording matters most. Knowing that a two-child family has at least one boy gives a one-third chance that both are boys; knowing that a specific child, such as the elder, is a boy gives one half. Our boy or girl paradox calculator works through both readings, because the two questions sound nearly identical and have genuinely different answers.
What all three share is that the observation carries information about how it was generated, not just about what it shows. Ignore the generating process and you get the intuitive answer; account for it and you get the correct one. Our probability calculator handles the underlying event arithmetic, and the birthday paradox calculator covers a different failure of intuition — underestimating how fast pairwise opportunities grow.
What Changing the Box Counts Reveals
Adding mixed boxes pushes the answer down towards zero, because every mixed box adds one gold coin whose partner is silver, without adding any gold coin whose partner is gold. Adding two-gold boxes pushes it up towards one, for the mirror reason. The two-silver boxes have no effect at all once gold has been observed, which surprises most people the first time they see it.
Two degenerate configurations are worth trying. With no two-gold boxes at all, the answer is exactly zero: every gold coin has a silver partner, so the second coin cannot match. With no mixed boxes, the answer is exactly one: every gold coin sits beside another gold coin. The tool states both cases in words rather than printing a bare number, because a bare 0 or 100 percent hides the reason.
The third degenerate case is having no coins of the observed colour anywhere — asking about a gold coin when no box contains one. That is not a probability of zero; it is a question that cannot be asked, because the condition you are conditioning on never occurs. The tool reports it as such instead of dividing by zero.
Arb Digital reads campaign and analytics data the way this puzzle demands: by asking what would have produced it.
Browse All Free Tools Talk To Our TeamCommon Mistakes to Avoid
- Counting boxes instead of coins — the two-gold box had two chances to produce the coin you saw, so the surviving boxes are not equally likely.
- Confusing it with Bertrand's paradox — that is the chord-length problem, it has three defensible answers, and it shares only the surname.
- Assuming the answer is always two thirds — that holds for one box of each type. Change the mix and the answer changes with it.
- Thinking the two-silver boxes matter — once a gold coin is observed they cancel out of the calculation entirely.
- Treating an impossible condition as a zero probability — if no gold coin exists, the question cannot be asked at all rather than having the answer zero.
Related Free Tools From Arb Digital
Solve the chord-length problem with the Bertrand's paradox calculator, work through the sibling version with the boy or girl paradox calculator, handle the general form with the conditional probability calculator and the Bayes theorem calculator, or start from first principles with the probability calculator. The full free online tools hub lists every statistics tool we publish.
Frequently Asked Questions
A puzzle with three boxes, one holding two gold coins, one two silver and one of each. A random coin drawn from a random box is gold, and the probability the other coin in that box is also gold turns out to be two thirds.
Because the two-gold box had two chances to produce the gold coin you saw and the mixed box had only one. Counting individual coins rather than boxes gives three equally likely possibilities, two of which have a gold partner.
No. Bertrand's paradox is the chord-length problem, which asks how likely a random chord of a circle is to exceed the side of an inscribed equilateral triangle and has three legitimate answers. They share only the name of their author.
Not once a gold coin has been observed. None of their coins could have been the one drawn, so they drop out of the calculation entirely no matter how many there are.
It is the same argument in a different setting. In both, one possibility had more ways to produce what you observed than another, so the surviving options are not equally likely.
With g boxes holding two gold and m holding one of each, the probability the partner of an observed gold coin is also gold equals 2g divided by 2g plus m.
The question cannot be asked. You would be conditioning on an event that never occurs, which is different from a probability of zero, and the tool reports it that way.
This page explains a standard probability puzzle for educational purposes. It assumes the box and the coin are each chosen uniformly at random.
