Bertrand’s Box Paradox Calculator
Three boxes: one holds two gold coins, one holds two silver, one holds one of each. You pick a box at random and draw a coin without looking at its partner: it is gold. What is the probability the other coin in the box is also gold? Intuition shouts one half — the box is either the gold-gold or the mixed one, surely a coin flip. The correct answer is two thirds, and the gap between those numbers is one of the great teaching moments in probability. This calculator generalises the classic: set how many double-gold, mixed and double-silver boxes are in play, and it computes the true posterior — and the explanation below names exactly where the one-half intuition goes wrong.
The Resolution: Count Faces, Not Boxes
The one-half argument says: “the box is GG or GS, two possibilities, equally likely.” The error hides in the last three words. The draw you made was not of a box but of a coin, and coins, not boxes, are the equally likely outcomes. There are three gold coins you might have drawn: both coins of the GG box, and the single gold coin of the GS box. Two of those three draws leave a gold partner; only one leaves silver. Hence 2/3. Joseph Bertrand published the puzzle in 1889 precisely to discipline this instinct — conditional probability conditions on what could have produced the evidence, weighted by how likely each producer was to produce it. The GG box produces a gold draw with certainty; the GS box only half the time; so the evidence “gold” arrives twice as often from GG as from GS, and the posterior odds GG : GS are 2 : 1, not 1 : 1. Bayes’ theorem, wearing a parlour costume — the calculator’s note performs this count for whatever box census you enter.
Playing With the Census
The general formula the widget computes is P = 2·GG / (2·GG + GS): every double-gold box contributes two gold faces whose partners are gold; every mixed box contributes one whose partner is not. Set GG = 2, GS = 1 and the probability climbs to 4/5 = 80%; set GG = 1, GS = 3 and it falls to 2/5 = 40% — below a half, because now most gold faces live in mixed boxes. The SS count never appears in the formula at all: given that a gold coin was drawn, the all-silver boxes are refuted by the evidence and only their absence from the numerator and denominator alike records their irrelevance — until you ask the mirror question about a silver draw (the note answers it), where SS plays GG’s role and GG becomes the irrelevant one. This asymmetry is the paradox’s second lesson: evidence does not merely include, it excludes, and the excluded possibilities stop voting entirely.
The Family of Two-Thirds
Bertrand’s box belongs to a famous family of problems whose correct answers all have this same shape — a naive 1/2 dissolved by counting the hidden multiplicities — and the hub is collecting them. The Monty Hall problem (the hub’s own calculator) is the closest relative: the host’s door-opening is evidence concentrated 2-to-1 on the unchosen door, exactly as the gold coin’s evidence concentrates 2-to-1 on GG. The boy or girl paradox (also on this hub’s list) turns on whether “at least one boy” was evidence gathered like a coin draw or like a census. The two envelopes and the sleeping beauty debates extend the family’s quarrels into philosophy. What unites them is the discipline Bertrand was teaching: before computing, enumerate the equally likely micro-outcomes — the faces, not the boxes — and let no summary description (“it’s one of two boxes”) smuggle in an equiprobability that the sampling process never granted.
Why This Matters Beyond Parlours
The counting error the paradox exposes is the daily bread of real inference, at higher stakes. A medical test’s positive result is a “gold draw”: its meaning depends not on the two boxes (disease, no disease) being equally likely, but on how often each produces positives — the base rate and the false-positive rate playing the roles of GG’s two faces and GS’s one, which is why the hub’s false-positive calculator tells the same story in clinical dress. Forensic likelihood ratios, spam filters, quality sampling and every Bayesian update in machine learning are the census-counting of this page industrialised: evidence weighted by its probability under each hypothesis, never by the number of hypotheses standing in a row looking symmetric. Bertrand built a toy in 1889 that still catches graduate students; the calculator’s general form is the toy with its working parts exposed, ready to be aimed at any three-box situation the world presents.
The Card Version, and a Word About Simulation
The paradox’s most common modern costume replaces the boxes with three cards: one red on both sides, one black on both sides, one red on one side and black on the other. A card is drawn and set down showing red: the probability the underside is red is again 2/3 — the three red faces (two from the double-red card, one from the mixed) vote, two of them for a red underside. If the abstract counting ever feels slippery, simulation is the honest broker: shuffle the classic setup a few thousand times (three coins in three boxes, blind draw, record the partner whenever gold appears) and the gold-partner share settles near 66–67% with the stubbornness of the law of large numbers. Paradoxes of this family never survive enumeration; they survive only in the gap between the situation and its description — which is why this page, like Bertrand, insists on describing it face by face.
The Mapping to Monty Hall, Made Explicit
The kinship claimed above can be drawn exactly. In Monty Hall, the car sits behind your door with probability 1/3 and behind one specific other door with 2/3 after the host’s informed opening — evidence (the opened goat door) that the host could always produce for the other doors but never for yours in the same way. In Bertrand’s box, the GG box plays the unopened door: the gold draw is evidence the GG box produces with both its faces, the GS box with only one. Two puzzles, one skeleton — a probability mass that naive counting splits evenly and correct counting splits 2-to-1, because one hypothesis was always twice as fecund in producing what was seen.
Related Tools
More from the Statistics Calculators hub: the live Monty Hall Problem Calculator — the 2/3 that conquered television — the Shannon Entropy Calculator for measuring the information the draw delivered, and the live Coin Toss Streak Calculator for more counted surprises.
Frequently Asked Questions
Why isn’t it 1/2? There are only two possible boxes.
Two possible boxes, but not two equally likely explanations of the draw. You sampled a coin, and the sampling process reaches the GG box’s gold twice as often as the GS box’s gold (two faces versus one). Probability conditions on the process that produced the evidence, and that process votes 2-to-1. The 1/2 answer would be correct if the box had been identified by a process indifferent between boxes given a gold draw — for instance, if someone who knew the contents announced “the box is GG or GS” without the draw being how you learned it.
Does it matter that I drew the coin randomly rather than being shown it?
Entirely — it is the whole paradox. If a knowing host deliberately shows you a gold coin (choosing which to display), the inference changes with the host’s policy; if the coin is drawn blind, the face-counting above is exact. Many published “solutions” quarrel precisely here, by quietly changing the protocol. This calculator assumes the blind draw of the classic setup, and its census formula generalises that protocol faithfully.
What if there are no silver coins at all?
Then set GS = 0 and SS = 0 and the formula returns 100% — with only double-gold boxes in the world, a gold draw’s partner is certainly gold, and the paradox dissolves into bookkeeping. The interesting region is where the formula straddles the naive answer: any mix where 2·GG and GS are comparable produces a posterior noticeably different from both 1/2 and 1, which is where intuition needs the escort this page provides.
Is this the same as Bertrand’s paradox about chords?
No — Bertrand published two famous paradoxes, and they share only an author and a moral. The chord paradox (also on this hub, as Bertrand’s Paradox) shows that “a random chord” is undefined until the randomising mechanism is specified; the box paradox shows that conditional probability must weight by the mechanism’s micro-outcomes. Both are lessons in saying exactly how the randomness operates before asking what it implies.
How does this connect to Bayes’ theorem?
It is Bayes’ theorem, pre-computed in counts. Prior odds GG : GS = 1 : 1 (three boxes equally likely, one of each in the classic). Likelihood ratio: P(gold | GG) / P(gold | GS) = 1 / (1/2) = 2. Posterior odds = 1 × 2 = 2 : 1, i.e. probability 2/3. The widget’s face-counting and the theorem’s odds form are the same arithmetic in two notations — use whichever your audience finds harder to argue with.
Can the trick be run in reverse — draw silver?
Symmetrically, yes, and the calculator’s note prints it: P(other coin silver | drew silver) = 2·SS / (2·SS + GS), which in the classic census is also 2/3. The two answers coexist peacefully because the mixed box contributes its gold face to one question and its silver face to the other — the same box voting in both elections with different coins is the entire mechanism, stated once more because it cannot be stated too plainly.