Bertrand's Box Paradox Calculator

Given the color of a drawn coin, find the probability the other coin in the same box matches — solved with Bayes' theorem. In the classic three-box setup the answer is 2/3, not 1/2.

Quick Facts

Method
Bayes' theorem: P(other matches | drawn color) = P(box is all-that-color) ÷ P(drew that color)
Boxes are equally likely; each coin in the chosen box is equally likely to be drawn.

Your Results

Calculated
P(other coin matches)
-
Posterior probability
As a fraction
-
Exact odds
P(drawing that color)
-
Bayes denominator
Naive guess
50%
The tempting wrong answer

Ready

Set the three boxes and the drawn color, then calculate.

What Bertrand's box paradox is

Bertrand's box paradox, posed by Joseph Bertrand in his 1889 Calcul des probabilités, is a classic exercise in conditional probability. There are three identical-looking boxes, each holding two coins:

  • Box GG: two gold coins.
  • Box SS: two silver coins.
  • Box GS: one gold coin and one silver coin.

You pick a box at random, reach in, and draw one coin at random. It turns out to be gold. What is the probability that the other coin in that same box is also gold? The intuitive answer is 1/2 — but the correct answer is 2/3. That gap is the paradox.

Why the answer is 2/3, not 1/2

The tempting reasoning goes: "The coin came from either box GG or box GS (the all-silver box is ruled out), those two are equally likely, so it's 50–50." The flaw is assuming the two remaining boxes are equally likely after seeing a gold coin. They are not.

Think in terms of individual coins, not boxes. There are three gold coins across all boxes — call them G1 and G2 (in box GG) and G3 (in box GS). Before you look, each of those three gold coins was equally likely to be the one you drew. If you drew G1 or G2, the other coin in the box is gold. If you drew G3, the other coin is silver. So two of the three equally-likely gold draws leave a gold partner: 2/3.

The formula (Bayes' theorem)

Formally, let the event be "drew a gold coin." By Bayes' theorem:

P(box is GG | drew gold) = P(drew gold | GG) × P(GG) ÷ P(drew gold)

Plugging in P(drew gold | GG) = 1, P(GG) = 1/3, and P(drew gold) = (1×1/3) + (0×1/3) + (½×1/3) = 1/2 gives P(GG | gold) = 1 × (1/3) ÷ (1/2) = 2/3. Because "the other coin is also gold" happens exactly when the box is GG, that posterior probability is the answer.

This calculator generalizes the setup: you choose the gold/silver split of all three boxes and the color you drew. It computes P(other coin matches | drawn color) = (number of coins in an all-matching box) ÷ (total coins of the drawn color). The classic configuration (2-gold, 1-gold-1-silver, 0-gold) with a gold draw returns exactly 2/3 ≈ 66.67%.

Reference values

  • Classic setup, drew gold: 2/3 ≈ 66.67%.
  • Classic setup, drew silver: also 2/3 — the problem is symmetric in color.
  • Two GG boxes and one GS box, drew gold: 4/5 = 80%. There are five gold coins; four of them sit in an all-gold box.
  • All three boxes GS (one gold, one silver each), drew gold: 0% — no box is all gold, so the partner is always silver.

Where the same trap appears

Bertrand's box is structurally identical to the Monty Hall problem and the "boy or girl" (two-child) paradox: in each, an observation changes the relative likelihood of the underlying cases in a way that ambushes the "just count the remaining options" instinct. Recognizing the pattern — weight each hypothesis by how likely it was to produce what you observed — is the practical takeaway.

Frequently Asked Questions

Why isn't the answer just 1/2?
Because seeing a gold coin is stronger evidence for the two-gold box than for the mixed box. The two-gold box has two gold coins that could have been drawn; the mixed box has only one. So among all the ways you could have drawn a gold coin, twice as many originate in the two-gold box. That 2:1 weighting, not a 1:1 split, is what makes the answer 2/3.
Is the paradox different if I drew a silver coin instead?
No. The classic three-box setup is perfectly symmetric between gold and silver, so drawing silver also gives a 2/3 chance the other coin is silver. This calculator handles both — just pick the color you drew.
What's the difference between this and the Monty Hall problem?
They share the same mathematical core: an observation shifts the posterior probabilities of the underlying cases, and naive counting misleads you. In Monty Hall the host's forced reveal concentrates probability on the unopened door (switching wins 2/3 of the time); in Bertrand's box the drawn coin's color concentrates probability on the matching box. Both resolve cleanly with Bayes' theorem.

Related calculators