How this instrument works
Bertrand's Box Paradox, first posed by Joseph Bertrand in 1889, starts with three boxes: one holding two gold coins (call it GG), one holding two silver coins, and one holding one of each metal (call it GS). You pick a box at random, draw one coin without looking at the other, and it turns out to be gold. The all-silver box is now impossible — but does that leave a 50/50 chance between GG and GS for what the hidden second coin turns out to be? Almost everyone's first instinct says yes. The correct answer is 2/3, not 1/2, and the reasoning is worth sitting with.
The trap is treating 'the drawn coin was gold' as if it points equally to either remaining box. It doesn't, because GG offers two separate opportunities to produce that outcome — either of its two matching coins — while GS offers only one, its single non-silver coin. Conditioning on the draw makes GG twice as likely to be the box you actually picked, so its partner coin is twice as likely to match as GS's partner coin is, which is silver.
This instrument generalizes the classic 1-and-1 setup to any number of GG and GS boxes: P(other coin matches | first coin drawn is gold) = 2·GG / (2·GG + GS). Each GG box contributes two ways to produce a first draw of that kind, while each GS box contributes only one, which is exactly why GG is doubled in the formula. The classic case (one box of each kind) is simply GG=1, GS=1, giving 2/(2+1) = 2/3.
- Enter how many gold-gold boxes are in the mix into Gold-gold boxes.
- Enter how many gold-silver (mixed) boxes are in the mix into Gold-silver (mixed) boxes — the classic paradox uses 1 of each.
- Read P(other coin is gold | first coin drawn is gold) for the resulting conditional probability.
- Try changing the two counts to see how adding more gold-gold boxes pushes the probability toward 1, and more gold-silver boxes pushes it toward 0.
Worked example — the classic three-box setup
Three identical-looking boxes sit on a table: one holds two gold coins, one holds two silver coins, and one holds one of each metal. Entering GG=1 (one box of matching coins) and GS=1 (one mixed box), P(other coin matches | first coin drawn is gold) reads 0.666667 — exactly 2/3.
The arithmetic is P = 2×1 / (2×1 + 1) = 2/3. Intuition says 1/2, since only two boxes could plausibly be the source of that first draw once the all-silver box is ruled out — but that overlooks that GG had two chances to produce the coin you actually saw (either of its two matching coins), while GS had only one. Weighting by those chances is exactly what turns a naive 1/2 into the correct 2/3.
Questions
Why isn't the answer just 1/2 once the silver-silver box is ruled out?
Because 'the box is ruled out' and 'the two remaining boxes are equally likely to be the one you picked' are different claims. Conditioning on having actually drawn a gold coin favors GG, since it had two coins that could have produced that outcome (either one) while GS had only one candidate coin — so GG is twice as likely, in hindsight, to be the box you drew from.
What happens with more than one GG or GS box?
The same weighting logic scales directly: each additional GG box adds two more ways to have produced that first draw, and each additional GS box adds one more. The formula P = 2·GG/(2·GG+GS) captures that ratio for any mix — more GG boxes relative to GS ones pushes the probability of a matching partner coin closer to 1, and the reverse pushes it toward 0.
Is this the same puzzle as the Monty Hall problem?
They're closely related but distinct puzzles that both trip up the same naive 'equally likely once one option is ruled out' intuition. Monty Hall involves a host who knowingly reveals a losing door, actively giving you information through that choice; Bertrand's Box Paradox involves no such active choice — the paradox comes purely from how the coins were distributed across the boxes to begin with, not from any decision made after the draw.
Does it matter that the silver-silver box exists at all?
Not for the final probability, no — once a gold coin has been drawn, the silver-silver box is completely eliminated from consideration and plays no further role in the arithmetic. It matters only in setting up the original scenario: without a silver-silver box to rule out, there would be nothing 'surprising' about which of the remaining boxes the coin came from.
How is this different from the site's Bertrand's Paradox (random chord) calculator?
Different puzzle entirely, despite the shared name — both were posed by the 19th-century mathematician Joseph Bertrand, but this one is about conditional probability with a specific, well-defined sample space of boxes and coins. This site's separate Bertrand's Paradox calculator instead concerns the probability of a random chord on a circle, which turns out to have no single well-defined answer at all without first specifying exactly how the chord is chosen at random — a fundamentally different kind of paradox.