Bertrand Box Paradox Calculator
The Bertrand box paradox shows why conditional probability is counter-intuitive. Given you drew a gold coin from a randomly chosen box, the probability the other coin in that box is also gold is 2/3 — not 1/2. Adjust the box counts to see how the probability changes.
2 gold-from-GG draws vs 1 gold-from-GS draws — only GG guarantees gold second
- 1
Gold-first draws from GG boxes
2 × 1 = 2Each GG box contributes two possible gold-first draws — one for each gold coin. - 2
Total gold-first draws
2 + 1 = 3 - 3
P(2nd gold | 1st gold)
2 ÷ 3 × 100 = 66.67
How does this calculator work?
In the Bertrand box paradox, P(2nd coin gold | 1st coin gold) = 2·n_GG / (2·n_GG + n_GS). With one box of each type, this is 2/3 — not 1/2 — because GG boxes produce twice as many gold-first draws as GS boxes. The paradox is resolved by Bayes' theorem: counting draws, not boxes.
Formula
How this is calculated
The classical Bertrand Box Paradox (Joseph Bertrand, 1889) uses three boxes: one with two gold coins (GG), one with one gold and one silver (GS), and one with two silver (SS). You pick a box at random and draw a coin at random — it is gold. What is the probability the other coin in the same box is gold?
Intuition says 1/2: you know you're not in the SS box, so it must be GG or GS — a 50/50 guess. But this is wrong. The correct approach is to count equally-likely gold-first draws, not equally-likely boxes. There are three possible gold-first draws: coin 1 from GG, coin 2 from GG, or the gold coin from GS. In two of the three, the other coin is gold (both from GG). So P = 2/3.
This generalises to n_GG, n_GS, n_SS boxes of each type. There are 2·n_GG gold-first draws that come from a GG box and n_GS that come from a GS box — giving P = 2·n_GG / (2·n_GG + n_GS). The paradox illustrates Bayes' theorem: the evidence (you drew gold) updates your belief about which box you picked — GG boxes are twice as likely to produce a gold draw as GS boxes.
Frequently asked questions
The mistake in the 1/2 answer is treating the two remaining boxes (GG and GS) as equally likely after seeing gold. They are not: a GG box would have given a gold coin on both possible draws, while a GS box would have given gold on only one of its two draws. The gold observation is twice as probable if you're in a GG box, so the GG box gets a higher posterior weight — giving P(GG | gold) = 2/3.
Both are conditional probability paradoxes where intuition suggests 1/2 but the correct answer differs. In both, the prior probability of each option is equal, but the observed evidence (gold coin drawn / host reveals a goat) is not equally likely under all options, shifting the posterior. Both are solved cleanly with Bayes' theorem.
If there are no mixed (GS) boxes, then drawing a gold coin guarantees you are in a GG box, so P(2nd gold | 1st gold) = 1. Conversely, with n_GG = 0 there are no GG boxes, so P = 0.
Also known as
TG we-Calculate Editorial Team. (2026). Bertrand Box Paradox Calculator [Online calculator]. TG we-Calculate. https://we-calculate.com/calculator/bertrand-box-paradox-calculator
TG we-Calculate Editorial Team. "Bertrand Box Paradox Calculator." TG we-Calculate. 2026. https://we-calculate.com/calculator/bertrand-box-paradox-calculator.
TG we-Calculate Editorial Team, "Bertrand Box Paradox Calculator," TG we-Calculate, 2026. [Online]. Available: https://we-calculate.com/calculator/bertrand-box-paradox-calculator
@misc{wecalculate_bertrand_box_paradox_calculator, title = {Bertrand Box Paradox Calculator}, author = {{TG we-Calculate Editorial Team}}, howpublished = {\url{https://we-calculate.com/calculator/bertrand-box-paradox-calculator}}, year = {2026}, note = {TG we-Calculate} }
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