Calculate conditional probability using Bayes' Theorem. Explore how evidence (drawing a gold coin) updates the probability of which box you selected.
Last updated: March 2026
| GG Boxes | GS Boxes | SS Boxes | P(Other Gold) |
|---|---|---|---|
| 1 | 1 | 1 | 66.67% (2/3) |
| 2 | 1 | 1 | 80.00% (4/5) |
| 1 | 2 | 1 | 50.00% (2/4) |
| 1 | 0 | 1 | 100% (2/2) ← Certain |
| 0 | 1 | 1 | 0% (0/1) ← Impossible |
Bertrand's Box Paradox is a classic probability problem that demonstrates the importance of careful conditional reasoning and Bayesian inference. The paradox often surprises people because the intuitive answer differs significantly from the correct mathematical result.
The Setup: You have three indistinguishable boxes. One contains two gold coins (GG), one contains one gold and one silver coin (GS), and one contains two silver coins (SS). You randomly select a box and draw one coin from it—the coin is gold. What is the probability that the remaining coin in the box is also gold?
The key insight is that drawing a gold coin provides strong evidence about which box you selected. The GG box is twice as likely to produce a gold coin as the GS box (2 gold coins vs. 1 gold coin). This asymmetry is what makes the posterior probability 2/3, not 1/2.
Specify how many boxes of each type (GG, GS, SS) are in your scenario. Start with 1 of each for the classic problem.
The calculator uses Bayes' Theorem to compute P(other coin is gold | drew gold). This accounts for the different probabilities of drawing gold from each box type.
Try changing the box counts to see how the posterior probability changes. More GG boxes increase the probability; more GS boxes decrease it.
Why it's not 50-50 (the classic error):
Intuition says two equally likely boxes (GG or GS), so 50%. But GG produces gold twice as often, making it twice as likely given that we drew gold.
We update our belief about which box was selected (prior: 1/3 each) using evidence that gold was drawn (likelihood differs by box type).
The formula adjusts automatically. More GG boxes increase probability; more GS boxes decrease it; SS boxes are irrelevant (can't produce gold).
Both involve conditional probability and how new information changes probabilities. Both confound intuition in similar ways.
After drawing one coin from the box (which is gold), one coin remains inside. We ask: what's the probability that remaining coin is gold?
No, the result is always 0–100%. Maximum (100%) occurs with only GG boxes; minimum (0%) occurs when GG = 0 and GS > 0.
It's paradoxical because outcomes seem equally likely at first glance, but careful analysis reveals asymmetry. The result contradicts naive intuition.
With n GG, n GS, n SS boxes: P(other gold) = 2/(2+1) = 2/3 regardless of n. This is Bertrand's original problem.
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