Table of Contents
The Bertrand Box Paradox
Proposed by Joseph Bertrand in 1889, this paradox involves three boxes: one with two gold coins, one with one gold and one silver, and one with two silver coins. You pick a box at random and draw one coin. If it is gold, what is the probability the other coin is also gold?
Most people intuitively answer 1/2, reasoning the gold coin came from either the gold-gold or gold-silver box. But the correct answer is 2/3, because drawing a gold coin is twice as likely from the gold-gold box.
Solution Using Conditional Probability
Box Contents
| Box | Coin 1 | Coin 2 | P(draw gold) |
|---|---|---|---|
| Box 1 | Gold | Gold | 1 |
| Box 2 | Gold | Silver | 1/2 |
| Box 3 | Silver | Silver | 0 |
FAQ
Why isn't the answer 1/2?
Because there are three equally likely gold coins you could have drawn (two from box 1, one from box 2). Two of the three gold coins have a gold companion. So P = 2/3.
How does this relate to Bayes' theorem?
It is a direct application. The prior probability of each box is 1/3, the likelihood of drawing gold differs by box, and Bayes' theorem gives the posterior probability of being in the gold-gold box.