Bertrand's Box Paradox
Why 1/2 is the wrong answer (and 2/3 is right)
The Puzzle
Joseph Bertrand posed this deceptively simple problem in 1889. You have three boxes:
- Box 1: Two gold coins (GG)
- Box 2: Two silver coins (SS)
- Box 3: One gold, one silver (GS)
You randomly pick a box and randomly draw one coin from it. The coin is gold.
Question: What is the probability that the other coin in the box is also gold?
Most people say 1/2. After all, you either picked Box 1 (other coin gold) or Box 3 (other coin silver)āa 50/50 split, right?
Wrong. The answer is 2/3.
Play the Game
Why 2/3?
The key insight: you didn't just pick a box, you picked a specific coin. There are 6 coins total, 3 of which are gold:
The Fallacy
The intuitive answer of 1/2 comes from thinking:
Wrong reasoning: "I drew gold, so I'm in either Box 1 or Box 3. That's two equally likely boxes, and one has gold, one has silver. So it's 50/50."
The error is treating the boxes as equally likely after drawing gold. But they're not!
- Box 1 (GG): 2 gold coins = 2 ways to draw gold
- Box 3 (GS): 1 gold coin = 1 way to draw gold
Box 1 is twice as likely to produce a gold draw as Box 3. So after drawing gold:
Connection to Monty Hall
Bertrand's Box is mathematically identical to the Monty Hall Problem! The structure is the same:
3 doors: 1 car, 2 goats
You pick a door
Monty reveals a goat
Should you switch? Yes! (2/3)
3 boxes: GG, SS, GS
You pick a box
You draw gold
Is other coin gold? Yes! (2/3)
In both cases, the initial random selection (picking a door/box) gets "weighted" by the revealing event (Monty's choice / your coin draw), making one option twice as likely as the other.
Bayes' Theorem Proof
For those who want the formal proof using Bayes' Theorem:
Where:
- P(gold | GG) = 1 (certain to draw gold from Box 1)
- P(gold | SS) = 0 (impossible to draw gold from Box 2)
- P(gold | GS) = 1/2 (50% chance from Box 3)