Bertrand's Box Paradox

by aoum, Mar 14, 2025, 8:23 PM

Bertrand's Box Paradox: A Counterintuitive Probability Puzzle

Bertrand's Box Paradox is a classic probability puzzle that demonstrates how our intuition about random selection can often be misleading. It was introduced by the French mathematician Joseph Bertrand in 1889 and remains a popular example of how probability can produce surprising results.

https://upload.wikimedia.org/wikipedia/commons/thumb/4/43/Three_mystery_boxes.jpg/260px-Three_mystery_boxes.jpg

1. What Is Bertrand's Box Paradox?

Imagine you have three boxes, each containing two coins:
  • Box A: Two gold coins (GG)
  • Box B: One gold coin and one silver coin (GS)
  • Box C: Two silver coins (SS)

You randomly pick a box and draw a coin. If the coin is gold, what is the probability that the other coin in the same box is also gold?

At first glance, you might think the answer is \( \frac{1}{2} \), but the correct probability is actually \( \frac{2}{3} \)!

2. Breaking Down the Probability

To understand why, let's carefully work through the situation:
  • There are three boxes, each equally likely to be chosen.
  • If you draw a gold coin, you could be in either Box A or Box B (you cannot be in Box C because it only contains silver coins).

However, not all outcomes are equally likely. Since Box A contains two gold coins, you are twice as likely to draw a gold coin from Box A as from Box B.

Step 1: Possible Outcomes

If we randomly select a box and then draw a coin, the following outcomes are equally likely:
  • Box A, Gold coin (probability \( \frac{1}{3} \times \frac{2}{2} = \frac{1}{3} \))
  • Box B, Gold coin (probability \( \frac{1}{3} \times \frac{1}{2} = \frac{1}{6} \))
  • Box C, Silver coin (probability \( \frac{1}{3} \times \frac{2}{2} = \frac{1}{3} \))

Since we know a gold coin was drawn, we can ignore the outcome involving Box C.

Step 2: Update Probabilities Using Conditional Probability

The total probability of drawing a gold coin is:

\[
P(\text{Gold}) = P(\text{Gold | Box A}) \times P(\text{Box A}) + P(\text{Gold | Box B}) \times P(\text{Box B})
\]
\[
P(\text{Gold}) = \frac{2}{2} \times \frac{1}{3} + \frac{1}{2} \times \frac{1}{3} = \frac{1}{3} + \frac{1}{6} = \frac{1}{2}
\]
The probability that the other coin is gold (meaning we are in Box A) is:

\[
P(\text{Box A | Gold}) = \frac{P(\text{Gold | Box A}) \times P(\text{Box A})}{P(\text{Gold})}
\]
\[
P(\text{Box A | Gold}) = \frac{\frac{1}{3}}{\frac{1}{2}} = \frac{2}{3}
\]
3. Why Does the Intuition Fail?

Our intuition fails because we overlook the fact that drawing a gold coin from Box A is twice as likely as from Box B. Since Box A contains two gold coins, it gets "counted" more in the total outcomes.

4. A Different Perspective: Counting the Coins

Instead of thinking about the boxes, imagine the individual gold coins:
  • Two gold coins from Box A
  • One gold coin from Box B

If you draw a gold coin, the chance it came from Box A is \( \frac{2}{3} \) because 2 of the 3 gold coins are in Box A.

5. Simulating Bertrand's Box Paradox with Python

We can confirm the result through simulation:

import random

def bertrand_simulation(trials=100000):
    boxes = [('G', 'G'), ('G', 'S'), ('S', 'S')]
    count_box_A = 0
    gold_drawn = 0

    for _ in range(trials):
        box = random.choice(boxes)
        coin = random.choice(box)
        
        if coin == 'G':
            gold_drawn += 1
            if box == ('G', 'G'):
                count_box_A += 1
    
    return count_box_A / gold_drawn

print(f"Probability the other coin is gold: {bertrand_simulation():.4f}")


This simulation should produce a value close to \( \frac{2}{3} \).

6. Applications of Bertrand’s Box Paradox

The paradox has implications beyond simple puzzles:
  • Medical Testing: Understanding false positives and interpreting conditional probabilities.
  • Cryptography: Analyzing probabilities in complex systems.
  • Bayesian Statistics: Applying conditional probability in decision-making.
  • Game Theory: Evaluating outcomes with unequal likelihoods.

7. Similar Paradoxes and Puzzles

Bertrand’s Box Paradox is related to other famous probability paradoxes:
  • The Monty Hall Problem: Another counterintuitive probability puzzle involving conditional probability.
  • The Two-Envelope Paradox: A paradox involving expectations and switching strategies.
  • The Three Prisoners Problem: A puzzle about updating probabilities with new information.

8. Conclusion

Bertrand’s Box Paradox shows how easily our intuition can be misled when dealing with conditional probabilities. By carefully analyzing the possible outcomes, we reveal that the likelihood of both coins being gold is \( \frac{2}{3} \), not \( \frac{1}{2} \). This paradox highlights the importance of precise reasoning in probability and serves as a powerful lesson for real-world applications.

References

Comment

2 Comments

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You can actually use Bays rule for conditional probability!

by rayliu985, Mar 16, 2025, 7:58 PM

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That's nice to know! I'll do a post on Bayes' Theorem, also known as Bayes' Law or Bayes' Rule.

by aoum, Mar 16, 2025, 10:25 PM

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