The mathematical foundation of Bertrand's Box Paradox provides insight into conditional probability theory and demonstrates the power of Bayesian reasoning in counterintuitive scenarios.
Formal Bayesian Analysis:
Let G represent observing a gold coin, and B₁, B₂, B₃ represent selecting box 1 (GG), box 2 (SS), and box 3 (GS) respectively. We seek P(other coin is gold | G observed).
Using Bayes' theorem: P(B₁|G) = P(G|B₁)P(B₁) / P(G), where P(G|B₁) = 1, P(G|B₂) = 0, P(G|B₃) = 1/2, and P(B₁) = P(B₂) = P(B₃) = 1/3.
Therefore: P(G) = P(G|B₁)P(B₁) + P(G|B₂)P(B₂) + P(G|B₃)P(B₃) = 1×(1/3) + 0×(1/3) + (1/2)×(1/3) = 1/2.
Thus: P(B₁|G) = 1×(1/3) / (1/2) = 2/3, and P(B₃|G) = (1/2)×(1/3) / (1/2) = 1/3. Since the other coin is gold only if we're in box 1, P(other coin gold | G) = 2/3.
Generalized Formula:
For n₁ boxes with two gold coins, n₂ boxes with two silver coins, and n₃ boxes with mixed coins, the probability that the other coin is gold given observing a gold coin is: P = (2n₁) / (2n₁ + n₃).
This formula shows that the probability depends only on the ratio of pure gold boxes to mixed boxes, weighted by their gold coin contributions. Silver-only boxes don't affect the calculation since they cannot produce the observed gold coin.
Connection to Other Paradoxes:
Bertrand's paradox shares mathematical structure with the Monty Hall problem, where switching doors yields 2/3 probability rather than the intuitive 1/2. Both problems involve conditional probability updates based on revealed information.
The paradox also relates to the boy-girl paradox and other problems where conditioning on partial information leads to counterintuitive results. These problems collectively demonstrate the importance of careful probabilistic reasoning in scenarios involving conditional events.