>>75761404One classic example of a paradox created by the principle of indifference in classical probability theory is the Bertrand's box paradox. Here's how it goes:
Suppose there are three boxes:
- Box A contains 2 gold coins
- Box B contains 1 gold coin and 1 silver coin
- Box C contains 2 silver coins
You choose a box at random and draw out a coin at random. The coin turns out to be gold. What is the probability that the other coin in the box is also gold?
Applying the principle of indifference naively in two different ways leads to contradictory answers:
1) There are two possibilities for the remaining coin: gold or silver. Applying the principle of indifference suggests the probability of each is 1/2. So the probability the other coin is gold appears to be 1/2.
2) We can also reason as follows: the gold coin could have come from Box A or Box B. The principle of indifference suggests Box A and Box B are equally likely. If it came from Box A, the probability the other coin is gold is 1. If it came from Box B, the probability the other coin is gold is 0. So the probability appears to be 1/2 * 1 + 1/2 * 0 = 1/2.
Both lines of reasoning apply the principle of indifference, but to different possibilities, resulting in the same probability of 1/2. However, a more careful Bayesian analysis shows that the correct probability the other coin is gold is actually 2/3.
The paradox arises because the principle of indifference is applied in a simplistic way without considering the full space of possibilities and the information provided by the observation. It shows the limitations and pitfalls of applying the principle of indifference indiscriminately.
A Bayesian analysis takes into account both the prior probabilities of each box being chosen, and the evidence provided by drawing a gold coin. Here's how we can calculate the probability step by step:
Let's define some events:
- A: The chosen box is A
- B: The chosen box is B
- C: The chosen box is C
- G: The drawn coin is gold
We want to find P(A|G), the probability that the box is A given that the drawn coin is gold.
Step 1: Prior probabilities
Before drawing a coin, the probability of each box being chosen is:
P(A) = P(B) = P(C) = 1/3
Step 2: Likelihood of drawing a gold coin from each box
If the chosen box is A, the probability of drawing a gold coin is 1.
If the chosen box is B, the probability of drawing a gold coin is 1/2.
If the chosen box is C, the probability of drawing a gold coin is 0.
So, P(G|A) = 1, P(G|B) = 1/2, P(G|C) = 0
Step 3: Use Bayes' theorem
P(A|G) = P(G|A) * P(A) / P(G)
where P(G) = P(G|A) * P(A) + P(G|B) * P(B) + P(G|C) * P(C)
= 1 * 1/3 + 1/2 * 1/3 + 0 * 1/3
= 1/2
So, P(A|G) = 1 * 1/3 / (1/2) = 2/3
Therefore, the probability that the box is A (and hence the other coin is also gold) given that a gold coin was drawn is 2/3.
This analysis takes into account the evidence provided by drawing a gold coin, which eliminates Box C and makes Box A more likely than Box B. The principle of indifference is applied to the prior probabilities, but the posterior probabilities are updated based on the evidence.