Bayesian Probability in a Dice Scenario: A Man’s Truthfulness and Die Outcome

Imagine a scenario where a man claims that he rolled a 6 on an unbiased die, yet you wonder if this statement is credible given his history of honesty. How can we apply Bayesian probability to determine the likelihood that his statement is true? This article will explore the problem using Bayes' theorem and discuss the underlying principles.

Scenario and Problem Setup

A person is known to speak the truth 75% of the time. If this individual announces that he has rolled a 6 on a fair six-sided die, what is the probability that he is actually telling the truth?

Defining the Events

Let's define the following events:

A: The die shows a 6. B: The man says the die shows a 6.

We want to find P(A | B), which is the probability that the die actually shows a 6 given that the man says it is a 6.

Bayes' Theorem

According to Bayes' theorem, we have:

P(A | B) (P(B | A) * P(A)) / P(B)

Step 1: Calculating P(A)

Since the die is unbiased:

P(A) 1/6

Step 2: Calculating P(B | A)

With a 75% chance of telling the truth:

P(B | A) 0.75

Step 3: Calculating P(B)

The total probability P(B) can be computed through the following two scenarios:

The die shows a 6, and the man tells the truth. The die shows any other number, and the man lies.

The probability that the die shows a 6 and the man tells the truth:
P(B | A) * P(A) 0.75 * (1/6) 0.125

The probability that the die shows any other number (5/6) and the man lies (25% chance):
P(B | A^c) * P(A^c) 0.25 * (5/6) 0.2083

Thus, the total probability P(B) is the sum of these two probabilities:

P(B) 0.125 0.2083 0.333

Applying Bayes' Theorem

Substituting the given values into Bayes' theorem:

P(A | B) (0.75 * (1/6)) / 0.3333 ≈ 0.375

Hence, the probability that the die actually shows a 6 given that the man says it is a 6 is approximately 37.5%.

Mathematical Notation and Generalization

Let's define S as the event that a 6 is rolled, and R as the event that the person says it's a 6. Given that we're working with a fair six-sided die, we know that:

mathbb{P}(S) 1/6

mathbb{P}(R|S) 4/5

Assuming the event of interest is mathbb{P}(S|R), we apply Bayes' rule as follows:

mathbb{P}(S|R) frac{mathbb{P}(R|S)cdotmathbb{P}(S)}{mathbb{P}(R|S)cdotmathbb{P}(S) mathbb{P}(R|S^c)cdotmathbb{P}(S^c)}

Substituting the known values:

mathbb{P}(S|R) frac{(4/5)cdot(1/6)}{(4/5)cdot(1/6) mathbb{P}(R|S^c)cdot(5/6)}

However, the exact value of mathbb{P}(R|S^c) is unknown, as it relies on the man's behavior when he lies. This value can range from 0 to 1, impacting the final probability.

Given the man's behavior and the assumption that he lies uniformly among other numbers when not rolling a 6, the probability:

mathbb{P}(S|R) in [4/29, 1]

Conclusion

The simplified interpretation based on the given conditions shows that the probability that the man is telling the truth when he says he rolled a 6 is approximately 37.5%. This problem showcases the power of Bayesian probability in assessing uncertain situations.