Is It Raining in Seattle?

Solution & Explanation

Problem Restatement

You have three friends in Seattle, each with a 2/3 chance of telling the truth and a 1/3 chance of lying. All three friends tell you it is raining. You need to determine the probability that it is actually raining.

Frequentist Approach

1. Understanding the Problem:

   • Each friend independently tells the truth with a probability of 2/3.
   • All three friends say it is raining.
   • We want to calculate the probability that it is indeed raining.

2. Calculate the Probability of All Friends Lying:

   • If all friends are lying, then it is not raining.

   • Probability that a single friend lies: 1/3.

   • Probability that all three friends lie:

     $P(\text{All Lie}) = \left(\frac{1}{3}\right)^3 = \frac{1}{27}$

3. Calculate the Probability of Raining:

   • The probability it is not raining is the probability that all friends are lying.

   • Thus, the probability it is raining is:

     $P(\text{Rain}) = 1 - P(\text{All Lie}) = 1 - \frac{1}{27} = \frac{26}{27}$

Bayesian Approach

1. Assumptions:

   • Let's assume the prior probability of rain in Seattle is 50% (though this is not given in the problem, it's a common assumption for such problems).
   • We define $F_i$ as the event that friend $i$ says it is raining.

2. Calculate Conditional Probabilities:

   • Probability all friends say it is raining given it is raining (all tell the truth):

     $P(F_1 \cap F_2 \cap F_3 | R) = \left(\frac{2}{3}\right)^3 = \frac{8}{27}$

   • Probability all friends say it is raining given it is not raining (all lie):

     $P(F_1 \cap F_2 \cap F_3 | R^c) = \left(\frac{1}{3}\right)^3 = \frac{1}{27}$

3. Apply Bayes' Theorem:

   • We use Bayes' theorem to find the probability it is raining given all friends say it is raining:

     $P(R | F_1 \cap F_2 \cap F_3) = \frac{P(F_1 \cap F_2 \cap F_3 | R) \cdot P(R)}{P(F_1 \cap F_2 \cap F_3)}$

4. Calculate Total Probability:

   • Total probability that all friends say it is raining:

     $P(F_1 \cap F_2 \cap F_3) = P(F_1 \cap F_2 \cap F_3 | R) \cdot P(R) + P(F_1 \cap F_2 \cap F_3 | R^c) \cdot P(R^c)$

     $= \frac{8}{27} \cdot 0.5 + \frac{1}{27} \cdot 0.5 = \frac{8}{54} + \frac{1}{54} = \frac{9}{54} = \frac{1}{6}$

5. Final Calculation:

   • Substitute back into Bayes' theorem:

     $P(R | F_1 \cap F_2 \cap F_3) = \frac{\frac{8}{27} \cdot 0.5}{\frac{1}{6}} = \frac{8}{9}$

Conclusion

The Bayesian approach yields a probability of $\frac{8}{9}$ that it is raining given all friends say it is raining, under the assumption of a 50% prior probability of rain. The frequentist approach, focusing on the probability that at least one friend is truthful, gives a probability of $\frac{26}{27}$. The difference arises due to the assumptions and focus of each method.