You have three friends in Portland, each of whom will lie to you 1/3 of the time; they all tell you that it's currently sunny in Portland... Full Question: > "You have three friends in Portland, each of whom will lie to you 1/3rd of the time; they all tell you that it's currently sunny in Portland what do you think is the chance that it is sunny, given it rains about 2/3rds of the time there?" My Solution: P(Sunny | all friends say it is sunny) = [P(all friends say it is sunny | sunny) P(all friends say it is sunny)]/P(sunny) P(all friends say it is sunny | sunny) = P(all friends tell the truth) = (2/3)^3 = 8/27 P(all friends say it is sunny) = (1/2)^3 = 1/8 P(sunny) = p(it is not raining) = (1/3) P(sunny | all friends say it is sunny) = [(8/27) * (1/8)] / (1/3) = 1/9 My Issue: I believe that I calculated the prior for 'all friends say it is sunny' incorrectly. I think there might be more hidden inside, or something that I am missing.

Bayes' theorem is used as such.
There are two possibilities:

1) It is raining and they are lying

2) It is sunny and they are telling the truth.

\begin{equation} \frac{P_s}{P_r + P_s} = \frac{\frac{1}{3}(\frac{2}{3})^3}{\frac{1}{3}(\frac{2}{3})^3+\frac{2}{3}(\frac{1}{3})^3} = 0.8 \end{equation}