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Saw this question elsewhere online, couldn't quite get a grip on it: "You have three friends, who tell the truth with probability 2/3 and lie with probability 1/3. You ask them independently if it is raining in your city, and they all say yes. What is the probability that it is raining?" Now, if we have a prior for rain in our city this is a simple application of Bayes rule. But if we don't? I suspect it is then an ill posed/unsolvable problem, but one approach I considered was that the probability of them all lying is 1/27, so it should rain with probability 26/27. That doesn't seem right though.

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one approach I considered was that the probability of them all lying is 1/27, so it should rain with probability 26/27. That doesn't seem right though.

The reason that this doesn't work is that $1/27$ is your prior probability for them all to be lying. Once you've heard that they all say it's raining, then $1/27$ is no longer the correct number to use. In particular, the probability that they're all lying depends on your prior for wheather it's raining or not. If you initially think it's extremely unlikely to be raining (say $1:1\,000\,000$) then when they say it's raining you will think it's very unlikely that they're telling the truth. Conversely if you think it's very likely to be raining then you'll think it's very likely that they are telling the truth.

You do need your prior to answer the original question. Let $A$="they all say it's raining" and $B$="it rains" then $P(A|B)=8/27$ and $P(A|¬B)=1/27$ so (cancelling factors of 27) $$P(B|A)=\frac{8.P(A)}{8.P(A)+1.(1-P(A))}=\frac{8P(A)}{1+7P(A)}$$

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