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There are four people $X,Y,Z,K$ who speak the truth with probability $\frac{1}{3}$ independently of each other. Given $X$ claimed that $Y$ denied that $Z$ declared that $K$ lied. Find the probability that $K$ speaks the truth?

My attempt: I plan to imitate the solution given in this post - Eddington's controversy simplified - but in our case, we have a lot more cases ($16$). I wonder if anyone has any thoughts for a slicker solutions, rather than brute-forcing? Also, I am confused if the word "claimed" is definite (aka. does it imply $X$ always speak the truth?)

StubbornAtom
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ghjk
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1 Answers1

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You don't need to do the 16 options. Just the ones where $K$ is telling the truth.

If $X$ is lying the we don't know, don't care what $Y,Z$ say. So the option is $X$ lying; $K$ (independently) telling truth.

If $X$ is telling the truth and $Y$ is lying then we know don't know, don't care what $Z$ says. So the options is $X$ truth, $Y$ lying, and $K$ (independently) telling the truth.

If $X$ is telling the truth and $Y$ is telling the truth and $Z$ is lying, the $K$ must have the truth. So the option is $X$ truth, $Y$ truth, $Z$ lying.

Those are the only three options that matter. Calculate them and add them up.

All other options $X$ lying and $K$ independently lying; $X$ truth $Y$ lying and $K$ independently lying; ad $X$ truth, $Y$ truth, $Z$ truth so $K$ must be lying-- are all $K$ lying.

If $X$ is lying, it doesn't matter what $Y$ and $Z$ do. That removes $4$ cases. If $X$ is truth but $Y$ is lying it doesn't matter what $Z$ does. That removes $2$ cases. And if $X$ and $Y$ are telling the true the $K$ telling the truth is entirely on $Z$ so that removes $2$.

fleablood
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