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A man is known to speak truth 3 out of 4 times he throws a die and reports that it is a six. What is the probability that it's actually a six.

So I applied Bayes theorem and found the answer as 3/8. And I felt weird because of my common sense, If a man says truth 75% of the time, shouldn't the six on the dice probability be 3/4 as he said it's a six.

I am really confused. Am I missing something really obvious here? Or is there a flaw in my thought process please help me

Mihir
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  • The idea behind Bayes's theorem is that we compute the probability of all the possible ways the event can happen, and sum them up. In this case, there are two possibilities: the man rolled a six, and spoke the truth, or he rolled not a six, and lied. Thus, the answer is $\frac16\cdot\frac34+\frac56\cdot\frac14=\frac13$ – Rushabh Mehta Jun 05 '22 at 13:35
  • @RushabhMehta I don't think that is correct. You calculated the probability of the man claiming that he got a 6, but not the probability that it is actually a 6, given that he claimed it. – Andreas Tsevas Jun 05 '22 at 13:47
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    As a side note: It's often difficult to parse problems like this in which "lie" is not defined. What rules govern the lie here? Is the man equally likely to say "6" or "it's a unicorn" or "it's the fear of spiders"? If so, then then fact that he declared a possible number more or less proves that it really was a $6$. Here, one might guess that a "lie" means "he selects an inaccurate number uniformly from the $5$ possible incorrect values" but of course that is not stated. Perhaps the man really likes the number $6$, and always selects that value (if possible) when he lies. Who knows? – lulu Jun 05 '22 at 13:55
  • ... and if the man does choose one of the other five numbers at at random when lying, then the probability of the throw being a $6$ when reported as a $6$ is $75%$ or more verbosely $\dfrac{\frac16 \times \frac34}{\frac16 \times \frac34+ \frac56 \times \frac14 \times \frac15}$ – Henry Jun 08 '22 at 00:44
  • Ambiguity also discussed at https://math.stackexchange.com/questions/1084785/a-man-who-lies-a-fourth-of-the-time-throws-a-die-and-says-it-is-a-six-what-is-t and https://math.stackexchange.com/questions/528718/a-man-is-known-to-speak-truth-3-out-of-4-times-he-throws-a-die-and-reports-that and https://math.stackexchange.com/questions/2530837/intuitively-understanding-the-purpose-of-bayes-theorem-for-a-3-4-probability – Henry Jun 08 '22 at 00:48

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First off, your calculation is correct.

\begin{equation} \begin{aligned} P(\text{rolled 6}|\text{say 6}) &= \frac{P(\text{say 6}|\text{rolled 6})P(\text{rolled 6})}{P(\text{say 6})}\\ &=\frac{P(\text{say 6}|\text{rolled 6})P(\text{rolled 6})}{P(\text{say 6}|\text{rolled 6})P(\text{rolled 6}) + P(\text{say 6}|\text{rolled not 6})P(\text{rolled not 6})}\\ &=\frac{\frac{3}{4}\cdot\frac{1}{6}}{\frac{3}{4}\cdot\frac{1}{6}+\frac{1}{4}\cdot\frac{5}{6}}\\ &=\frac{3}{8} \end{aligned} \end{equation}

You are right that we trust the man $75\%$ of the time, in general. However, he claimed a really improbable event that only happens once every six times, so our expectations are below $75\%$. It is more probable that the lied to us than that he got a 6. That is the beauty of Bayes' theorem: we use the additional knowledge we have to update our beliefs rigorously.

Consider a more extreme example of the same man claiming to have won the lottery with odds $1:1,000,000,000$. You would definitely not think that the man actually won the lottery with probability $75\%$, would you? You would rather think that the man is lying to you this time, with huge confidence.

This makes intuitive sense: Your trust in the statements of a known liar decreases the more far-fetched these statements are.

(Edit: Excellent sidenote by lulu in the post comments! $3/8$ is the correct answer only if you assume that every statement of the man is correct with probability $75\%$. This is not equivalent to the statement by OP "The man speaks the truth in $75\%$ of the times.")

Andreas Tsevas
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