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
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.")