Intuitively understanding the purpose of Bayes' theorem for a $3/4$-probability truth-teller The problem states: The probability of a man telling the truth is = $3/4$. He rolls a dice and claims he got a six. What is the probability that he actually got a six?
And my textbook solves this using Bayes' theorem and gets $3/8$ as the answer.
The way I'm looking at it is: The probability that he really got what he's claiming (a six) should be equal to the probability of him telling the truth which = $3/4$. Where am I going wrong here?
 A: The missing link is this: ignoring whether the man lied or not, it's overwhelmingly likely that he didn't roll a 6. There are more scenarios in which he didn't roll a 6, but then lied and said he did, than that he just rolled a 6 and told the truth. 
Here's a similar scenario to illustrate the distinction: imagine that he was rolling a 100-sided die, and that he told the truth 3/4 of the time. If he rolled the die and claimed to have obtained a 100, it is overwhelmingly more likely that he didn't roll a 100 and lied about it than that he did and told the truth.
Computationally, if $A$ is the event that he gets a $6$, and $B$ is the event that he claimed to get a $6$, then you know that $P(B | A) = 3/4$, but you're interested in $P(A | B)$, which is why you need Bayes' theorem. Its use is to reverse the conditioning in this fashion.
A: There's actually a good argument for the answer being $\frac34$:
It hinges on what he says when he doesn't get a $6$ and lies.  Does he always lie by saying that he got a $6$, or does he distribute his answer (perhaps randomly) among five possible lies.
The book's answer works if you assume that if he doesn't get a $6$ and lies, he always says he got a $6$.  
Your answer is correct if when he doesn't get a $6$ he randomly chooses (with equal probability) one of the five other values to claim as his roll.  You can see this by looking at $24$ 'average rolls', in which we could expect to get four $6$s.  On three of these four $6$s (on average) he will say $6$; and the fourth time he will say some other value. For the other twenty (non-$6$) rolls, on fifteen of them he will tell the truth and not say $6$.  For the other five, if he distributes lies uniformly at random, he will give the lie $6$ once (on average).  Thus in the $24$ rolls he will say $6$ four times, of which three will be for actual $6$s.  This agrees with your answer of $\frac34$.  On the other hand if all his lies on non-$6$ rolls are 'I got a $6$', then he said $6$ eight times, of which three were for actual $6$s, which would agree with the book's answer. 
Exact wording of the problem is important.  For instance if the man is asked 'Did you get a $6$?' And he says yes or no with the given pattern of lying, then the book is correct.
A: Intuitively, Bayes' theorem makes allowance for to what degree the false positives (caused by lying) outweigh the true positives.  If the space of negatives is very large, (i.e. 5/6 of rolls are not a six) then the false negatives are more likely to outweigh the true negatives.
A more precise way of describing this is to say that Bayes Theorem makes allowance for the degree to which the sample is a biased estimator of the population.
