Probability related to truth and false This is not a homework question. I came across this question but then I thought about doing something different. 
Question: A person is known to speak the truth exactly 3 out of 4 times. He rolls a die and is asked what is the outcome and he says a six. What is the probability that it is a six?
Method 1:
I can give this question a straight approach saying that since the probability of his saying the truth is 3/4 then the probability that it is a six is also 3/4.
Method 2:
If I go by a more rigorous method, let E1 be the event that the man reports a six and E be the event that it is a six.
$P(E)=\frac{1}{6}$
$P(E')=\frac{5}{6}$
$P(E1|E)=\frac{3}{4}$ (When the man speaks the truth)
$P(E1|E')=\frac{1}{4}.\frac{1}{5}$ (When the man speaks the false, he could say any of the remaining 5, one of which is six. Since the probability, that the man is false 1/4 and that if he chooses fairly between the remaining options the probability of saying a six if he lied is 1/5).
From Baye's theorem,
$$P(E|E1)=\frac{P(E)P(E1|E)}{P(E)P(E1|E)+P(E')P(E1|E')}$$
which gives P(E|E1) as 3/4.
Doubt:
Now, the first method did not depend on my assumption that when the person lies he chooses fairly among the remaining options. Something could be possibly wrong in either or both the methods. What is it?
Trying to verify whether the same answer in both methods is only a coincidence or not, I framed this question 
Modified Question: A person is known to speak the truth exactly 3 out of 4 times. He rolls a die and he is asked whether it is a six or not and he says 'yes'. What is the probability that it is a six?
Method 1:
I still get the same answer as previous problem because it makes sense to again say that since the probability of his saying the truth is 3/4 then the probability that it is a six is also 3/4.
Method 2:
Let E1 be the event that the man reports a six and E be the event that it is a six.
$P(E)=\frac{1}{6}$
$P(E')=\frac{5}{6}$
$P(E1|E)=\frac{3}{4}$ (When the man speaks the truth)
$P(E1|E')=\frac{1}{4}$ (When the man speaks the false, he could say only a yes if it is not a six. The probability that the man speaks false is 1/4)
From Baye's theorem,
$$P(E|E1)=\frac{P(E)P(E1|E)}{P(E)P(E1|E)+P(E')P(E1|E')}$$
which gives P(E|E1) as 3/8.
Doubt:
Now, how does this answer come different. I cannot find any fallacy in my logic in each of the four methods. Where am I going wrong?
 A: Operating with the interpretation of the problem I provided in the comments (you ask the man a yes/no question about whether or not the value on the die is a six to which he must respond, the man's decision to tell the truth or lie is independent of the value shown on the die, and the die itself is a fair standard six-sided die.)
Let $A$ be the event that the die actually shows a six and let $B$ be the event that the man is telling the truth.
The event "The man responded 'yes it is a six'" is then expressible as $(A\cap B)\cup (A^c\cap B^c)$
We are asked to find $Pr(A\mid (A\cap B)\cup (A^c\cap B^c))$
We recognize that $A$ is independent of $B$.
Applying definition of conditional probability:
$Pr(A\mid (A\cap B)\cup (A^c\cap B^c))=\dfrac{Pr(A\cap ((A\cap B)\cup (A^c\cap B^c))}{Pr((A\cap B)\cup (A^c\cap B^c))}$
$=\dfrac{Pr(A)Pr(B)}{Pr(A)Pr(B)+Pr(A^c)Pr(B^c)}=\dfrac{\frac{1}{6}\cdot\frac{3}{4}}{\frac{1}{6}\cdot\frac{3}{4}+\frac{5}{6}\cdot\frac{1}{4}}=\dfrac{3}{3+5}=\dfrac{3}{8}$

"Since he tells the truth $\frac{3}{4}$ of the time, why isn't it $\frac{3}{4}$?"
Imagine if you will that the question didn't have to do with rolling of dice at all, but instead the man decides to spew out arbitrary trivia facts (or falsities).
If he tells you "The moon is made of cheese" you know he is lying.  If he tells you "$1+1=2$" then you know he is telling the truth.  Hearing him say one of these two things lets us know whether or not he was lying just then.  If I hear him say "The moon is made of cheese", even if I know that he lies only a quarter of the time ordinarily, I know that this most recent statement was indeed a lie.
Now... if he says something incredibly outlandish but still technically possible... such as "I poured out my piggy bank and miraculously all thousand pennies in it landed heads up" although we can't outright prove it is a lie, we should be incredibly skeptical of it being the truth as such an occurrence is incredibly unlikely.
In the same way, hearing him say "I rolled a die and it landed on a six", the event is unlikely and so we are skeptical since the event itself is unlikely, but give him some benefit of the doubt since we do know he tells the truth every so often.  Specifically how much benefit of the doubt we show him is outlined in the calculations above.
A: I think the first question has a reasonably natural interpretation (though JMoravitz is absolutely correct that one must outline a specific procedure to avoid ambiguity here). You will ask them 'what is on the die' and with $3/4$ probability they will say the true answer, and with $1/4$ probability, one of the other five possibilities, uniformly chosen. I agree with your calculation here, saying the answer is $1/4.$ 
But there must be some difference between these questions so that the naive line of reason gets the right answer in the first case and wrong in the second case. It's easier to look at what goes wrong in the second case (JMoravitz explained this in the last part of their answer but I'll reiterate). Note that when they answer 'yes', that conveys some information to you about whether or not they're lying. You've asked them a question about whether an event with $1/6$ prior probability is true. Them answering 'yes' means, all other things being equal, that they're more likely to be lying, cause they've answered 'yes' to a question that had a low probability of being true. So you've gotten information about whether they're lying, and it's no longer coherent to believe there's only a $1/4$ chance... as you worked out, there is actually a $5/8$ chance that they're lying now that you know they said 'yes'. 
On the other hand, in the first situation, them answering '6' gives you no information as to whether they're lying. The situation is completely symmetrical and they're just as likely to have said '6' as any of the other five numbers. So your belief about whether they're lying doesn't change and there's still a $1/4$ chance.
A: The probability that a subject would say "it is a six" is: the probability that either it is a six and they tell the truth, or that it is not a six, they lie, yet say it is a six.  Assuming that when they lie about the result by unbiasedly selecting another valid option then the probability that the subject tells the truth when saying "it is a six" is:
$$\mathsf P(T\mid S)=\dfrac{\tfrac 16\tfrac 34}{\tfrac 16\tfrac 34+\tfrac 56\tfrac 14\tfrac 15} = \dfrac 34$$
However, that is not the only way the subject could lie, so we cannot truely claim that is the probability.
Case in point: suppose you ask a different question...

Asking for a yes/no response to "is it a six?" is a diffeent situaltion to directly asking for the result, so obtaining a different probability should not be surprising.
The subject can only lie by saying "yes" when it is not a six, and "no" when it is.   So in this case$$\mathsf P(T\mid Y)=\dfrac{\tfrac 16\tfrac 34}{\tfrac 16\tfrac 34+\tfrac 56\tfrac 14}=\dfrac{3}{8}$$ 
Why is this lower than $3/4$?   Well, the event of a six is much rarer than the event of not a six.   Thus saying "yes" is more likely to be a lie .
