Q:

Probability question, explanation does not make sense?!

Q:  A man speaks truth $3$ out of $4$ times. He throws a die and reports
  it to be a $6$. What is the probability of it being a $6$?
Provided Answer: There are two cases 1) he is telling truth that the
  die reports $6$, its probability $= 3/4\cdot 1/6 = 1/8$. 2) he is telling lie
  that the die reports $6$, its probability $= 1/4\cdot 5/6 = 5/24$ So required
  probability $= (1/8)/(1/8)+(5/24) = (1/8)/(1/3) = 3/8$

What I don't understand is if we know that the man speaks truth $3$ out of $4$ times, and if he says there is a $6$ on the die, doesn't that mean that there is a $3/4$ probability that there actually is a $6$.
Why does reporting what is on the die makes the man less trustworthy than he is supposed to be. What am I missing here?

A:

Let X be the event where the man rolled a 6, !X be the event that the man did not roll a 6, and R be the event that the man said he scored a 6.
What you want is something like:
$$P(X|R) = \frac{P(R|X)}{P(R)} P(X)$$
I think you know $P(R|X)=3/4$, $P(X)=1/6$, $P(R) = P(R|X)P(X)+P(R|!X)P(!X)$. I am a bit bothered that $P(R|!X)=1/4$ seems to suggest that if the man doesn't roll a 6 he will lie and claim a 6 rather than claiming one of the other four possible values. This would lead to the answer that the text implies... $$P(X|R) = \frac{3/4}{8/24}1/6 = 3/8$$
This seems wrong because the man will have 5 options to lie with. It seems like it would give up the game if the man said more than one option. If instead we plug in $P(R|!X) = (1/4)(1/5)=1/20$ we have:
$$P(X|R) = \frac{3/4}{4/24}1/6 = 3/4$$

