Question about application of Bayes theorem? 
A person speaks the truth 80 percent of the time. A 6 sided dice is tossed. He reports there is a 6. What is the probability there is actually a 6? (Use Bayes' theorem)

I know that the answer according to Bayes' theorem will be (4/5)(1/6)/((4/5)(1/6)+(1/5)*(5/6))=4/9=44.4%
ie. The probability that he is lying is more than him telling the truth.
However I am confused as to the fact that the same result will come if we had asked about the other cases (1,2,3,4,5). This will be true every time (?).
So if the person has a higher probability of lying each time how come he speaks the truth 80 percent not the time?
Thank you
 A: Imagine performing the experiment "roll the die, and ask this person if the die is a $6$" a total of $600$ times. Then the average result (obviously there will be random fluctuation) is that:

*

*$100$ times, the die will land $6$. The person will tell the truth and say "the die landed $6$" $80$ times, and lie and say "the die did not land $6$" $20$ times.

*$500$ times, the die will land on another face. The person will tell the truth and say "the die did not land $6$" $400$ times, and lie and say "the die landed $6$" $100$ times.

Out of $600$ trials, this person has told the truth $480$ times, which is $80\%$.
However, if you only look at the trials in which the person said "the die landed $6$", then there will be $80$ such trials in which the person is telling the truth, but $100$ trials in which the person is lying. That's the $44.4\%$ probability you're getting. Even though this person is mostly truthful, they get many more opportunities to lie about the die being a $6$ than they do to tell the truth, so most such statements end up lies.
Imagine that every day you ask this person, "Did you win the lottery today?" They will lie and say yes about $20\%$ of the time, and maybe one in a million times, they will tell the truth and say yes. Even though this person is mostly truthful, if they tell you "Yes, I won the lottery", you can be almost certain they're lying. That's because winning the lottery is much much more unlikely than the $20\%$ chance of a lie.
The die rolling example is less extreme but similar.
A: There isn't enough information given to be able to solve the problem.  Let $S$ be the event that a $6$ is rolled and $C$ be the event that this person claims a $6$ was rolled.  We are told that $\Pr(C|S)=.8$ and asked to compute $\Pr(S|C)$.   By Bayes' rule, $$\Pr(S|C)=\frac{\Pr(C|S)\Pr(S)}{\Pr(C)}=\frac{.8}{6\Pr(C)}$$
But we don't know what $\Pr(C)$ is.  We know that when a $6$ is rolled, the person claims it's a $6$ $80\%$ of the time, but what about when a $6$ isn't rolled?  $80\%$ of the time he tells the true, but what about the $20\%$ of the time when he lies?  Maybe he always claims it's a $6$ when he lies.  Maybe he never claims it a $6$ when he lies.  Maybe he chooses one of the false rolls uniformly at random.  Your calculation seems to take the point of view that whenever the roll is not $6$ and he decides to lie, he says it's $6$.  That is he falsely claims it's $6$ with probability $$.2\cdot\frac56$$
