Probably that an $80\%$-truthful person actually rolled a $6$ A person, $A$, speaks the truth $4$ out of $5$ times. The person throws a die and reports that he obtained a $6$. What is the probability that he actually rolled a $6$?
I know there is a similar question like this 
but my doubts are different from it and also I want to identify and solve total probability theorem questions so I posted a side doubt also.
In my attempt, I defined the events
\begin{align*}
 E_1&: \text{The person tells the truth.} \\
 E_2&: \text{The person lies.} \\
 E_3&: \text{The person reports that the die landed on a 6.} 
\end{align*}
I noted that $P(E_1)=\frac{4}{5}$, $P(E_2)=\frac{1}{5}$, $P(E_3|E_1)=6^{-1}$ and
$P(E_3|E_2)=0$ and obtained
\begin{align*}
 P(E_3) = \frac{4}{5} \cdot \frac{1}{6} + \frac{1}{5} \cdot 0 = \frac{2}{15}.
\end{align*}
However, the correct answer is, $\frac{4}{9}$. What did I do wrong?
Side doubt: Even though the first experiment (truth and lying) is different from the second experiment, can we still apply total probability theorem? In my book the dependent experiment lies inside the sample space associated with the mutually and exhaustive events.
 A: There are two serious problems with your work:


*

*$P(E_3|E_2)$ is probably not zero. You don't provide the details about how $A$ lies, but it's certainly plausible, for example, that $A$ rolled a $4$ and lied saying he rolled a $6$.

*You were asked to find the probability of the event "$A$ rolled a $6$", but that event doesn't even appear in your model! You tried to find $P(E_3)$ instead.


As an aside, your answer key seems to assume $A$ lies in odd way — that when he lies about the die roll he will always say "I obtained a $6$" if possible, rather than any of the other four ways to lie.
A: There is some missing information, namely: When he lies, does he lie at random, or with some pattern? In particular, if the die shows up as a $2$, for example, is he equally likely to report $1,3,4,5$ or $6$ as the outcome?
If we assume that all of these are equally likely, then the probability that he reports a $6$, given that a number other than $6$ was rolled, would be $\frac15\cdot\frac15=\frac1{25}$. (The first $\frac15$ is the probability of lying; the second $\frac15$ is the probability that "$6$" is the chosen lie.) On the other hand, the probability that he reports a $6$ given that a $6$ was rolled, is $\frac45$. Thus, when he reports a $6$, the probability that a six was actually rolled is:
$$\frac{\frac16\cdot\frac45}{\frac16\cdot\frac45 + \frac56\cdot\frac1{25}}=\frac45$$
If that's not the answer, then there is some other assumption at work here that we don't know about. Perhaps he consistently lies by reporting “$6$” when non-$6$ numbers are rolled? If that's the case, then our calculation changes:
$$\frac{\frac16\cdot\frac45}{\frac16\cdot\frac45 + \frac56\cdot\frac15}=\frac49$$
The problem should really be asked in a way that makes it clear how lying actually works in this situation. If we ask the question, "is the outcome a $6$?", then we're in the latter case, because then it's clear exactly what a lie would look like.
A: Let $D_{6}$ denote the event that the die lands on a $6$ and $R_{6}$ denote the event that the person reports that it landed on a $6$. You know that
\begin{align*}
 P(R_{6}|D_{6})     &= 0.8 \\
 P(D_{6})           &= 6^{-1}
\end{align*}
In order to obtain the correct answer, you also need to assume that $P(R_{6}|D_{6}^{c})=0.2$, that is, if the person lies and doesn't obtain a $6$, then he says he obtained a $6$.
Since you wish to determine $P(D_{6}|R_{6})$, which is an inverse probability of $P(R_{6}|D_{6})$, a strategy that usually works is applying Bayes' theorem.
\begin{align*}
 P(D_{6}|R_{6}) 
 &= \frac{P(D_{6})P(R_{6}|D_{6})}
{P(D_{6})P(R_{6}|D_{6})+P(D_{6}^c)P(R_{6}|D_{6}^c)} \\
 &= \frac{6^{-1} \cdot 0.8}{6^{-1} \cdot 0.8 + 5 \cdot 6^{-1} \cdot 0.2}
 = \frac{4}{9}
\end{align*}
A: Several answers already show that the correct solution is 4/9, but I want to answer the question "What did I do wrong?"
Let's look at your calculation. Even if the person would tell the truth all the time, your probability would be 1/6. But if he is telling the truth, he MUST have thrown a 6.
You forgot to take into account that the person already told you that he has thrown a 6. Instead, you calculated the beforehand probability that the person will come up to you, telling you truthfully that he has thrown a 6. This is 2/15.
(You can check this easily: The same probability holds for any other number, so the probability that he will tell you any number and speak the truth is 12/15 or 4/5, which is the probability of him speaking the truth.)
A: I see that the question has been edited a lot; I'm answering the current version (revision 7), which reads like this: 

"A person, A, speaks the truth 4 out of 5 times. The person throws a die and reports that he obtained a 6. What is the probability that he actually rolled a 6?"

A rolls the die, looks at the result, and then states that the result is a 6. Since A knew what the roll was when stating "it is a 6", and it is given that A tells the truth 4/5 of the time, there are only two alternatives:


*

*A told the truth, with probability 4/5. The roll is a 6.

*A lied, with probability 1/5. The roll is not a 6. 


Therefore the probability that A rolled a 6 is 4/5, and the 4/9 you report as "the correct answer" is wrong. There is no reason to start by considering what the roll is, because if A is lying, the roll cannot be a 6.
A: There are 4 outcomes: speaks and 6, does not speak and 6, speaks and not 6, and does not speak and not 6. We know that he speaks so there are 2 outcomes: speaks and 6, and speaks and not 6. The probability of the first one is 1/6 and the other one is 5/6. So it is 1/6.
A: Your error would appear to be in P(E3|E2)=0
You are missing out the fact that the user may report a 6 being throw when in fact one has not
In fact your use of E3 is not helpful as it covers two scenarios.
What you are interested in is
User throws a six AND tells the truth
User lies AND reports that they threw a 6.
