TITLE: Truth and Lie Problem in Probability
QUESTION [3 upvotes]: A fair die is rolled once and the outcome is observed by person $A$. $A$ then reports the outcome to person $B$ and $B$ writes it on a paper slip which shows that $6$ appeared on the die. Given that $A$ speaks the truth $3$ out of $4$ times and $B$ writes the truth (i.e, what A told him) $7$ out of $8$ times, what is the probability that $6$ actually appeared on the die? Given that when $A$ and $B$ are lying they tell or write a random number from $1$ to $6$ other than the one which they have observed (in case of $A$) or been told (in case of $B$).
Since I'm not that good at math so I made a code that imitated the given scenario and iterated it a billion times. The probability, thus obtained, approached $0.1107$. But I want to solve this problem mathematically.

REPLY [3 votes]: Let $B$ be the event that $B$ writes $6$ and $R$ be the event that a $6$ was rolled.  We need to calculate $$P(R|B)=\frac{P(R\cap B)}{P(B)}$$

Let $p$ be the probability that $A$ tells the truth. The probability that $B$ is told that a $6$ was rolled is $$\frac16p+\frac56(1-p)\left(\frac15\right)=\frac16p+\frac16(1-p)=\frac16$$ since either a $6$ is rolled and $A$ tells the truth or a $6$ is not rolled, $A$ lies, and $A$ chooses to report a $6$ among all possible false values.  Therefore, the probability that $A$ reports $6$ to $B$ is $\frac16$ -- independent of the probability that $A$ tells the truth.
Obviously, the same calculation applies to the second half.  There is a $\frac16$ probability that $B$ is told to write a 6 and that is the probability that he will write a $6$ independently of how trustworthy he is.  Thus, $P(B)=\frac16$.

Now we need to find $P(R\cap B)$, the probability that both a $6$ was rolled and written.  For this, we do need the trustworthiness probabilities.  To have a 6 rolled and written, we need a six rolled AND (either A and B both tell the truth or A and B both lie and B chooses to write 6), which is $$\frac16(\frac34\cdot\frac78+\frac14\cdot\frac18\cdot\frac15)=\frac{53}{480}$$
Thus $$P(R|B)=\frac{{53}/{480}}{1/6}=\frac{53}{80}$$
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