Let there be two people $A$ and $B$. Both have $0.5$ probability of speaking truth. I roll a die blindfolded. $A$ says $2$ has rolled and $B$ says $3$ has rolled over. Whats the probability of $2$ and $3$ being rolled over respectively
Now both of these have to be equal because of symmetry, but I want to understand the general concept of tackling these problems. I first thought $P(2)=0.5*(1-0.5)$ because it's the probability that $A$ spoke the truth and $B$ spoke a lie, but that means total probability becomes less than $1$ as $P(2)=P(3)=P(\neq2,3)=0.25$
In general even if there is no symmetry how do you go about tackling these problems
I am making the assumptions that the die is fair,
and that a lying person will randomly choose one of the five wrong answers.
Taking the specific problem first, as there are $6$ equi-probable die outcomes,
I am leaving out the $\dfrac16$ factor for a particular outcome for simplicity.
P(A says 2 & B says 3 | die shows $2$) $= 0.5\times 0.5/5= 0.05$
P(A says 2 & B says 3 | die shows $3$) $= 0.5/5 \times 0.5= 0.05$
P(A says 2 & B says 3 | die shows $1,4,5,6$) $= 4\times 0.5/5 \times 0.5/5= 0.04$
Restricted probability sample space $= 0.05+0.05+0.04 = 0.14$
P(die actually shows 2) = P(die actually shows 3) $= \dfrac{5}{14}$
For the general case, taking the truth probabilities for A and B as $a$ and $b$,
P(A says 2 & B says 3 | die shows $2$) $= a(1-b)/5$
P(A says 2 & B says 3 | die shows $3$) $= b(1-a)/5$
P(A says 2 & B says 3 | die shows $1,4,5,6$) $= 4(1-a)(1-b)/25$
Restricted probability sample space $= [5(a+b-2ab) + 4(1-a)(1-b)]/25$
P(die shows 2) $= \dfrac{5a(1-b)}{5(a+b-2ab) + 4(1-a)(1-b)}$
Work out similarly for P(die shows 3)