Bayes Theorem Challenge Problem [Edited]

Question

You have a person who likes telling the truth 2/3 of the time. He rolls a fair six sided dice and claims that he rolled a 4. Calculate the probability that he actually rolled a 4.

My progress: I did $(2/3)(1/6) /[ (2/3)(1/6) + (1/3)(5/6)]$ and got this to be equal to $2/7$ but apparently this is wrong. Any help would be appreciated.

Edit: The answer apparently is 2/3 but I don't understand why?

Answer 1

We shall assume that when he lies, he is as likely to claim one face as any another.

Rolled a $4$ and tells the truth: $\overset{\substack{\text{rolled}\\4\\\downarrow}}{\frac16}\cdot\overset{\substack{\text{tells}\\\text{truth}\\\downarrow}}{\frac23}=\frac19\\$
Rolled something else and says $4$: $\overset{\substack{\text{didn't}\\\text{roll }4\\\downarrow}}{\frac56}\cdot\overset{\substack{\text{tells}\\\text{lie}\\\downarrow}}{\frac13}\cdot\overset{\substack{\text{says}\\4\\\downarrow}}{\frac15}=\frac1{18}\\$
Probability he says $4$: $\overset{\substack{\text{rolled $4$}\\ \downarrow}}{\frac19}+\overset{\substack{\text{didn't}\\\text{roll }4\\\downarrow}}{\frac1{18}}=\frac16\\$

Thus, the probability he rolled a $4$ if he says $4$: $$ P(\text{rolled $4$ }|\text{ says $4$})=\frac{P(\text{rolled $4$ and says $4$})}{P(\text{says $4$})}=\frac{1/9}{1/6}=\frac23 $$

Answer 2

Let $R=\{\text{rolled a }4\}$, $S=\{\text{claims they rolled a }4\}$, $T=\{\text{telling truth}\}$, $L=\{\text{telling a lie}\}$.

You want $P(R\mid S)$. You must assume that the person's lying tendencies are independent of their die rolls, and assume they pick a number at random when they lie.

\begin{align*} P(R\mid S) &= \frac{P(R \text{ and } S)}{P(S)}\\ &= \frac{\frac{1}{6}\cdot \frac{2}{3}}{P(S\mid T)P(T) + P(S\mid L)P(L)} \end{align*}

Now $P(S \mid T)$ is $1/6$, because if the person is telling the truth, there is $1/6$ chance they actually rolled a $4$.

$P(S \mid L) = 5/6 \cdot 1/5 = 1/6$ because given the person is lying, there is $5/6$ chance they did not roll a $4$ (which is necessary for them to lie and claim they rolled a $4$), and a $1/5$ chance they chose $4$ as the lie.

You know $P(T)$ and $P(L)$, so you can continue from here.