Eddington's controversy simplified I have been given the following probability problem:

A, B and C are three people, who each independently speak the truth one of three times.
A denies that B declares that C is lying.
What is the probability that C is telling the truth?

This problem is similar to Eddington's controversy

If A, B, C, D each speaks the truth 1 in 3 times (independently), and A affirms that B denies that C delcares that D is a liar, what’s the probability that D was speaking the truth?

I believe that the answer to the Eddington's problem was $\frac{25}{71}$ as given by Eddington himself. However, the other solution (based on different initial assumptions) is $\frac{13}{41}$.
My question is, how are the above solutions worked out? If I understand the logic behind Eddington's problem, I hope I can solve my probability problem.
 A: Assume that there is an obviously true external fact A, B, and C know about, say, the external fact that $x$ equals $x$. Also, assume that A, B hear each other and they hear C and choose to lie independently of what they hear. However, if they lie then they meticulously negate what they've heard.

If we don't know anything else then the  probability that C lies is $\frac13$.

The question is if we can learn more about the probability that C lied if we hear the following:

A says: "I deny that B declares that C is lying."

The meaning of the statement above is not ambiguous since it does not refer to the truth of the fact ($x=x$), it refers to what B said. Independently of C telling or not telling the truth, B either said that "C was lying" or she did not. 
We have then the following table in which we quote what our characters say. Color $\color{red}{\text{red}}$ denotes a lie and color $\color{green}{\text{red}}$ denotes the truth.
$$\begin{matrix}
\text{ A says }&\text{ B says }&\text{ C says }&\text{The probability}\\
\color{red}{\text{ I deny that B declares that C is lying }}&\color{green}{\text{ C lies }}&\color{red}{x\not=x}&\frac13\frac23\frac13=\frac2{27}\\
\color{red}{\text{ I deny that B declares that C is lying }}&\color{red}{\text{ C lies }}&\color{green}{x=x}&\frac23\frac23\frac13=\frac4{27}\\
\color{green}{\text{ I deny that B declares that C is lying }}&\color{red}{\text{ C does not lie }}&\color{red}{x\not=x}&\frac23\frac13\frac13=\frac2{27}\\
\color{green}{\text{ I deny that B declares that C is lying }}&\color{green}{\text{ C does not lie }}&\color{green}{x=x}&\frac23\frac23\frac23=\frac8{27}\\
\color{green}{\text{ I don't deny that B declares that C is lying }}&\color{green}{\text{ C lies }}&\color{red}{x\not=x}&\frac23\frac23\frac13=\frac4{27}\\
\color{green}{\text{ I don't deny that B declares that C is lying }}&\color{red}{\text{ C  lies }}&\color{green}{x=x}&\frac23\frac13\frac23=\frac4{27}\\
\color{red}{\text{ I don't deny that B declares that C is lying }}&\color{red}{\text{ C does not lie }}&\color{red}{x\not=x}&\frac13\frac13\frac13=\frac1{27}\\
\color{red}{\text{ I don't deny that B declares that C is lying }}&\color{green}{\text{ C does not lie }}&\color{green}{x=x}&\frac13\frac23\frac13=\frac2{27}\\
\end{matrix}$$
Having listed all the possibilities we need the following conditional probability
$$P(\text{ C is lying }\mid \text{A denies that B declares that C is lying})=$$
$$\frac{P(\text{ C is lying }\cap \text{A denies that B declares that C is lying})}{P(\text{A says that B denies that C is lying})}.$$
In the denominator we'll have to add the probabilities belonging to the first four rows of the table. For the numerator we'll have to add the first and the third probabilities. That is,
$$P(\text{ C is lying }\mid \text{A denies that B declares that C is lying})=\frac{\frac4{27}}{\frac{16}{27}}=\frac14.$$
So the probability that C lies is $\frac13$ but under the condition that A said what she said the same probability decreased to $\frac14$ "in our ears".
Note
It would be different if A said

"I claim that B is lying."

Even in that case a similar table could be built but one had to pay more attention to choose the colors.
