Truth or Lie? - Probability 
Probability that each of the four men $A$, $B$, $C$ and $D$ tells the truth is $1/3$. $A$ makes a statement. $D$ says that $C$ says that $B$ says that $A$′s statement is true. What is the probability that $A$'s statement is actually true? 

Here's what I thought: 
$1$. Probability of $A$ making a true statement independently is $1/3$.
$2$. The probability of $A$ making the true statement given that $B,C,D$ make their true/false statements is obviously different (how exactly, I'm not sure)
$3$. The favorable cases are:
(i) $A, B, C, D$ all are telling the truth.
(ii) $A$ is telling the truth, whereas exactly two of $B, C, D$ are lying. 
How do I go about it now? Is there anything I've missed? Please help with proper explanation.
P.S. Can we generalise this to $n$ persons, given that the probability of each person independently telling the truth is $p$?
P.P.S. The answer I have, to the above question is $1/5$.
 A: The probability that $D$ is telling the truth is $\frac 13$.  The probability that $D$ is lying is $\frac 23$.  If $D$ is lying then his statement has no value whatsoever.  (Note: it is NOT the case that he is lying that C said the opposite-- Anything could have happen.  C could have said nothing.  C could have said "ducks eat crackers", etc.)
So the probability that $A$ is telling the truth and $D$ is lying is $\frac 23*\frac 13 = \frac 29$.
If $D$ is telling the truth and then there is $\frac 23$ chance that $C$ was lying.  So the probability if $D$ telling truth, $C$ lying, and $A$ telling truth is $\frac 13*\frac 23*\frac 13 =\frac 2{27}$.
If $D$ and $C$ are telling the truth then then probability of $B$ lying and $A$ telling truth is $\frac 13*\frac 13 *\frac 23*\frac 13 = \frac 2{81}$.
If $D$ and $C$ and $B$ is telling the truth then $A$ must be telling the truth (because that is what $B$ honestly said).  So that is $\frac 13*\frac 13*\frac 13 = \frac 1{27}$.
So the probability that $A$ telling the truthe is $\frac 29 + \frac 2{27} + \frac 2{81} + \frac 1{27} = \frac {29}{81}$.
A: Here's one possible model for this set-up:
$A$ says a statement, which is either a truth or a lie.
$B$ either says "$A$ is telling the truth" or "$A$ is lying"
$C$ either says "$B$ says that $A$ is telling the truth" or "$B$ says that $A$ is lying"
$D$ says either "$C$ says that $B$ says that $A$ is telling the truth" or "$C$ says that $B$ says that $A$ is lying"
Each statement is independently either a truth (probability $1/3$) or a lie (probability $2/3$). 
The question is: GIVEN that $D$ says "$C$ says that $B$ says that $A$ is telling the truth", what is the conditional probability that $A$ is actually telling the truth?
Given that $D$ says "$C$ says that $B$ says that $A$ is telling the truth", what are the possible configurations of Truth/Lies that are possible? Let's call the two statements that $B$ and $C$ can make $B1$,$B2$ and $C1,C2$, in the order I've written them.
We look at a few cases: If $D$ is telling the truth, $C$ said $C1$. If $C$ is telling the truth, $B$ said $B1$. If $B$ is telling the truth, $A$ is telling the truth. So the sequence "TTTT" (read from $D$ to $A$) is possible.
If $D$ is telling the truth, $C$ said $C1$. If $C$ is lying, $B$ said $B2$. If $B$ is lying, $A$ is telling the truth. So "TLLT" is also possible. 
The pattern- which you should convince yourself is true- is this: Only sequences with an even number of L's are possible. 
There is one sequence with zero L's: TTTT. This has probability $\frac{1}{81}$
There are 6 sequences with two L's. Each has probability $\frac{4}{81}$. In half of these six (total probability $\frac{12}{81}$) $A$ is lying, and in the other half $A$ is telling the truth.
Finally, there is one sequence with four L's. This has probability $\frac{16}{81}$.
So the answer is:
$$\frac{\text{Probability A is telling the truth given D's statement}}{\text{Probability of D's statement}} = \frac{\frac{1}{81} + \frac{12}{81}}{\frac{1}{81} + \frac{12}{81} + \frac{12}{81} + \frac{16}{81}} = \frac{13}{41}$$
EDIT: I believe that the general answer for $n$ people is:
$$\frac{3^{n-1} + (-1)^{n-1}}{3^{n} + (-1)^{n}}$$
A: Consider just two people $A$ and $B$ and $A$ made a statement. 
A said true statement (1/3)
A said false statement (2/3)
B said A said true statement = $1/3 * 1/3 = 1/9$ (A told truth, B told truth)
B said A said true statement = $2/3 * 2/3 = 4/9$ (A told lie, B told lie)
So probability = $\frac{1/9}{5/9} = 1/5$
Extending it to $3$ people, we will get 
C said B said A said true statement = $1/3 * 1/3*1/3 = 1/27$ (A told truth, B told truth, C told truth)
C said B said A said true statement = $2/3 * 2/3 *1/3= 4/27$ (A told lie, B told lie, C told truth)
Probability is again =$\frac{1/27}{5/27} = 1/5$
For 4:
A told lie, B told lie, C told truth, D told truth = 4/81
A told truth, B told truth, C told truth, D told truth = 1/81
Prob = $1/5$
