Truth or Lie? - Probability

Question

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 $1/5$.

Answer 1

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 = $1/5$

Answer 2

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 $\frac 29 + \frac 2{27} + \frac 2{81} + \frac 1{27} = \frac {29}{81}$.

Answer 3

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: $$\frac{3^{n-1} + (-1)^{n-1}}{3^{n} + (-1)^{n}}$$