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One of four people got an information. That person sends an information as a signal $1$ or $0$ to the second person, second person sends an information to third person, third person to fourth, and fourth person sends an information farther. It is known that each person speaks the truth in $1/3$ cases. What is the probability that the first person told the truth, if the fourth person told the truth?

I got the equation:

$$\frac{1}{3}\cdot\frac{2}{3}x+\frac{1}{3}\cdot\frac{2}{3}y+\frac{1}{3}\cdot\frac{2}{3}z+\frac{1}{3}\cdot\frac{2}{3}\cdot1=\frac{1}{3}\cdot\frac{2}{3}p$$

where $x,y,z$ are probabilities that the first, second and third person is telling the truth.

From this equation we can't get any unknown.

How to solve this problem?

Note: Please don't suggest Markov chain method.

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    Not sure this is clear. I suspect you mean "what is the probability that the fourth person passes on the original datum.", yes? The point being that this could happen as a result of $\#4$ lying or telling the truth, depending on what $\#3$ passed on. – 2017-01-10
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    I think drawing a tree might help. All you need to find is when person 1 and person 4 speak truth together, which can happen in a few ways, depending on what person 2 and person 3 do. A tree might be fastest way to do this. – 2017-01-10
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    Just to be clear, if by "the fourth person told the truth" you mean the "fourth person accurately repeated what the third person said" then the answer is $\frac 13$ because person one's truth telling is independent of everyone else's. But, again, I do not believe that this is what you meant. – 2017-01-10

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Only way 4 and one are both telling the truth is

1. Either both 2 and 3 tell the truth in which case everyone spoke truth and you get $(\frac{1}{3})^{4}$ = $\frac{1}{81}$

  1. 2 and 3 can both not tell the truth in which case probability is : $(\frac{1}{3})^{2}$$\times$$(\frac{2}{3})^{2}$ = $\frac{4}{81}$.

So you sum them up and your answer should be $\frac{5}{81}$

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    it's conditional on 4 telling the truth though – 2017-01-10
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    well hasn't 4 told the truth already though, as in that the conditional based on – 2017-01-10
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Let $T_i$ be the event that i tells the truth. Then $$P(T_1|T_4) = \frac{P(T_4|T_1)P(T_1)}{P(T_4)} = P(T_4|T_1)$$ Since $P(T_1) = P(T_4) = 1/3.$ So the probability that the 1st person told the truth given that the 4th did is equal to the probability that the 4th tells the truth given that the 1st did.

So, given the 1st tells the truth, between the 2nd, 3rd and 4th person there must be an even number of lies for the message to be transmitted. So 0 or 2 lies. There is one way to have zero lies, which has probability $(2/3)^3$ and there are three ways to have two lies, each with probability $(1/3)^2(2/3)$ So we get $$(2/3)^3 + 3(1/3)^2(2/3)=14/27$$

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I believe, possibly incorrectly, that the question intends to ask "what is the probability that the first person spoke truly conditioned on the fact that the fourth person correctly passed along the correct datum". In any case, I'll answer that question.

Suppose the correct datum is $1$. We consider the paths in which person four says "$1$". They are $$(1,1,1,1),(1,1,0,1),(1,0,1,1),(1,0,0,1),(0,1,1,1),(0,1,0,1),(0,0,1,1),(0,0,0,1)$$

We note that person one speaks truly in the first four paths, and lies in the final four.

We remark that the probability of a specific path is $\frac {2^L}{3^4}$ where $L$ is the number of lies told along that path. We compute $L$ for each of the paths we care about: $$0,2,2,2,2,4,2,2$$

Thus the total probability of our universe is $$\frac {1+2^2+2^2+2^2+2^2+2^4+2^2+2^2}{3^4}=\frac {41}{81}$$

The paths along which both persons one and four speak the truth have probability $$\frac {1+2^2+2^2+2^2}{3^4}=\frac {13}{81}$$

Therefore the conditonal probability is $$\frac {13/81}{41/81}=\boxed {\frac {13}{41}}$$

Once again, it is possible I am not interpreting the question correctly.