I came across this task in an exam a few days ago:
There are 4 men. The first man receives a signal (a "YES" or a "NO"), and tells it to the second man, the second to the third and the third to the fourth. Each of the men tells the truth in 1/3 of cases. If you know that the fourth man told the truth (the truth being the original signal), what is the probability that the first man told the truth?
I tried this approach:
- calculate the probability that the fourth man told the truth
- express the probability that the third man told the truth (P3) via P4 (the probability that the fourth one did, which I calculated), then P3 via P2, and finally P2 via P1, which would give me an equation with only one unknown - P1.
However, that led me to the wrong result of an 80% probability.
Note that "telling the truth" is not the same as "giving the right answer" - if a person is told the wrong answer and he means to tell the truth, then he will actually pass on the wrong answer. So I believe that the question you want to ask should be properly stated as "if you know that the fourth man gave the correct answer, what is the probability that the first man gave the correct answer?"
Assuming I have correctly interpreted your question, let $A$ be the event that person A gave the right answer, and similarly for $B,C,D$. Note that for B,C,D, this is not the same as saying that they told the truth. The probability we need is $$P(A\,|\,D)=\frac{P(A\cap D)}{P(D)}\ .$$ Now D gives the correct answer if and only if $0,2$ or $4$ of the men lie. The probability of this is $$\Bigl(\frac{1}{3}\Bigr)^4+C(4,2)\Bigl(\frac{1}{3}\Bigr)^2\Bigl(\frac{2}{3}\Bigr)^2+\Bigl(\frac{2}{3}\Bigr)^4=\frac{41}{81}\ .$$ Note that there has to be an independence assumption here. Next, A and D both give the correct answer if A and either $1$ or $3$ of the others tell the truth. The probability is $$\Bigl(\frac{1}{3}\Bigr)\left(3\Bigl(\frac{1}{3}\Bigr)\Bigl(\frac{2}{3}\Bigr)^2+\Bigl(\frac{1}{3}\Bigr)^3\right)=\frac{13}{81}\ .$$ Therefore $$P(A\,|\,D)=\frac{13}{41}\ .$$
This question is very similar to Eddington's probability problem which you will certainly find through a Google search.