How do you solve Eddington's Problem- conditional probability. this was in a book on astronomy-
"A,B,C,D each lie 1/3 of the time (independently).
D makes a statement.
A affirms that B denies that C declares that D is lying.
What is the probability D spoke the truth??
the answer is 25/71"
Is that answer correct?  How is it obtained?  Where does the 71 come from?  It seems that the denominator should be some power of 3.
 A: Unless I've made a mistake, either you've misquoted the statement of the problem or the solution is wrong.
Assumptions:


*

*D made a statement, which had probability $1/3$ of being a lie.

*C either declared that D is lying or declared that D is not lying.  This statement had probability $1/3$ of being a lie.

*B either denied or confirmed that C made that declaration.  B's statement had probability $1/3$ of being a lie.

*A affirmed that B denied ...  This statement had probability $1/3$ of being a lie.


If all tell the truth, C must declare that D is not lying, B must indeed deny that C declared D was lying, and A must affirm this.  This case has probability $(2/3)^4 = 16/81$.
Each switch of one person's statement from truth to lie would change A's statement.  So A makes the affirmation if and only if an even number of A,B,C,D are telling the truth.  That has probability $16/81 + {4 \choose 2} \cdot 4/81 + 1/81 = 41/81$.  The cases where an even number of A,B,C are telling the truth but D is lying have total probability ${3 \choose 2} \cdot 4/81 + 1/81 = 13/81$.  So the conditional probability that D is lying, given that A makes the affirmation, should be $13/41$, not $25/71$.
EDIT: Ah, Eddington's assumptions were different.  See A New Look at Eddington's Liar Problem.
A: {* Eddington's probability problem-  }
"A,B,C,D each tell the truth 1/3 of the time (independently). 
D makes a statement. A affirms that B denies that C declares that D is lying.
What is the probability D spoke the truth??
Let's generalize the problem to any number of speakers,  each having
a probability $p_j$ of telling the truth.  (j=0,.... j=n)
Let's make a Table of (n+1) rows,  where each row has 2 columns. 
Row 0  has no previous speakers. Row 0 -  $T_0$ is the probability that $Speaker_0$ told the truth.
 Row j, Column T contains the probability that the speaker j says that the previous speaker told the truth. 
-$\qquad \qquad \qquad \qquad \qquad   T $
    $\qquad \qquad \qquad \qquad \qquad   F $
The table starts with     
$ Stage_0 \qquad \qquad \qquad     T_0 = p_o  \qquad \ \  and \qquad \ \qquad F_0 = (1-p_0) = q_0 $
$ Stage_1 \qquad \qquad \qquad    T_1= (p_1 * p_0 + q_1 * q_0)\qquad  F_1 = (1-T_1)$
...
...
$Stage_n \qquad  \qquad  T_n=p_n * T_{n-1}  + q_n *F_{n-1}  \qquad  \     F_n = (1 - T_n) $   

Now let's make a similar table 
for the $\mathbf {Conditional Probability} $
 for the later stages if $Speaker_0$ told the truth.
$Stage_0 \qquad \qquad \qquad    T_0 = 1 \qquad \qquad \qquad  F_0 = 0$
$ Stage_1 \qquad \qquad \qquad     T_1 = p_1  \qquad \qquad  \qquad \ \qquad F_1 = (1-p_1) = q_1 $
$ Stage_2 \qquad \qquad \qquad    T_2= (p_2 * p_1 + q_2 * q_1)\qquad  F_2 = (1-T_2)$
...
and continues like this
...
$Stage_{n-1} \qquad  \qquad  T_n=p_n * T_{n-1}  + q_n *F_{n-1}  \qquad  \     F_n = (1 - T_n) $
$T_j$ is the probability that Speaker j said that speaker (j-1) said speaker (j-2).... etc ... said that speaker 0 told the truth
$Stage_n$ of this conditional table is the same as $Stage_{n-1}$ - of the unconditional table, except that the Conditional table starts with  $T_1 = p_1 $ instead of $p_0$**
The EDDINGTON problem seeks the 'opposite' conditional probability-
i.e. it seeks the probability of $Stage_0$  given the   condition that
'$Speaker_n$ said that $Speaker_{n-1}$ said that ... etc..$Speaker_1$ said that $Speaker_0$ told the truth'.
The opposite conditional probabilities are related to each other via $\mathbf {Bayes Theorem} $  
P(A | $B_n$)   = [ P(A) * P($B_n$ | A) ] /  P ($B_n$) 
We have all the needed quantities on the right hand side, so we can plug in
specific probabilities for each stage,  and get the solution to Eddington's puzzle.
P(A) is $p_0$, the probablity that $Speaker_0$ tells the truth.
 P($B_n$ | A) is the conditional probability that we showed is the
    content of the T column of the (n-1)th  row of our table that starts with $p_1$
P ($B_n$) is the unconditional probability that $Speaker_n$ says what he says,
   and it is found in the T column of the (n)th row of our table that starts with $p_0$  
Noticing that a denial with probability p  is the same as an affirmation with a probability q=(1-p)
we can make all the statements in Eddingtons puzzle be affirmations, but two of the speakers 'truthiness'
 must be 2/3 instead of the 1/3 given in the puzzle.  We also note from the construction of the table 
that swapping $p_j$ for $q_j$ in any row will switch theT and F columns in subsequent rows , so
the two 2/3 entries we got from changing denials to affirmations, cancel each others effect, and we
can take all the probabilities in the puzzle to be 1/3  as in the original formulation
so, with that,  our table becomes
$Stage_0 \qquad \qquad \qquad    T_0 = 1 \qquad \qquad \qquad  F_0 = 0$
$ Stage_0 \qquad \qquad \qquad     1/3  \qquad \qquad \qquad \qquad  2/3 $
$ Stage_1 \qquad \qquad \qquad     5/9  \qquad \qquad \qquad \qquad  4/9 $
$ Stage_2 \qquad \qquad \qquad     13/27  \qquad \qquad \qquad  14/27 $
$ Stage_3 \qquad \qquad \qquad     41/81  \qquad \qquad \qquad  40/81 $
..
so
$P_{eddington}$ = P(A | $B_n$)   = [ P(A) * P($B_n$ | A) ] /  P ($B_n$) =        $\frac {[\frac13 * \frac{13}{27} ]}{\frac{41}{87}}$  =  $\frac{13}{41}$
This is the same result that Feller gives for this problem in his textbook on Probablity. He derives it after developing the theory of Markov chains of Bernoulli trials, and several pages of algebra.
Comment- In all cases, as the number of intervening speakers increases, the T and F column of succeeding stages tends to  being .5 each.
  This makes the solution to Eddington's problem, i.e. the conditional probability of $Speaker_0$ speaking the truth become 
equal to  $Speaker_0$' 's original unconditional likelihood of speaking the truth,  i.e. $p_0$,  independent of the details like how many stages and
what the 'truthiness' probability is at each stage.  At the 3rd stage in Eddingtons puzzle, the T and F columns are already practically .5 each, and the 
conditional probability  is .317 which is already quite close to the unconditional probability $p_0$ = $\frac13$.
