Conditional Probability: Problem with 6 persons 
There are $6$ persons, let's name them $ A, B, C, D, E$ and$ F $. They are ordered according to their popularity, but without letting them know their order, so they guess all ways they can be ordered are equally possible. 
  Taking as a fact that those persons learn that $A$ is more popular than $B$, what do they think the probability for $ A$ to be more popular than $C$ is?

This question confuses me quite a lot, especially the way it's stated. All I know is that it's about conditional probability. Even a small hint would be a great help.
 A: I am going to make use of Bayes' Theorem:
$$P({A > C | A > B})=\frac{P(A>C \cap A>B)}{P(A>B)}$$
Consider lining up the letters as follows:
_ _ _ _ _ _
where popularity is rated from left to right.
By symmetry $$P(A>B)=\frac{1}{2}$$
We must solve for $$P(A>C \cap A>B)$$
which is the probability that $A$ is bigger than both $B$ and $C$.
If $A$ is in the leading position with probability $\frac{1}{6}$ then it will be greater than $B$ and $C$ with probability $1$
If $A$ is in the second position with probability $\frac{1}{6}$ then it will be greater than $B$ and $C$ with probability $\frac{4}{5}\cdot\frac{3}{4}$
If $A$ is in the third position with probability $\frac{1}{6}$ then it will be greater than $B$ and $C$ with probability $\frac{3}{5}\cdot\frac{2}{4}$
If $A$ is in the fourth position with probability $\frac{1}{6}$ then it will be greater than $B$ and $C$ with probability $\frac{2}{5}\cdot\frac{1}{4}$
If $A$ is in the fifth of final position then it is impossible for it to be greater than both $B$ and $C$.
Thus we have
$$\begin{align*}
P({A > C | A > B})
&=\frac{P(A>C \cap A>B)}{P(A>B)}\\\\
&=\frac{\left(\frac{1}{6}\cdot1\right)+\left(\frac{1}{6}\cdot\frac{4}{5}\cdot\frac{3}{4}\right)+\left(\frac{1}{6}\cdot\frac{3}{5}\cdot\frac{2}{4}\right)+\left(\frac{1}{6}\cdot\frac{2}{5}\cdot\frac{1}{4}\right)}{\frac{1}{2}}\\\\
&=\frac{2}{3}
\end{align*}$$
A: By the definition of conditional probability,
$$P(A>C | A>B) = \frac{P((A>C) \cap (A>B))}{P(A>B)}$$
To find $P(A>B)$, note that we can ignore all the other persons' popularities and consider only $A$ and $B$.  There are two possible orderings of $A$ and $B$, which are equally likely, and in only one of those orderings is $A>B$; so
$$P(A>B) = \frac{1}{2}$$
To find $P((A>B) \cap (A>C))$, we may ignore all the persons' popularities except those of $A$,$B$, and $C$.  There are $3!$ possible orderings of $A$, $B$, and $C$, all of which are equally likely, and in only two of those orderings is $A>B$ and $A>C$, namely $A>B>C$ and $A>C>B$.  So
$$P((A>B) \cap (A>C)) = \frac{2}{3!}$$
Combining these results, we see
$$P(A>C | A>B) = \frac{2/3!}{1/2} = \frac{2}{3}$$
