Answer: $\frac{5}{16}$
Solution: Suppose $A>B$ means that $A$ finished earlier than $B$. Then, the criteria imply the following:
(a) either $B>A>C$ or $C>A>B$
(b) either $E>D>F$ or $F>D>E$
(c) either $D>B>C$ or $C>B>D$

Combining Criteria 1 and 2 together implies either $D>B>A>C$ or $DB>A>C$, there's one place before $D$, and four after $D$ (between the letters). Furthermore, there are two ways to permute $E$ and $F$. Thus, there are $(1)(4)(2)=8$ possible sequences, among which only 1 has $F$ being last. On the other hand, when $C>A>B>D$, there are again 8 possibilities. If $F$ is last, $E$ still has 4 possible places before $D$.
Thus, the required probability is $\frac{1+4}{8+8}=\frac{5}{16}$.
7. In a race with six runners, $A$ finished between $B$ and $C, B$ finished between $C$ and $D$, and $D$ finished between $E$ and $F$. If each sequence of winners in the race is equally likely to occur, what is the probability that $F$ placed last?