Arrangements in a circle such that A is closer to F than B is to F There are six things labelled $A$ to $F$. I need to calculate the number of distinct circular arrangements such that $A$ is closer to $F$ than $B$ is.
Here's what makes sense to me: fix $F$. Either $A$ is closer, further, or equidistant (than/with $B$). By symmetry the probabilities of 'closer' and 'further' are equal, say $p$. Let the probability of 'equidistant' be $q$. Then $1=q+2p$. Upon fixing $F$ there are four arrangements in which $A,B$ are equidistant, so $q=\frac 4{5!}$. Hence $p=\frac 12(1-\frac 4{5!})$, and $5!\cdot p=58$ which is my answer.
Is this okay? If not, what are my mistakes?
 A: The number of ways to choose the positions for A and B is $5\cdot4=20$, 
so this gives $p=\frac{1}{2}(1-\frac{4}{20})=\frac{2}{5}$.

Alternate solution:
1) If A is one place away from F, then there are 2 choices for A and 3 choices for B.
2) If A is two places away from F, then there are 2 choices for A and only 1 choice for B.
Therefore $ p=\frac{2\cdot3+2\cdot1}{5\cdot4}=\frac{8}{20}=\frac{2}{5}$.
A: You have the right idea, but you calculated the number of arrangements in which $A$ and $B$ are equidistant from $F$ incorrectly. There are indeed $4$ placements of $A$ and $B$, but for each of them there are $3!$ ways to fill the $3$ positions not occupied by $A,B$, or $F$ with $C,D$, and $E$. Thus, there are actually $4\cdot3!$ arrangements in which $A$ and $B$ are equidistant from $F$, and 
$$p=\frac12\left(1-\frac{4\cdot3!}{5!}\right)=\frac25\;.$$
You can keep the numbers smaller by adopting user84413’s approach and looking only at the $20$ possible locations for $A$ and $B$ instead of at the $5!$ possible arrangements of $A,B,C,D$, and $E$: no matter where $A$ and $B$ are, there are $3!$ possible arrangements of $D,E$, and $F$.
