smc

Let $ACEDFB$ be the octahedron such that points $CDEF$ are coplanar, $C/D$ are opposite each other and $E/F$ are opposite each other. The set of points that the ants from $A/B$ can travel to is $\{C,D,E,F\}$, the set of points that the ants from $C/D$ can travel to is $\{A,B,E,F\}$ and the set of points that the ants from $E/F$ can travel to is $\{A,B,C,D\}$. So the problem is analogous to finding the number of ways to take 2 elements from each of these sets to make the set $\{A,B,C,D,E,F\}$. Case 1: Ants from $E/F$ travel to $A/B$ If this occurs then the ants from $C/D$ must travel to $E/F$ and therefore the ants from $E/F$ must travel to $C/D$. So this gives 1 case. Case 2: Ants from $E/F$ travel to $A/C$ If this happens then one of the ants from $C/D$ must go to $B$ and one of the ants from $A/B$ must go to $D$ and the remaining ant from $C/D$ can choose to go to $E$ or $F$ and then the remaining ant from $A/B$ has only 1 choice. So this gives 2 cases. The ants from $E/F$ traveling to $A/B$ is equivalent to them traveling to $C/D$ (both remove 2 elements from another set). And the ants from $E/F$ traveling to $A/C$ is equivalent to them traveling to $A/D$ or $B/C$ or $B/D$ (both remove 1 element from the other 2 sets). Also, for each of the 3 pairs of ants there are 2 ways the pair of ants can go to 2 points (for example, $E\mapsto A$ and $F\mapsto B$ or $E\mapsto B$ and $F\mapsto A$) so there are $2^3(2\cdot 1+4\cdot 2)=80$ possibilties. There are a total of $2^{12}$ ways the ants can move so the answer is $\frac{80}{2^{12}}=\boxed{\frac{5}{256}}$.