We only care about the relative positions of A, B, and C. By symmetry, A will be in the middle position of these three letters in $1/3$ of the permutations.
If this seems too simple, consider arranging the letters A, B, C, D, E, F. Since they are distinct, this can be done in $6!$ ways. As for the favorable cases, we can place D in six ways, place E in five ways, and place F in four ways. Once we have done so, A must be placed in the middle open position. That leaves us with two ways to place B and one way to place C, which gives $6 \cdot 5 \cdot 4 \cdot 1 \cdot 2 \cdot 1$ favorable cases, so the probability that A is somewhere between B and C is
$$\frac{6 \cdot 5 \cdot 4 \cdot 1 \cdot 2 \cdot 1}{6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1} = \frac{1}{3}$$
which confirms the symmetry argument given above.