There are $n\geq 3$ cities in a country and between any two cities $A$ and $B$, there is either a one way road from $A$ to $B$, or a one way road from $B$ to $A$ (but never both). Assume the roads are built such that it is possible to get from any city to any other city through these roads, and define $d(A,B)$ to be the minimum number of roads you must go through to go from city $A$ to $B$. Consider all possible ways to build the roads. Find the minimum possible average value of $d(A,B)$ over all possible ordered pairs of distinct cities in the country.
1. **Graph Construction for Odd \( n \):**
   - Arrange the \( n \) vertices in a circle.
   - Orient the edges clockwise.
   - From each vertex, orient the edges to the rest of the vertices in an alternating manner.
   - For any vertex \( A \), half of the other vertices are reachable with a path of length 1, and the other half with a path of length 2.
   - Calculation of \( d(G) \):
     \[
     d(G) = \sum_{A \neq B} d(A, B) = 3 \binom{n}{2}
     \]
     This is because for each pair of vertices, the sum of the distances in both directions is 3.

2. **Graph Construction for Even \( n \):**
   - Construct the graph for \( n-1 \) vertices as described above.
   - Add a new vertex and connect it to two consecutive vertices in the \( n-1 \) polygon.
   - Ensure that the new vertex has paths of length at most 2 to all other vertices.
   - This construction ensures that the sum of distances \( d(G) \) remains optimal.

3. **Optimality Proof:**
   - Consider any pair of vertices \( (A, B) \). There will be a path from \( A \) to \( B \) of length 1 and a path from \( B \) to \( A \) of length at least 2.
   - Therefore, the total sum of distances for all pairs is at least \( 3 \binom{n}{2} \).
   - The average distance is then:
     \[
     \frac{d(G)}{\binom{n}{2}} = \frac{3 \binom{n}{2}}{\binom{n}{2}} = 3/2
     \]

4. **Special Case for \( n = 4 \):**
   - For \( n = 4 \), the construction does not hold as one vertex will not have a path to the new one.
   - This case needs to be proven separately to have a larger average distance.

\(\blacksquare\)

The final answer is \( \boxed{ \frac{3}{2} } \)