This might be hard to explain in text but I'm going to do my best.
The players place puzzle pieces on the Conspiracy portion of the board in an effort to connect one or more of the four sides of the center puzzle piece to the outer rim of the puzzle that connects to something that will earn them VP points for a certain condition they have or think they will meet during the game.
Lets say there are two paths starting from two different sides of the center puzzle piece that both lead to the same 'innocent hunches earn +2 VP per puzzle piece'. In order for this to happen these two paths that both start from different sides of the center conspiracy puzzle piece must at some point over lap and share a portion of the same path to the outer 'innocent hunches earn +2 VP per puzzle piece'.
Now, here comes my question.
When counting the puzzle pieces of each path do you count the puzzle pieces that are shared by both paths once only, or once per path?
Let's say path 1 is made up of 4 puzzle pieces to go from the center to the outer edge.
And path 2 takes 5 puzzle pieces to reach the same destination of 'innocent hunches earn +2 VP per puzzle piece'.
Both paths share the outer most puzzle piece. The piece that actually touches the outer edge of the puzzle board at 'innocent hunches earn +2 VP per puzzle piece'
Does this last piece that belongs to both paths get counted once or twice?
If it's counted once per path your score would look like:
Path 1 is 4 x 2 VP for 8 VP Points
Path 2 is 5 x 2 VP for 10 VP Points
Grand Total 18 VP Points for anyone who got their innocent hunch.
But if it counts only once your score would look like:
4 pieces for path 1 and 4 pieces for path 2. 4 for path 2 because the 5th piece in path 5 was already counted when adding up Path 1.
So, Path 1 -> 4 x 2 for 8 VP Points
And Path 2 -> 4x 2 for 8 VP Points
Grand Total is 16 VP Points.
I hope my question makes sense. This was hard to describe without being able to demonstrate on the board.