Arrow switching
The mathematical 'proof' that you arrow-switch just over one round in eight to get close to the ideal average level of competition between two pairs, playing in the same direction is well known.
https://www.ebu.co.uk/documents/media/bridge-movements-the-maths.pdf
However: what happens if you compare two pairs playing in opposite directions.
If T = tables, R = Rounds and Q = Rounds arrowswitched. (We will assume one board/ round and Single matchpoint/ board)
Given that the total matchpoints is R.(T-1) and the number of opposing pairs is (2T-1) we have.
Average competition = R.(T-1)/(2T-1) which is a bit less than R/2
However for our two pairs.
When they play each other (1 round) they are competing for (R-1) matchpoints.
On (R-1)-2Q rounds they are co-operating but for the 2Q rounds they are again competing - as they are now playing in the same direction. (2Q because each arrowswitched round affects two boards).
This gives us (R-1) + 2Q - (R-1)+ 2Q as the level of competition. The (R-1)s cancel leaving us with, unsurprisingly 4Q.
But this time 4Q is LESS than R/2 so you should arrowswitch less than one round in 8. The opposite.
Is this correct or is my argument (ignoring anomalous cases) flawed?