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The Berger tables for round-robin tournaments state:
Where there is an odd number of players, the highest number counts as a bye.
As far as I understand this statement correctly, this means that, for example, in an 8-player round-robin, the 8th seeded player will always be the bye player and hence will not play any game.
What else could this mean?
It means that in the pairings given in the tables, if there are an odd number of players (so not an 8-person round-robin for instance), the player playing against the highest number receives a bye.
For instance, for 5 players (use the 5-6 player table):
Its all documented here: https://www.fide.com/fide/handbook.html?id=184&view=article
Example: Where there is an odd number of players, the highest number counts as a bye.
9 or 10 players:
Rd 1: 1-10, 2-9, 3-8, 4-7, 5-6.
Rd 2: 10-6, 7-5, 8-4, 9-3, 1-2.
Rd 3: 2-10, 3-1, 4-9, 5-8, 6-7.
Rd 4: 10-7, 8-6, 9-5, 1-4, 2-3.
Rd 5: 3-10, 4-2, 5-1, 6-9, 7-8.
Rd 6: 10-8, 9-7, 1-6, 2-5, 3-4.
Rd 7: 4-10, 5-3, 6-2, 7-1, 8-9.
Rd 8: 10-9, 1-8, 2-7, 3-6, 4-5.
Rd 9: 5-10, 6-4, 7-3, 8-2, 9-1.
As result - who ever plays highest number - has bye, for example round 5 Nr3 gets bye as he plays against 10.
