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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?

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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):

  • Round 1: #1 gets a bye
  • Round 2: #4 gets a bye
  • Round 3: #2 gets a bye
  • Round 4: #5 gets a bye
  • Round 5: #3 gets a bye
  • Well, that makes sense, thank you! – Wais Kamal May 24 at 8:22
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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.

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