I would like to know how the Buchholz tiebreak system works, and how it is calculated, taking into consideration other factors such as bye, no show (default win), etc.


The information is hidden away in Standards of Chess Equipment and tournament venue for FIDE Tournaments section 13.4.

13.4. Buchholz System
13.4.1. The Buchholz System is the sum of the scores of each of the opponents of a player.
13.4.2. The Median Buchholz is the Buchholz reduced by the highest and the lowest scores of the opponents.
13.4.3. The Median Buchholz 2 is the Buchholz score reduced by the two highest and the two lowest scores of the opponents.
13.4.4. The Buchholz Cut 1 is the Buchholz score reduced by the lowest score of the opponents.
13.4.5. The Buchholz Cut 2 is the Buchholz score reduced by the two lowest scores of the opponents.

Now it gets complicated.

how it is calculated, taking into consideration other factors such as bye, no show (default win)

Since these pose a problem for all tie breaks which take into account the performance of opponents, in 2012 FIDE introduced the concept of a "virtual opponent" to be used in these cases. The details are spelt out in section 13.14.2

13.14. Tie-Break Systems using Ratings in Individual Tournaments (where all the players are rated) ...
13.14.2. For tie-break purposes a player who has no opponent will be considered as having played against a virtual opponent who has the same number of points at the beginning of the round and who draws in all the following rounds. For the round itself the result by forfeit will be considered as a normal result.

This gives the formula:

Svon = SPR + (1 – SfPR) + 0.5 * (n – R)

where for player P who did not play in round R:
n = number of completed rounds
Svon = score of virtual opponent after round n
SPR = score of P before round R
SfPR = forfeit score of P in round R

Since I, like a lot of arbiters, use the Swiss Manager pairing program I though I'd see what the Swiss Manager help has to say about this.

4 Buchholz-Tie-Break Calculation On account of many requests about the Buchholz-Tie-Break I publish the following Excel file with suitable examples.

Some quite complicated examples are given there but with minimal explanation.

One last final twist which helped me understand the examples is this one regarding the score of a player which is going to be used in the Buchholtz calculations for other players.

If a player has received a pairing bye for which he has received one point (as opposed to an elective bye for which the player receives only half a point) then this is treated as an elective bye for Buchholtz calculations, so only counts as a half point bye.

So, if I play player A who has a total score of 3 and he had a one point bye (meaning he scored 2 points from playing games and 1 point from the bye) then when calculating my Buchholtz score I don't add 3 points for this opponewnt, I only add 2.5 points because his one point bye only counts as a half point bye.

  • Thanks for the prompt reply. I would like to see an example with calculations in order to understand. I would also like to know what happens when two players are tied but the other one has a soft point (bye or no show). Does that mean that the player with a soft point will be disadvantaged on the tiebreak? – Phemelo Khetho Jul 18 '19 at 14:04
  • 1
    @PhemeloKhetho OK, I've done more digging and found out some more about these cases which I've added to the answer. – Brian Towers Jul 19 '19 at 12:12

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