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Treatment of unplayed games
for Buchholz tie break:
the FIDE rule
by Luigi Forlano
www.vegachess.com
Why we need adjustments
A Swiss tournament is a statistical system in which an equal,
limited number of games is played by each participant
Tiebreak systems based on the sum of opponents’ scores are
not fair to the players who did not play one or more of the
games – particularly when those players were not responsible
for the unplayed game!
Each unplayed game has consequence on the player himself
and on all of his opponents
Hence, it is fair to assign to each unplayed game a reasonable
result, to produce a standing not distorted by unfair side effects
caused by unplayed games
The new result should turn out to be neither an advantage nor a
punishment for those players whose standing have been
affected by the unplayed game, but just an adjustment to let
the tiebreak system work normally.
The FIDE rule for Buchholz
For tiebreak purposes, if a Player A does not
play at a given round it is necessary to
determine:
(a) the result of the unplayed game, and
(b) the tiebreak points got by his opponents
for the unplayed game.
(a) Results of unplayed games
The purpose is achieved by means of a virtual opponent V.
1. at the beginning of the round, V has as many points as
player A.
2. The result by default of player A is treated as a normal
result:
If player A is absent then V wins:
A–V=0–1
If player A wins (because of a BYE or due to the absence
of his opponent) then V loses:
A–V=1–0
If player A draws (1/2 point BYE) then V too draws:
A – V = 1/2 – 1/2;
For each subsequent round, V gains half a point.
(b) Points got by A’s opponents
In order to reducing the consequences for the
opponents, each result by default of a player is
counted as a draw.
This means that the total score of player A to be used by his
opponents for Buchholz calculation needs to be adjusted:
all the wins and losses by default should be
considered as draws.
Example 1
In a 9 round Swiss, player A achieves 6
points, including a default win in round
3.
After round 2, A had a 2 points score.
His virtual opponent gets:
2 + 0 + 6 x 0.5 = 5 points Buchholz
Draw for the next 6 rounds
The contribution of A for his opponents’
Buchholz is 5.5
Example 2
In a 9 round Swiss, player A
was absent in round 7 and scored 6
points after round 9.
After round 6, A had 4 points.
His virtual opponent get :
4 + 1 + 2 x 0.5 = 6 points Buchholz
Draw for the next 2 rounds
The contribution of A for his opponents’
Buchholz is 6.5
A complete case study
ID NAME
Pts |
1
2
3
4
5
------------------------------------------------------------------------1 ANDREA
2.5 | +W5
=B4
-W2
-B9
+W8
2 BRUNO
3.5 | +B6
=W9
+B1
-W3
+B7
3 CARLO
4.0 | +W7
-B5
+BYE
+B2
+W4
4 DARIO
2.0 | +B8
=W1
-B9
=W5
-B3
5 FRANCESCO
1.5 | -B1
+W3
-BYE
=B4
-W6
6 GIORGIO
2.0 | -W2
-B7
+BYE
-BYE
+B5
7 LUIGI
2.0 | -B3
+W6
-BYE
+BYE
-W2
8 MARIO
2.0 | -W4
+BYE
-BYE
+BYE
-B1
9 ROBERTO (withdrawn)
3.5 | +BYE
=B2
+W4
+W1
--
We must adjust all these results
First let’s consider all of them DRAW and calculate the new scores…
A complete case study
ID NAME
Pts |
1
2
3
4
5
-----------------------------------------------------------
New Pts
-----
1 ANDREA
2.5 | +W5
=B4
-W2
-B9
+W8
2.5
2 BRUNO
3.5 | +B6
=W9
+B1
-W3
+B7
3.5
3 CARLO
4.0 | +W7
-B5
=
+B2
+W4
3.5
4 DARIO
2.0 | +B8
=W1
-B9
=W5
-B3
2.0
5 FRANCESCO
1.5 | -B1
+W3
=
=B4
-W6
2.0
6 GIORGIO
2.0 | -W2
-B7
=
=
+B5
2.0
7 LUIGI
2.0 | -B3
+W6
=
=
-W2
2.0
8 MARIO
2.0 | -W4
=
=
=
-B1
1.5
9 ROBERTO (withdrawn)
3.5 |
=B2
+W4
+W1
=
3.5
=
Now let’s calculate the score of the virtual opponent…
A complete case study
ID NAME
Pts |
1
2
3
4
5
-----------------------------------------------------------
New Pts
Buc Tot
-------
-------
1 ANDREA
2.5 | +W5
=B4
-W2
-B9
+W8
2.5
12.5
2 BRUNO
3.5 | +B6
=W9
+B1
-W3
+B7
3.5
13.5
3 CARLO
4.0 | +W7
-B5
+2.0
+B2
+W4
3.5
11.5
4 DARIO
2.0 | +B8
=W1
-B9
=W5
-B3
2.0
13.0
5 FRANCESCO
1.5 | -B1
+W3
-3.0
=B4
-W6
2.0
13.0
6 GIORGIO
2.0 | -W2
-B7
+1.0
-2.5
+B5
2.0
11.0
7 LUIGI
2.0 | -B3
+W6
-3.0
+1.5
-W2
2.0
13.5
8 MARIO
2.0 | -W4
+1.5
-3.0
+1.5
-B1
1.5
10.5
9 ROBERTO (withdrawn)
3.5 | +2.0
=B2
+W4
+W1
-4.5
3.5
14.5
Now we can calculate the Buchholz as usual
A complete case study
ID NAME
Pts |
1
2
3
4
5
-----------------------------------------------------------
New Pts
Buc Tot
-------
-------
1 ANDREA
2.5 | +W5
=B4
-W2
-B9
+W8
2.5
12.5
2 BRUNO
3.5 | +B6
=W9
+B1
-W3
+B7
3.5
13.5
3 CARLO
4.0 | +W7
-B5
+2.0
+B2
+W4
3.5
11.5
4 DARIO
2.0 | +B8
=W1
-B9
=W5
-B3
2.0
13.0
5 FRANCESCO
1.5 | -B1
+W3
-3.0
=B4
-W6
2.0
13.0
6 GIORGIO
2.0 | -W2
-B7
+1.0
-2.5
+B5
2.0
11.0
7 LUIGI
2.0 | -B3
+W6
-3.0
+1.5
-W2
2.0
13.5
8 MARIO
2.0 | -W4
+1.5
-3.0
+1.5
-B1
1.5
10.5
9 ROBERTO (withdrawn)
3.5 | +2.0
=B2
+W4
+W1
-4.5
3.5
14.5
For example for player 7 that had two virtual opponents we have
3.5 + 2 + 3 + 1.5 + 3.5 = 13.5
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