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Treatment of unplayed games for Buchholz tie break: the FIDE rule

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

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Treatment of unplayed games for Buchholz tie break: the FIDE rule

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  1. Treatment of unplayed games for Buchholz tie break: the FIDE rule by Luigi Forlano www.vegachess.com

  2. 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 beneither 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.

  3. 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.

  4. 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. (a) Results of unplayed games

  5. In order to reducing the consequences for the opponents, each result by default of a player is counted as a draw. (b) Points got by A’s opponents 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.

  6. Draw for the next 6 rounds 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 The contribution of A for his opponents’ Buchholz is 5.5

  7. Draw for the next 2 rounds 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 The contribution of A for his opponents’ Buchholz is 6.5

  8. We must adjust all these results 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 -- First let’s consider all of them DRAW and calculate the new scores…

  9. 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 = +B2 +W4 4 DARIO 2.0 | +B8 =W1 -B9 =W5 -B3 5 FRANCESCO 1.5 | -B1 +W3 = =B4 -W6 6 GIORGIO 2.0 | -W2 -B7 == +B5 7 LUIGI 2.0 | -B3 +W6 == -W2 8 MARIO 2.0 | -W4 === -B1 9 ROBERTO (withdrawn) 3.5 | = =B2 +W4 +W1 = New Pts ----- 2.5 3.5 3.5 2.0 2.0 2.0 2.0 1.5 3.5 Now let’s calculate the score of the virtual opponent…

  10. 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 +2.0 +B2 +W4 4 DARIO 2.0 | +B8 =W1 -B9 =W5 -B3 5 FRANCESCO 1.5 | -B1 +W3 -3.0 =B4 -W6 6 GIORGIO 2.0 | -W2 -B7 +1.0-2.5 +B5 7 LUIGI 2.0 | -B3 +W6 -3.0+1.5 -W2 8 MARIO 2.0 | -W4 +1.5-3.0+1.5 -B1 9 ROBERTO (withdrawn) 3.5 | +2.0 =B2 +W4 +W1 -4.5 New Pts ------- 2.5 3.5 3.5 2.0 2.0 2.0 2.0 1.5 3.5 Buc Tot ------- 12.5 13.5 11.5 13.0 13.0 11.0 13.5 10.5 14.5 Now we can calculate the Buchholz as usual

  11. 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 +2.0 +B2 +W4 4 DARIO 2.0 | +B8 =W1 -B9 =W5 -B3 5 FRANCESCO 1.5 | -B1 +W3 -3.0 =B4 -W6 6 GIORGIO 2.0 | -W2 -B7 +1.0-2.5 +B5 7 LUIGI 2.0 | -B3 +W6 -3.0+1.5 -W2 8 MARIO 2.0 | -W4 +1.5-3.0+1.5 -B1 9 ROBERTO (withdrawn) 3.5 | +2.0 =B2 +W4 +W1 -4.5 New Pts ------- 2.5 3.5 3.5 2.0 2.0 2.0 2.0 1.5 3.5 Buc Tot ------- 12.5 13.5 11.5 13.0 13.0 11.0 13.5 10.5 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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