Pierre Dénommée
12-15-2018, 02:50 PM
The FIDE Technical Commission has decided to reintroduce the Cumulative tiebreak in competitive chess. The date of this reintroduction is not specified. Formerly, this tiebreak has been removed based on this argument: if two players did met the same opponents and got the same results against each opponent, the cumulative tiebreak would declare a winner in a situation in which, clearly, both players are indeed equal.
This argument was lame because under normal Swiss Pairing rules, the probability of such case is infinitesimal.
Example
Player 1
Round 1 Opponent number 45 result 1
Round 2 Opponent number 25 result 1
Round 3 Opponent number 15 result 1
Round 4 Opponent number 10 result 0
Round 5 Opponent number 2 result draw
Player 2
Round 1 Opponent number 10 result 0
Round 2 Opponent number 15 result 1
Round 3 Opponent number 25 result 1
Round 4 Opponent number 45 result 1
Round 5 Opponent number 1 result draw
How can you explain the first round pairing 2-10 when it should have been 2-46 ? The only possible explanation is that player 46 and the opponent of player 10 did not show within the default time and that those players have been paired together after the default time.
In the general case, having the same opponents and the same results against each opponent while complying with all Swiss pairing rules will occur rarely. No tiebreak system is perfect and apart from the cumulative, none has been withdrawn based on a single hypothetical case.
8.3
(3) In an L2, L3 or L4 tournament: If, after the round has started two players do not have a game, then they can be paired against each other. This is only allowed when the arbiter and both players agree and they have not already played in this tournament. The arbiter shall adjust the clock times in an equitable manner
http://www.fide.com/images/stories/NEWS_2018/FIDE_NEWS/Congress/Annexes/Annex_41.pdf
This argument was lame because under normal Swiss Pairing rules, the probability of such case is infinitesimal.
Example
Player 1
Round 1 Opponent number 45 result 1
Round 2 Opponent number 25 result 1
Round 3 Opponent number 15 result 1
Round 4 Opponent number 10 result 0
Round 5 Opponent number 2 result draw
Player 2
Round 1 Opponent number 10 result 0
Round 2 Opponent number 15 result 1
Round 3 Opponent number 25 result 1
Round 4 Opponent number 45 result 1
Round 5 Opponent number 1 result draw
How can you explain the first round pairing 2-10 when it should have been 2-46 ? The only possible explanation is that player 46 and the opponent of player 10 did not show within the default time and that those players have been paired together after the default time.
In the general case, having the same opponents and the same results against each opponent while complying with all Swiss pairing rules will occur rarely. No tiebreak system is perfect and apart from the cumulative, none has been withdrawn based on a single hypothetical case.
8.3
(3) In an L2, L3 or L4 tournament: If, after the round has started two players do not have a game, then they can be paired against each other. This is only allowed when the arbiter and both players agree and they have not already played in this tournament. The arbiter shall adjust the clock times in an equitable manner
http://www.fide.com/images/stories/NEWS_2018/FIDE_NEWS/Congress/Annexes/Annex_41.pdf