Suppose LC or AS win both of their last games. All games in the Quad are drawn. Now all 4 + one of the LC or AS have 6.5…Does the non quad player get to be co champ?
Also suppose 5 had 5 points and the 5th man left out won out and the quad 4 drew…that would mean the 5th man finishes with 7 and the quad 4 only 6.5…Who wins the championship?
It seems inherently unfair to have four players battle it out for the title when two of them will play with the White pieces twice, while the other two will only have one White out of three games. The only way to insure fairness would be to have a double round robin.
Michael Bacon
It almost is a double round robin. Of the six pairings in the final three rounds, five of them appeared in an earlier round with the colors reversed. Only the Shulman-Kamsky pairing in round 10 will be a pairing between players that didn’t meet in an earlier round.
And, interestingly enough, a decisive result in the Shulman-Kamsky pairing will determine the champion. The players with two whites each in the final quad (Nakamura and Onischuk) are out of the running (with Nakamura losing with white to Shulman and Onischuk losing with black to Kamsky). Shulman will finally get his only white in the quad.
It’s appropriate that the final round of the event brings the two leaders in a head to head match to decide it.