Am I the only one who thinks it's weird that a 3-way tie is solved by hth record among the 3 players as the first tiebreak? I think hth is a reasonable tiebreak for a 2-way tie, in which case player A beats player B because A has a positive record against B. But in the potential 3-way tie scenario in the Candidates', Caruana would beat Karjakin and Anand because he has a positive score against Anand (makes sense so far) ... and an equal score against Karjakin? Essentially Caruana would beat Karjakin because the former has a better score against Anand. It seems a bit odd to me... although you could say that the 3-way hth tiebreak is a natural continuation of the 2-way hth tiebreak, so... I don't know. Maybe tiebreaks are by nature arbitrary to a degree.
Possible Three-way Tie in the Candidates
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Hello there, I've been following the Candidates Tournaments this year and I realized something, currently Caruana and Karjakin are tied with 7 and a 1/2, while Anand has 7 going into the last round. Caruana faces Karjakin in the last round, what if they draw the game and Anand wins his game?
That would result in a three-way tie, but how would be solved? The first tiebreak is direct encounter, Karjakin-Caruana would be 1-1, 2 draws, Anand-Karjakin would aso be 1-1, a win each, but Caruana-Anand would be 1+1/2-1/2, so that means that Caruana would win the tournament?
If we use the second tiebreak, number of wins, to solve the other 2 draws we would have a situation where Caruana>Anand, Anand>Karjakin and Karjakin>Caruana. So what do you think, does Anand even has a chance to win the tournament or the tiebreaks give the win to Caruana, even tho Anand has more wins than anyone?
It would be funny if somehow Anand winning his last game gave Caruana the tournament.