Who will win the FIDE Women's Candidates Tournament 2024?
13
140
640
resolved Apr 22
100%98.9%
Tan Zhongyi
0.7%
Lei Tingjie
0.1%
Kateryna Lagno
0.1%
Aleksandra Goryachkina
0.0%
Nurgyul Salimova
0.0%
Anna Muzychuk
0.0%
R Vaishali
0.1%
Koneru Humpy

The FIDE Women's Candidates Tournament 2024 will be an eight-player chess tournament held to determine the challenger for the Women's World Chess Championship 2025. It is scheduled to be held from 3 April to 22 April 2024 in Toronto, Canada, alongside the Candidates Tournament 2024.

The tournament is an eight-player, double round-robin tournament, meaning there are 14 rounds with each player facing the others twice: once with the black pieces and once with the white pieces. Players get 1 point for a win, ½ point for a draw and 0 points for a loss. The player with most points at the end of the tournament wins. The tournament winner will qualify to play Ju Wenjun for the World Championship in 2024.

In case of 2 or more players having the same score at the end of the tournament, tiebreaks will decide the winner of the tournament. Tiebreaks for the first place are addressed as follows:

  • Players would play two rapid chess games at 15 minutes plus 10 seconds per move. If a three- to six-way tie had occurred, a single round-robin would be played. If seven or eight players had been tied, a single round-robin would be played with a time limit of 10 minutes plus 5 seconds per move.

  • If any players had still been tied for first after the rapid chess games, they would play two blitz chess games at 3 minutes plus 2 seconds per move. In the case of more than two players being tied, a single round-robin would be played.

  • If any players were still tied for first after these blitz chess games, the remaining players would play a knock-out blitz tournament at the same time control. In each mini-match of the proposed knock-out tournament, the first player to win a game would win the mini-match.

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