FIDE Chess World Cup 2025 in India from October 30

India will host this year's chess World Cup from October 30 to November 27 with a host city for the event to be named in "due course", the sport's global governing body FIDE announced on Monday.

The tournament that will see 206 players battling for the coveted title and qualification spots in the 2026 FIDE Candidates Tournament. India last hosted the event in 2002 in Hyderabad where Viswanathan Anand won the title.

The players will compete in a knockout format, where the losing player in each round is eliminated.


"The event has used several formats over the years, but since 2021, it has followed a single-elimination format. Each round spans three days: two classical games on the first two days, followed by tie-breaks on the third day, if necessary," FIDE stated.

In the first round, the top 50 players receive byes, while players seeded from 51 to 206 compete, with pairings based on the principle of the top half versus the reversed lower half.

"The top three finishers of the 2025 World Cup will earn direct qualification for the 2026 Candidates Tournament, which determines the challenger for the World Chess Championship," FIDE said.

Reigning world champion D Gukesh, R Praggnanandhaa, the runner-up of the 2023 World Cup, and Arjun Erigaisi, currently ranked world number five, are among the stars who would be seen in action. World number one Magnus Carlsen has also qualified to compete in the biennial event that he won in 2023.

India has recently hosted major events, including the 2022 Chess Olympiad, Tata Steel Chess India, the 2024 World Junior U20 Championships 2024, and the 5th leg of the FIDE Women's Grand Prix in April this year.

"We are thrilled to bring the FIDE World Cup 2025 to India, a country with a deep-rooted passion and support for chess. Indian chess fans' enthusiasm has always been remarkable, and we anticipate great interest in the event among local chess lovers, both on-site and online," FIDE CEO Emil Sutovsky said in a press release. PTI