Goa has been announced the official host for the FIDE World Cup 2025. This knockout tournament will commence in Goa on October 30 and will end on November 27. The FIDE World Cup 2025 has marked its return to India after 23 years. It was last held in 2002 when the legendary Viswanathan Anand had won the title while Uzbekistan’s Rustam Kasimdzhanov finished as the runner-up.
Details about the FIDE World Cup 2025
A total of 206 players will take part in FIDE World Cup 2025. The participants will leave no-stone-unturned to clinch a prize fund of USD 2,000,000 and the three spots in the 2026 candidates tournament. The World Cup is known for its win-or-go-home format. It will be an eight round knockout tournament and the top 50 seeds will be entering from round 2. The players will get to play two classical games in each match, a rapid and blitz playoff in the event of a tie.
Why GOA was selected for the FIDE World Cup 2025?
According to a release, FIDE has stated that Goa was chosen because of its blend of natural beauty and cultural appeal. According to FIDE, the combination of these factors offer the fans and players a vibrant experience in the high-stakes competition. The release added that India’s recent rise as a chess powerhouse has added to the case for staging the world cup in the country.
The release said, “Indian chess has reached new heights in the past year with D Gukesh becoming the world champion, the national teams triumphing at the Chess Olympiad in both the Open and Women’s categories, and teenager Divya Deshmukh claiming the women’s world cup in July.”
The release also stated that India has become one of the strongest chess nations, with outstanding players and passionate fans. As per the release, post the success of the FIDE Women’s World Cup held in Georgia earlier this year, the International Chess Federation is proud to bring the FIDE World Cup to Goa.
