The final match of the FIDE World Cup Chess tournament was played between Praggnanandhan and Carlsen, which ended in a draw. In the final, 18-year-old Rameshbabu Praggnanandha will take on world number one Magnus Carlsen. There was a tough fight between the two. Now after both the matches are drawn, the tie-break will be decided on August 24 (today). Let’s see what the tiebreak rules say.
Tiebreak rules are as follows
- There are two classical games in the final of the FIDE World Cup chess tournament. If both games end in a draw, a tiebreak is decided.
- A tiebreak consists of two rounds of 25 minutes each. If both these bets also remain a draw, then the decision will be made by 10-10 bets. If a winner is still not decided after this, the game can be reduced to two games of 5 minutes each and then to games of 3 minutes each.
- Through the FIDE World Cup Chess Tournament, three players qualify for the Candidates Tournament, with Praggnananda qualifying for the Candidates by reaching the final of the World Cup Chess Tournament.
- There are a total of 8 players in the candidates, the winner of which will challenge next year’s World Cup champion Ding Liren of China and become the world champion if won.
Praggnanand’s journey to the finals had come
Let us tell you that Pragyanand defeated world number three Fabiano Caruana 3.5-2.5 in the semi-finals and made it to the final. Remarkably, Praggnanandhaa became only the second Indian to reach the final. Before him, the great Vishwanath Anand was successful in making it to the finals of the World Cup. Now it will be interesting to see whether he can win the final title in tiebreak or not.
