The final match of the FIDE World Cup Chess Tournament was played between Praggnanandhan and Carlsen, both of whom ended in a draw. In the final, 18-year-old Rameshbabu Praggnanandhaa meets world number one Magnus Carlsen. There was fierce competition between the two. Now that both games have been drawn, the decision will be made in the tie-break on August 24 (today). Let us know what the tiebreak rules say.
- There are two classical games in the final of the FIDE World Cup Chess Tournament. If both games end in a draw, the tie-break decides.
- In the tie-break there are two rounds of 25-25 minutes each. If both bets are also tied, the decision is made by 10-10 bets. If the winner is still not determined after this, the game may be extended to two games of 5-5 minutes and then to games of 3-3 minutes.
- The FIDE World Cup Chess Tournament qualifies three players for the Candidates Tournament, where Praggnanandhaa qualified for the Candidates Tournament by reaching the final of the World Cup Chess Tournament.
- There are 8 players in total among the candidates, the winner of which will challenge next year’s World Champion, China’s Ding Liren, and in case of victory will be crowned World Champion.
Please let me know that Praggnanandha defeated World No. 3 Fabiano Caruana 3.5:2.5 in the semifinals to advance to the final. Significantly, Pragyananand is only the second Indian to reach the final. Before him, the great player Vishwanath Anand had made it to the final of the World Championship. Now it will be interesting to see if Pragyananand can win the final title in the tiebreak or not.