India's 18-year-old Rameshbabu Praggnanandhaa is competing with World No.1 Magnus Carlsen in the tie-breaks of the FIDE Chess World Cup final in Baku, Azerbaijan on Thursday, August 24. The event has a $1,834,000 prize fund (Rs 15.14 crore), with $110,000 (Rs 90.83 lakh) going to the winner. The runner-up earns $ 80,000 (Rs 66.06 lakh).
The tiebreaker is taking place on Thursday after the first two games between Praggnanandhaa and Carlsen ended in a draw. The players will play shorter, fast-paced games in the tiebreaker to decide who wins the World Cup. If he wins, Carlsen will claim his first World Cup title while Prggnanandhaa is hoping to become the youngest winner of the World Cup by claiming his first title.
The tiebreaker will involve rapid chess games with the players having 25 minutes each with an increment of 10 seconds per move in the first two games. In case the tie is still not resolved, they will play two more games with shorter time control - 10 minutes for each player with 10 seconds added to the clock after every move. If the tie persists, they will next play two more tiebreak games with five minutes for each player with three seconds added to their clock after each move. In case, the issue is not resolved even then, they will play unlimited games with each player with three minutes each on the clock with two seconds added to their respective clock after every move till one of them wins the mini-match.
(With IANS inputs)