The 2019 Indian Open was a professional snooker tournament. It was due to take place between 18 and 22 September 2018 at the Grand Hyatt Kochi Bolgatty in Kochi, India but was postponed due to the 2018 Kerala floods. The rescheduled Indian Open was played in Kochi from 27 February to 3 March 2019.[1] It was the fifteenth ranking event of the 2018/2019 season.[2]
Qualifying took place on 15 and 16 August 2018 in Preston, England.
Selt went on to win his first ranking title, beating Lyu Haotian 5–3 in the final.[5]
Zhou Yuelong made the first maximum break of his career in the fourth frame of his first round loss to Lyu Haotian. It was the 150th maximum in professional events.[6]
Prize fund
The breakdown of prize money for this year is shown below:
