| Tournament Name: | 2019 Indian Open |
| Venue: | Grand Hyatt Kochi Bolgatty |
| Location: | Kochi |
| Country: | India |
| Organisation: | World Snooker |
| Format: | Ranking event |
| Total Prize Fund: | £323,000 |
| Winners Share: | £50,000 |
| Highest Break: | (147) |
| Score: | 5–3 |
| Previous: | 2017 |
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.
John Higgins was the defending champion, having beaten Anthony McGill 5–1 in the 2017 final,[3] but he lost to Matthew Selt in the semi-finals.[4]
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]
The breakdown of prize money for this year is shown below:
The "rolling 147 prize" for a maximum break: £10,000
| Final: Best of 9 frames. Referee: Colin Humphries Grand Hyatt Kochi Bolgatty, Kochi, India, 3 March 2019. | |||
| Matthew Selt | 5–3 | Lyu Haotian | |
| 57–48, 89–6,,,,,, | |||
| 102 | Highest break | 115 | |
| 1 | Century breaks | 1 | |
These matches were held between 15 and 16 August 2018 at the Preston Guild Hall in Preston, England. All matches were best of 7 frames.
There were a total of 33 century breaks made during the tournament.[9] Zhou Yuelong made a century in a held over match.
There were a total of 10 century breaks made during the qualifying matches preceding the event.[10]
Asutosh Padhy
