| Tournament information | |
|---|---|
| Dates | 27 February – 3 March 2019 |
| Venue | Grand Hyatt Kochi Bolgatty |
| City | Kochi |
| Country | India |
| Organisation | World Snooker |
| Format | Ranking event |
| Total prize fund | £323,000 |
| Winner's share | £50,000 |
| Highest break |  Zhou Yuelong (CHN) (147) |
| Final | |
| Champion |  Matthew Selt (ENG) |
| Runner-up | Lyu Haotian (CHN) |
| Score | 5–3 |
← 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]
- ^ "2018/19 Calendar Amendments – New Indian Open Dates Confirmed". World Snooker. 3 September 2018. Retrieved 27 February 2019.
- ^ "Indian Open Draw and Preview - SnookerHQ". SnookerHQ. 26 February 2019. Retrieved 27 March 2019.
- ^ "Indian Open: John Higgins beats Anthony McGill to take title". BBC Sport. 16 September 2017. Retrieved 27 February 2019.
- ^ "Indian Open - World Snooker". World Snooker. Retrieved 27 March 2019.
- ^ "Indian Open: Matthew Selt ends 13-year wait, bags first world-ranking snooker title". Sportstar. Retrieved 27 March 2019.
- ^ "Zhou Joins 147 Club". worldsnooker.com. World Professional Billiards and Snooker Association. 28 February 2018.