| 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]
Prize fund
The breakdown of prize money for this year is shown below:
|
|
The "rolling 147 prize" for a maximum break: £10,000
Main draw
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Final
| Final: Best of 9 frames. Referee: Colin Humphries Grand Hyatt Kochi Bolgatty, Kochi, India, 3 March 2019. | ||
| Matthew Selt England |
5–3 | Lyu Haotian China |
| 57–48, 89–6 (84), 0–115 (115), 12–78 (56), 21–72 (66), 102–0 (102), 67–49, 96–41 (72) | ||
| 102 | Highest break | 115 |
| 1 | Century breaks | 1 |
| 3 | 50+ breaks | 3 |
Qualifying
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.
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- Notes
Century breaks
Main stage centuries
There were a total of 33 century breaks made during the tournament.[8] Zhou Yuelong made a century in a held over match.
- 147, 102 – Zhou Yuelong
- 140 – Mark Davis
- 140 – Zhao Xintong
- 137 – Yuan Sijun
- 136 – Graeme Dott
- 133 – Andrew Higginson
- 132 – Scott Donaldson
- 130, 116, 109 – Li Hang
- 127, 113 – Anthony Hamilton
- 125 – Joe Perry
- 123 – Stuart Bingham
- 121, 104, 104 – John Higgins
- 120, 109, 101 – Luca Brecel
- 115, 114, 106 – Lyu Haotian
- 108 – Peter Ebdon
- 108 – Shaun Murphy
- 106, 103 – Hossein Vafaei
- 104 – Sam Craigie
- 103 – Sam Baird
- 103, 102 – Matthew Selt
- 102 – Michael Holt
Qualifying stage centuries
There were a total of 10 century breaks made during the qualifying matches preceding the event.[9]
- 135 – Hamza Akbar
- 122 – Rod Lawler
- 118 – Joe Swail
- 115 – Lyu Haotian
- 114 – Ross Muir
- 110 – Oliver Lines
- 102 – Gerard Greene
- 102 – Michael Holt
- 101 – Stuart Bingham
- 101 – Elliot Slessor
References
- ↑ "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.
- ↑ Årdalen, Hermund. "Indian Open (2019) - snooker.org". snooker.org (in Norwegian). Retrieved 27 March 2019.
- ↑ "Indian Open 2019 – Centuries". World Snooker. Archived from the original on 2 April 2019. Retrieved 11 December 2021.
- ↑ "Indian Open 2018 Qualifiers – Centuries". World Snooker. Archived from the original on 16 March 2019. Retrieved 11 December 2021.