2017 Indian Open
Snooker tournament
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The 2017 Indian Open was a professional ranking snooker tournament that took place between 12 and 16 September 2017 in Vishakhapatnam, India.[1] It was the fourth ranking event of the 2017/2018 season.
| Tournament information | |
|---|---|
| Dates | 12–16 September 2017 |
| Venue | Hotel Novotel Varun Beach |
| City | Vishakhapatnam |
| Country | India |
| Organisation | World Snooker |
| Format | Ranking event |
| Total prize fund | £323,000 |
| Winner's share | £50,000 |
| Highest break |  Zhou Yuelong (CHN) (141) |
| Final | |
| Champion |  John Higgins (SCO) |
| Runner-up | Anthony McGill (SCO) |
| Score | 5–1 |
← 2016 2019 → | |
Qualifying took place between 1 and 2 August 2017 in Preston, England.
Anthony McGill was the defending champion, having beaten Kyren Wilson 5–2 in the 2016 final. McGill reached the final again but was beaten by John Higgins, who won his 29th ranking event.[2]
Prize fund
The breakdown of prize money for this year is shown below:[3]
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The "rolling 147 prize" for a maximum break stood at £25,000.
Main draw
Final
| Final: Best of 9 frames. Referee: Terry Camilleri Hotel Novotel Varun Beach, Vishakhapatnam, India, 16 September 2017. | ||
| Anthony McGill Scotland |
1–5 | John Higgins Scotland |
| 2–70, 43–84, 78–28, 0–71 (71), 35–71, 1–104 | ||
| 49 | Highest break | 71 |
| 0 | Century breaks | 0 |
| 0 | 50+ breaks | 1 |
Qualifying
These matches were held between 1 and 2 August 2017 at the Preston Guild Hall in Preston, England. All matches were best of 7 frames.[4]
- Notes
Century breaks
Qualifying stage centuries
Total: 18
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Televised stage centuries
Total: 14
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