| 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 → | |
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.
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]
- ^ "2017/18 World Snooker Calendar" (PDF). World Professional Billiards and Snooker Association. 14 July 2017. Retrieved 14 July 2017.
- ^ "Indian Open: John Higgins beats Anthony McGill to take title". BBC Sport. 16 September 2017. Retrieved 5 February 2018.