Snooker - Robertson to face Higgins in Wuxi final

Neil Robertson avenged his defeat to Robert Milkins at the World Championships to move into the Wuxi Classic final.

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Neil Robertson

Robertson suffered a surprise 10-8 first round defeat to the Englishman in Sheffield in April but always looked in control in the Chinese ranking event during a comfortable 6-2 victory.

Robertson won a long first frame on the back of some outstanding tactical play and breaks of 71 and 51 in the second saw him go 2-0 in front.

A run of 59 put him in front again in the third but he did let Milkins back in with a chance to nick it. However, the Englishman broke down on 23 and Robertson finished the job.

A break of 77 then made sure he had a 4-0 lead at the interval and it looked like being a short day at the office for the former world champion.

Milkins did win the next two frames though to make a game of it, with a 72 in the sixth frame reducing the deficit to 4-2.

However, a couple of mini-breaks got Robertson back on track in the next and the victory was secured with a match best break of 92.

Robertson will play John Higgins after the four-time world champion's 6-2 win over Matthew Stevens in the second semi-final.

The Scot blasted into a 5-0 lead with four 50+ breaks in a row after taking the opener on the final colours.

But when Higgins needed just one more to seal victory, Stevens mounted a small resistance with breaks of 55 and 70.

A 101 clearance by Higgins in the eighth was the only century of the match and finally sealed his place in the final where Robertson awaits.

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