New Delhi: World No.1 Neil Robertson suffered a shock defeat against China's Ding Junhui to crash out of the Indian Open World Ranking snooker tournament on Thursday.
Robertson did not look in his elements and lost 2-4 to World No.4 Junhui in the first quarter-final of the 300,000 pound event.
It was Junhui who set the ball rolling with a break of 72 to pocket the first frame 96-4. He went on to win the second frame with a 52-clearance to take the 2-0 lead.
Robertson regrouped his resources quickly and reigned supreme on the green baize with fluent breaks of 86 and 104 in the next two frames to level the issue at 2-2.
When it looked like Robertson would walk away with the honours, Junhui turned the game on its head and came up with breaks of 85 and 91 in the fifth and sixth frames respectively to cap off an impressive win over his much-fancied opponent.
Talking about the match, Junhui, who had defeated Robertson in the Shanghai Open recently, said, "I played well today. I took my chances against him and came up with some good clearances.
"The conditions are good here. The tables are not at all sticky. I am looking forward to my semi-final tomorrow," Junhui added.
In the other quarter-final clashes, Robbie Williams beat Anthony McGill 4-0, while Stephen McGuire won a hard-fought battle against Michael White 4-3.
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