Indian Open snooker: Aditya Mehta defeats former world No. 1 Ebdon
Mehta asserted his supremacy with the day's highest break of 127 in the decisive seventh frame to clinch a 4-3 victory over his Scottish rival.
New Delhi: India's top player Aditya Mehta constructed two century breaks to defeat former world No. 1 Peter Ebdon in the opening round of the Indian Open World-ranking snooker tournament on Monday.
Mehta, ranked 72nd on the international roster, asserted his supremacy with the day's highest break of 127 in the decisive seventh frame to clinch a 4-3 victory over his fancied rival from Scotland.
Mehta came close to wrapping up the match in the sixth frame itself, but allowed Ebdon to bounce back through a late error.
Ebdon, whose erratic form in recent years has seen his world ranking slump to No. 30, was in a tight spot when trailing by 40 points with just two red balls remaining on the table. An error by Mehta allowed him to bounce back and level scores at three frames each.
The young Indian star then grabbed his chance with a 127-point break, surpassing his fourth frame score of 122, to secure a momentous 66-48, 69-51, 8-91, 122-1, 32-81, 73-85, 131-0 victory.
"Peter is one of the top players I've defeated," said Mehta, who led 2-0 and 3-1 on frames but was taken the full distance by Ebdon.
"He (Ebdon) is just phenomenal, very strong, and that showed in his comeback. It was a very difficult match. My safety play saved me today. I could have on 4-2," Mehta said.
"I have been in such situations many times in my career, where I have thrown away the lead and come back to win the match," he added.
Mehta began the contest on a tentative note and even conceded an early foul, but settled down to take the first frame 66-48.
Both players showed plenty of nerves in the second frame before some lucky breaks came Mehta's way and helped him take a 2-0 lead against Ebdon.