Arabic version: فوز روبرتسون يربط رقم كروسابل للاعبين المصنفين

Neil Robertson of Australia triumphed over China’s Pang Junxu with a score of 10-6, achieving a significant milestone at the 2026 World Snooker Championship. This victory marks the equalization of a Crucible record, as 15 seeded players successfully advanced past the first round of the tournament. This is only the third instance in history where nearly all seeds progressed, with the last occurrence dating back to 1983.

According to BBC News, the only qualifier remaining in the competition is world number 32 Hossein Vafaei from Iran, who defeated China’s Si Jiahui 10-3 earlier in the day. Vafaei’s victory was notable, as he managed to win nine consecutive frames after trailing 3-1.

Robertson, who previously won the World Championship in 2010, expressed his surprise at the struggles experienced by the qualifiers, attributing it to a lack of experience among some debutants. He noted the absence of closely contested matches this year, suggesting that the competition has been dominated by higher-ranked players.

In his match against Pang, Robertson demonstrated strong performance, leading 5-4 overnight and finishing with impressive breaks of 77, 80, and 100 in the final frame. He will face Chris Wakelin in the last 16, a rematch of their previous encounter in the first round last year.

Meanwhile, 2005 champion Shaun Murphy took a commanding 6-2 lead against Xiao Guodong in their last-16 matchup. Murphy’s earlier round was a nail-biter that ended in a 10-9 victory over Fan Zhengyi, but he appeared more in control in this session. Northern Ireland’s Mark Allen also leads his match against Kyren Wilson, holding a 5-3 advantage as the tournament progresses.