Pang Junxu defeats Murphy

In the round of 32 to 16 at the 2026 Wuhan Snooker Open, Pang Junxu secured a place in the last 16 by defeating renowned player Shaun Murphy 5-3, with one break of 102 and five breaks over 50.

At the start of the match, both players were locked in a defensive battle, with the first frame alone lasting more than 40 minutes. Pang Junxu seized an opportunity and won the opening frame 75-65. Riding on his momentum, Pang took the next three frames in a row with breaks of 75, 94, and 62, reaching the interval with a commanding 4-0 lead.

After the break, in the fifth frame, Pang potted the last black ball but then the cue ball was potted, allowing Murphy to pull one back with a score of 54-52. In the sixth and seventh frames, Murphy achieved consecutive comeback wins, narrowing the gap to 3-4. However, in the eighth frame, Pang gave Murphy no more chances, producing a break of 102 to seal a 5-3 victory and advance to the last 16. In the next round, he will play against the winner of He Guoqiang versus Selby.