Stan Moody Eliminates John Higgins in British Open

Stan Moody, a 19-year-old snooker player, achieved a notable victory by defeating four-time champion John Higgins 4-2 in the last 16 of the British Open held in Cheltenham.

Moody took an early lead, going up 1-0 and then 2-1, but Higgins managed to equalize both times. A break of 46 allowed Moody to pull ahead 3-2, and he sealed the match with a clearance of 69 in the sixth frame, despite Higgins initially leading 61-0.

With this win, Moody advances to the quarter-finals for the second consecutive year, where he will face fellow English player Liam Highfield. Other matches saw Judd Trump winning 4-0 against David Gilbert, while Jack Lisowski defeated Lyu Haotian 4-1.