Lei Peifan
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| Born | 31 May 2003 Ankang, Shaanxi, China |
|---|---|
| Sport country |  China |
| Professional | 2019–2023, 2024–present |
| Highest ranking | 29 (December 2025) |
| Current ranking | 30 (as of 6 April 2026) |
| Tournament wins | |
| Ranking | 1 |
Lei Peifan (Chinese: 雷佩凡; born 31 May 2003) is a Chinese professional snooker player.[1] While ranked 84th in the world, he won his first ranking event in December 2024, the 2024 Scottish Open, by defeating Wu Yize 9–5 in the final.[2]
In May 2019, Lei came through Q School on the overall Order of Merit to earn a two-year card on the World Snooker Tour for the 2019–20 and 2020–21 seasons.[3][4] He didn't retain his tour card after the end of the 2020–21 season but immediately gained a new two-year card for the 2021–22 and 2022–23 seasons by coming through Q School Event 3.[5][6] In March 2024, by beating Australian Vinnie Calabrese from 5–3 down to win 6–5, Lei won the 2024 APSBF Asia Pacific Open Snooker Championship to earn another two-year card on the World Snooker Tour, starting from the 2024–25 season.[7]
In December 2024, after never having advanced beyond the Last 16 of a professional tournament, Lei won the 2024 Scottish Open, defeating compatriot Wu Yize 9–5 in the final. It was only the fourth ever all-Chinese ranking event final and the second of the 2024–25 season (after the 2024 Wuhan Open). Ranked 84th in the world, he became the lowest-ranked player to win a ranking event since 93rd-ranked Dave Harold won the 1993 Asian Open.[2][8] On his way to the final, he defeated Shaun Murphy, Stuart Bingham, Tom Ford and Mark Allen, all in deciding frames.[9] He advanced to a career high ranking of 43 in the world due to this win.[2][8]
He defeated defending world champion Kyren Wilson in a deciding frame in the first round at the 2025 World Snooker Championship, before facing eventual champion Zhao Xintong in the second round.[10]
