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Billiards: Hoang Sao impresses, 4 Vietnamese players enter knockout round of US tournament

Four Vietnamese players have qualified for the knockout round (64 players) at the Florida Open Pool Championship 2025 taking place in the US. Among them, Duong Quoc Hoang made a strong impression with a series of consecutive wins.

Báo Thanh niênBáo Thanh niên08/08/2025

The 2025 Florida Open Pool Championship takes place from August 5 to 10, bringing together 256 players from around the world . The world's top pool billiards players such as Fedor Gorst, Shane Van Boening, Jayson Shaw and current 9-ball world champion Carlo Biado are naturally present. Vietnam has 8 representatives participating in the tournament. Of which, Duong Quoc Hoang (nickname Hoang Sao) is the Vietnamese representative who performed most impressively in the qualifying round.

Hoang Sao won all 3 qualifying matches

Hoang Sao is the first Vietnamese player to qualify for the knockout round with a record of 3 wins. The player born in 1987 defeated Robert Dowling (USA, 9-4), John Souders (USA, 9-1) and Patric Gonzales (Philippines, 9-3).

Billiards: Hoàng Sao gây ấn tượng, 4 cơ thủ Việt Nam vào knock-out giải Mỹ - Ảnh 1.

Hoang Sao is the first Vietnamese player to win a ticket to the round of 64 of the Florida Open Pool Championship 2025, with a record of 3 wins in all matches.

PHOTO: WNT

Besides Hoang Sao, the round of 64 of the Florida Open 2025 also has the participation of 3 other Vietnamese players: Luong Duc Thien, Pham Phuong Nam and Le Quang Trung. Duc Thien, Phuong Nam and Quang Trung all received 1 defeat in the qualifying round and won tickets to the knockout when they passed the loser's bracket.

The matches of round 64 will take place from the evening of August 8. At 10:00 p.m. (Vietnam time): Luong Duc Thien meets Jesus Atencio (Venezuela), Hoang Sao faces Chang Yu Lung (Taiwan), Pham Phuong Nam faces Mustafa Alnar (Türkiye).

At 23:30 (Vietnam time): Le Quang Trung will face Alexander Kazakis (Greece).

Florida Open Pool Championship 2025 Prizes

The Florida Open Pool Championship 2025 has a total prize money of 200,000 USD (about 5.24 billion VND). Of which, the champion receives 40,000 USD (equivalent to 1 billion VND), the runner-up receives 16,000 USD (about 419 million VND), and the third-place winner receives 262 million VND each. In addition, players who reach the quarterfinals receive 7,000 USD, the round of 16 receive 4,000 USD, the round of 32 receive 2,000 USD, and the round of 64 receive 1,000 USD.

Source: https://thanhnien.vn/billiards-hoang-sao-gay-an-tuong-4-co-thu-viet-nam-vao-knock-out-giai-my-18525080811023706.htm


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