Alex de Minaur of Australia is set to compete against Russia's Andrey Rublev in the quarterfinals of the ATP tournament in Beijing. As of now, de Minaur holds a singles ranking of 16, while Rublev is ranked 24 in the world.
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Both players have established themselves with impressive career statistics. De Minaur was born on February 17, 1999, and turned professional in 2015. He stands at 6'0" (183 cm) and plays right-handed. Rublev, born on October 20, 1997, is taller at 6'2" (188 cm) and also plays with his right hand. He embarked on his professional journey in 2014.
In terms of match statistics, de Minaur has a total of 1 ace and 7 double faults, while Rublev registered 2 aces and 6 double faults. De Minaur converted 68% of his first serves successfully, winning 67% of those points, whereas Rublev recorded a first serve success rate of 63%, winning 68% of the points he served. On the second serve, de Minaur won 44% of the points, while Rublev had a stronger showing with 75%.
When it comes to break points saved, de Minaur managed to save 64% of 11 break points he faced, while Rublev did slightly better, saving 75% of his break points. In terms of return statistics, de Minaur won 33% of first return points, compared to Rublev's 32%. The players have faced each other multiple times on the ATP circuit, with their most recent meeting taking place at the Australian Open on January 21, 2024.