Andrey Rublev suffers shock exit to home favourite
Andrey Rublev ended his Indian Wells campaign at the hands of an American player ranked outside the top 50, his second such loss in 2021.
The Russian came into Indian Wells seeded fourth and with two Masters 1000 finals to his name in 2021, at Monte Carlo and Cincinnati.
His third round opponent was American world number 60 Tommy Paul. The pair had met three times previously, with Rublev triumphing on all three occasions.
However, this time Paul dug deep to overcome the 23-year-old 6-4, 3-6, 7-5 for the biggest win of his career. It is a second 2021 loss for Rublev to a player ranked outside the top 50. The last was Kei Nishikori, then ranked 69th, in the first round of the Tokyo Olympics.
It is a second ever top-10 win for Paul, having defeated Alexander Zverev at Acapulco 2020, when Zverev was ranked seventh. World number five Rublev represents a first top five victory for the American.
Paul defeated tour veteran Feliciano Lopez in the first round, before dispatching 28th seed Dusan Lajovic in the second.
Now, the 24-year-old will face 21st seed Cameron Norrie in the fourth round. The American had never previously been beyond the second round of a Masters 1000 event.
Paul is the second lowest ranked player left in the men’s draw at Indian Wells, with former world number five Kevin Anderson (ranked 69th) the lowest ranked.
Should Paul overcome a third seed consecutively and defeat Norrie, he would secure a new career-high of at least 49th in the world. His current career-high is 50th, achieved in June of this year.
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