Taylor Fritz vs Holger Rune – Breakdown and Prediction

Susan Mullane-USA TODAY Sports/Sipa USA

Taylor Fritz vs Holger Rune: ATP Shanghai – Round of 16

The highest-ranked American player Taylor Fritz will take on Holger Rune in the round of 16 of the ATP Shanghai.

 

Taylor Fritz Profile

Taylor Fritz is currently seventh in the ATP rankings, two spots lower than his career high. The American is 47-19 for the season and 20-9 on hard surface.

He is coming off a 6-3, 6-4 win over Yosuke Watanuki in the previous round after beating Terence Atmane in two tie breaks in the opener.

A couple of weeks back, Fritz made it to his first ever Grand Slam final but lost to Jannik Sinner at the US Open, 6-3, 6-4, 7-5.

Fritz won the titles in Eastbourne and Delray Beach this season.

In 2022, he won the ATP Tokyo defeating Frances Tiafoe in the final.

 

Holger Rune Profile

Holger Rune is 14th in the ATP rankings, ten spots lower than his career high. The Danish is 39-20 for the season and 22-10 on hard surface.

Rune is coming off a 6-4, 7-5 win over Jiri Lehecka in the third round after beating Matteo Berrettini in the second.

His best result of the season was the final in Brisbane where he lost to Grigor Dimitrov.

Last year in Shanghai, Rune lost in the second round to Brandon Nakashima.

 

Head-to-Head Tied 1-1

  • Rune vs Fritz 2-6, 7-6, 6-3 (2024 ATP Indian Wells, round of 16)
  • Fritz vs Rune 6-3, 6-4 (2023 ATP Miami, round of 16)

 

Prediction and Conclusion

Taylor Fritz is playing on top level and after a while is back in the top 10 of the rankings, and now Holger Rune wants to take the same road. The Danish has been struggling all year long, but it seems like he is back in full form in Shanghai. Rune beat Fritz earlier this year, which proves that he can be competitive in this match. Fritz’s main advantage is his serve on this surface, which will put him in a situation to stay on the baseline since Rune is one of the fastest players on the tour. This will be a tough fight, so we go with over 23.5 games in total.

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