Carreno Busta, Querrey To Play Pivotal Matches In Paris
Carreno Busta, Querrey To Play Pivotal Matches In Paris
With two spots remaining at the Nitto ATP Finals, Pablo Carreno Busta and Sam Querrey begin their final campaign for qualification on Tuesday in Paris. Carreno Busta, who currently occupies the last London spot, faces French wild card Nicolas Mahut in one of three second-round matches. The Spaniard broke into the Top 8 of the Emirates ATP Race to London after defeating a record four qualifiers en route to the US Open semi-finals, one of whom was the 35-year-old Mahut.
Also in second-round action, Querrey meets Serbian qualifier Filip Krajinovic. If Querrey and Carreno Busta both win on Tuesday, they will meet in the third round with the Spaniard’s Race position on the line. If Querrey wins and Carreno Busta loses on Tuesday, they will be tied with 2,615 points. Carreno Busta owns the tiebreaker because he has earned more points than Querrey at Grand Slams and mandatory ATP World Tour Masters 1000 events.
View FedEx ATP Head2Head matchups for Day 2 and vote for who you think will win!
Mahut vs. Carreno Busta | Cuevas vs. Khachanov | Verdasco vs. Rublev
Nitto ATP Finals qualifier Dominic Thiem plays in the other second-round match against German lucky loser Peter Gojowczyk. Thiem is ranked a career-high No. 6, but he has yet to reach a semi-final off of clay courts this season. The Austrian has struggled throughout his career during the fall. Thiem is 12-20 lifetime in October and November with one quarter-final and no semi-final finishes.
The last eight matches of the first round are also on Tuesday’s schedule, including a pair of France vs. Spain showdowns. French wild card Pierre-Hugues Herbert faces Feliciano Lopez, who beat him 7-6(3), 6-4 in the 2016 Rolex Paris Masters first round. Then, Adrian Mannarino looks to end his sensational season on a high note against David Ferrer. Mannarino has earned 24 of his career-high 32 wins this year since June 25, when he began his run to the inaugural Antalya final.