Quentin Halys is about to go ahead with his campaign in the draw in the Maia Challenger.
Draw Prediction with head to head records
This is Quentin Halys’s projected draw with the presumable opposition at the the Maia Challenger. Quentin has a a favorable record (2-0) with the plausible opponents.
R1 – Geoffrey Blancaneaux’s rankings and performance.
Head to head record 0-0 in Challenger and Futures
Geoffrey Blancaneaux has a 23-15 win-loss record in 2020, 5-7 on clay. During the last 5 years, Geoffrey has a 97-51 record on clay (Click here to see full career record). Blancaneaux won 3 of his last 10 matches.
The French player most significant result of this year was getting to the final in M25 Rancho Santa Fe and M15 Cancun.
Last year in the Maia Challenger Blancaneaux didn’t compete in this event.
H2H Quentin Halys vs. Geoffrey Blancaneaux
There is no head to head record between Quentin Halys and Geoffrey Blancaneaux since this would be the first time that they will fight against each other in the main tour.
R2 – Henri Laaksonen’s rankings and performance.
Head to head 2-0 for Quentin Halys
Henri Laaksonen has a 14-14 win-loss record in 2020, 7-5 on clay. In the course of the previous 5 years, the Swiss has a 164-138 record on clay (Click here to see full career record). The Swiss won 6 of his last 10 matches.
Last year in the Maia Challenger Laaksonen didn’t compete in this event.
H2H Quentin Halys vs. Henri Laaksonen
This match would represent the 3rd time that Quentin Halys and Henri Laaksonen square off. The head to head is 2-0 for Halys (see full H2H stats), but they have never competed against each other on clay at challenger or futures level.
However, Quentin Halys could face Zdenek Kolar. Their head to head record is 1-1.
1/4 – Paolo Lorenzi’s rankings and performance.
Head to head record 0-0 in Challenger and Futures
Paolo Lorenzi has a 7-14 record in 2020, 2-6 on clay. Regarding the last 5 years, Paolo has a 457-312 record on clay (Click here to see full career record). Paolo won 2 of his last 10 matches.
Last year in the Maia Challenger Lorenzi capitulated to Jozef Kovalik in the quarter 7-5 7-5.
H2H Quentin Halys vs. Paolo Lorenzi
There is no head to head record between Quentin Halys and Paolo Lorenzi since this would be the first time that they will play each other in the main tour.
On the other hand, Quentin Halys could play Bernabe Zapata Miralles. Their H2H is 0-0.
1/2 – Pedro Sousa’s rankings and performance.
Head to head record 0-0 in Challenger and Futures
Pedro Sousa has a 17-16 record in 2020, 16-13 on clay. As for the previous 5 years, the Portuguese has a 431-240 record on clay (Click here to see full career record). Pedro won 5 of his last 10 matches.
Sousa’s best result of this current season was getting to the final in Buenos Aires, the Split Challenger and the Lisbon Challenger.
Last year in the Maia Challenger Pedro lost to Riccardo Bonadio in the 3rd round 4-6 6-2 7-5.
H2H Quentin Halys vs. Pedro Sousa
There is no head to head record between Quentin Halys and Pedro Sousa since this would be the first time that they will face off in the main tour.
Nonetheless, Quentin Halys could compete against Nikola Milojevic. Their head to head record is 1-1.
Final – Pedro Martinez Portero’s rankings and performance.
Head to head record 0-0 in Challenger and Futures
Pedro Martinez Portero has a 35-12 win-loss record in 2020, 29-7 on clay. Regarding the last 5 years, Martinez Portero has a 260-124 record on clay (Click here to see full career record). Pedro will try to keep a positive momentum since he is on a 5 matches winning streak.
Martinez Portero’s most significant result of the year was reaching the final in the Alicante Challenger and the Marbella Challenger.
Last year in the Maia Challenger the Spaniard didn’t compete in this event.
H2H Quentin Halys vs. Pedro Martinez Portero
There is no head to head record between Quentin Halys and Pedro Martinez Portero since this would be the first time that they will face off in the main tour.
On the other hand, Quentin Halys could take on Jozef Kovalik. Their H2H is 0-0.
