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• Ho Ray|Bu Yunchaokete x Christopher Patzanovsky/Frederic Schlossmann Statistik

Ho Ray|Bu Yunchaokete vs Christopher Patzanovsky/Frederic Schlossmann statistik 01.11.2022

• 01/11/22 09:00
• 

  Yunchaokete Bu/ Ray Ho
  • 2 : 0
  • Selesai
  

  Christopher Patzanovsky/ Frederic Schlossmann

T

1

2

Ho Ray|Bu Yunchaokete

2

6

6

Christopher Patzanovsky/Frederic Schlossmann

0

3

2

Ho Ray|Bu Yunchaokete

Н2Н

Christopher Patzanovsky/Frederic Schlossmann

Semua turnamen All matches

Tidak ada yang cocok

Ini adalah pertandingan pertama dari para pemain ini

Hasil terakhir: Ho Ray|Bu Yunchaokete

ATP - Doubles Shanghai 2

WP 06/10/25 06:40

Ho Ray|Bu Yunchaokete

• 7 ⁷
• 6 ⁴
• 8

1 Harri Heliovaara/Henry Patten

• 6 ³
• 7 ⁷
• 10

2

WP 03/10/25 05:10

Marcelo Melo/Alexander Zverev

• 6 ³
• 6
• 5

1 Ho Ray|Bu Yunchaokete

• 7 ⁷
• 4
• 10

2

ATP - Doubles Wimbledon 1

WP 03/07/25 06:10

Ho Ray|Bu Yunchaokete

• 6 ⁵
• 3

0 Marcel Granollers/Horacio Zeballos

• 7 ⁷
• 6

2

Challenger Men - Doubles Shenzhen 2 1

WP 11/10/23 03:50

Ho Ray|Bu Yunchaokete

• 6 ⁵
• 3

0 Piotr Matuszewski/Matthew Christopher Romios

• 7 ⁷
• 6

2

ITF Men - Doubles M25 Monastir 11 1

WP 03/12/22 08:00

Arthur Bouquier/Neil Oberleitner

• 6
• 6

2 Ho Ray|Bu Yunchaokete

• 1
• 4

0

Hasil terakhir: Christopher Patzanovsky/Frederic Schlossmann

ITF Men - Doubles M15 Überlingen 1

WP 13/08/24 13:10

Christopher Patzanovsky/Frederic Schlossmann

• 3
• 3

0 Maximilian Homberg/Neo Niedner

• 6
• 6

2

ITF Men - Doubles M25 Monastir 5 2

WP 07/09/23 10:00

Ostapenkov Daniil|van Wyk Kris

• 3
• 6
• 10

2 Christopher Patzanovsky/Frederic Schlossmann

• 6
• 3
• 6

1

WP 05/09/23 08:30

Christopher Patzanovsky/Frederic Schlossmann

• 6
• 6

2 Bird S./Loccisano B.

• 2
• 4

0

ITF Men - Doubles M15 Allershausen 1

WP 31/08/23 08:30

Christopher Patzanovsky/Frederic Schlossmann

• 4
• 3

0 Naw H./Wallner M.

• 6
• 6

2

ITF Men - Doubles M25 Monastir 9 1

WP 01/11/22 09:00

Ho Ray|Bu Yunchaokete

• 6
• 6

2 Christopher Patzanovsky/Frederic Schlossmann

• 3
• 2

0