Jeremy Gschwendtner vs Nemanja Malesevic: Where to Watch, Predictions, Stats

Bet nowGamble Responsibly 18+
Tennis ITF M15 Opatija Men Jeremy Gschwendtner - Nemanja Malesevic
Result
2:1
26/03/2025 at 08:45 (Time Zone – GMT-4 (America/New_York))

Additional information

  • ITF MEN - SINGLES: M15 Opatija (Croatia), clay - 1/16-finals
Jeremy Gschwendtner - Nemanja Malesevic - Gamble Responsibly 18+
00:00 / 90:00
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Match Stats

Last 10 Balls
55

Chances of winning


Jeremy Gschwendtner 50.8%Nemanja Malesevic 49.2%

Jeremy Gschwendtner - Nemanja Malesevic Odds

1 X 2 Marga
Avarage Now
(Initially)
1.81
(1.81)

1.87
(1.87)
9%
(9%)
Preview Facts
  • Gschwendtner is ranked ATP: 1124.
  • Malesevic could have problems with physics compare with it opponent - last days they played more matches than their opponents.
  • The chances to win of two players are almost the same in this match.
  • Recently, the players did not play each other.
Latest results of Nemanja Malesevic
Draw
2025

1/16-finals
1Mikrut L. (1)Raquillet L.2 : 0
2Gschwendtner J.Malesevic N.2 : 1
3Rybkin S.Simundza J.1 : 2
4Madarasz G.Obradovic D. (8)2 : 1
5Oberleitner N. (4)Fuele M.2 : 1
6Voljavec A.Perez Contri S.0 : 2
7Hallquist Lithen J.Battiston A.2 : 0
8Kucera J.Nedic A. (7)0 : 2
9Popovic S. (5)Milic O.1 : 2
10Ljubicic L.Kupcic J.0 : 2
11Ivanisevic E.Krivoshchekov M.2 : 0
12Basic N.Vrbensky M. (3)0 : 2
13Bielinskyi V. (6)Blokhin N.2 : 0
14Ivanovski K.Bonsach G. K.2 : 0
15Sorger S.Maksimovic M.2 : 0
16Mashtakov N.Kolar Z. (2)0 : 2

1/8-finals
1Mikrut L. (1)Gschwendtner J.2 : 0
2Simundza J.Madarasz G.2 : 1
3Oberleitner N. (4)Perez Contri S.0 : 2
4Hallquist Lithen J.Nedic A. (7)0 : 2
5Milic O.Kupcic J.2 : 0
6Ivanisevic E.Vrbensky M. (3)0 : 2
7Bielinskyi V. (6)Ivanovski K.1 : 2
8Sorger S.Kolar Z. (2)0 : 2

Quarter-finals
1Mikrut L. (1)Simundza J.2 : 0
2Perez Contri S.Nedic A. (RET.) (7)1 : 0
3Milic O.Vrbensky M. (3)0 : 2
4Ivanovski K.Kolar Z. (2)2 : 0

Semi-finals
1Mikrut L. (1)Perez Contri S.2 : 0
2Vrbensky M. (3)Ivanovski K.1 : 2

Final
1Mikrut L. (1)Ivanovski K.2 : 0
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