Peers/Kontinen Face Herbert/Mahut In Australian Open 2019 Doubles Finals | ATP Tour

John Peers and Henri Kontinen will look to capture their second Australian Open doubles crown on Saturday when they face Pierre-Hugues Herbert and Nicolas Mahut in the final. Herbert and Mahut will be attempting to become the eighth team in the sport’s history to complete the career Grand Slam of doubles titles.

Kontinen and Peers, the No. 12 seeds and 2017 champions, knocked out Leonardo Mayer and Joao Sousa 6-1, 7-6(8) in 87 minutes. They will now look to improve upon their 13-1 record in tour-level finals, two years on from beating Bob Bryan and Mike Bryan in the final at Melbourne Park.

The Finnish-Australian team broke at 3-3 in the second set, but were unable to convert one match point at 5-3, with Mayer serving at 30/40, prior to their opponents breaking the Peers serve in the next game. Kontinen and Peers got off to a 4/1 lead in the tie-break, but then lost four straight points. They finally sealed victory on their third match point.

Later in the day, fifth-seeded Frenchmen and 2015 finalists Herbert and Mahut cruised past Americans Ryan Harrison and Sam Querrey 6-4, 6-2 in 70 minutes. After a nervous start, featuring three straight service breaks, Herbert and Mahut dominated and broke the serves of Harrison and Querrey in the fifth and seventh games of the second set.

The Frenchmen will be bidding to become the eighth men’s team in history to complete the set of four major championship doubles titles on Saturday, having won the 2015 US Open (Murray/Soares), 2016 Wimbledon (d. Benneteau/Roger-Vasselin) and last year’s Roland Garros (d. Marach/Pavic).

Herbert and Mahut, who are 13-6 lifetime in tour-level finals, are hoping to follow in the footsteps of Frank Sedgman and Ken McGregor, Ken Rosewall and Lew Hoad, Neale Fraser and Roy Emerson, John Newcombe and Tony Roche, Jacco Eltingh and Paul Haarhuis, Mark Woodforde and Todd Woodbridge, and also the Bryan brothers.

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