2014 ATP World Tour Finals – Doubles
David Marrero and Fernando Verdasco were the defending champions, but did not qualify this year.
Bob and Mike Bryan won the title, defeating Ivan Dodig and Marcelo Melo in the final, 6–7(5–7), 6–2, [10–7].
Seeds
Alternates
Draw
Key
Finals
Group A
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Bryan
Bryan |
Peya
Soares |
Rojer
Tecău |
Kubot
Lindstedt |
RR W–L |
Set W–L |
Game W–L |
Standings |
1 |
Bob Bryan
Mike Bryan |
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7–6(7–3), 7–6(7–2) |
6–7(4–7), 6–3, [10–6] |
6–7(3–7), 3–6 |
2–1 |
4–3 (57.1%) |
36–35 (50.7%) |
2 |
3 |
Alexander Peya
Bruno Soares |
6–7(3–7), 6–7(2–7) |
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6–3, 3–6, [12–10] |
4–6, 6–3, [6–10] |
1–2 |
3–5 (37.5%) |
32–33 (49.2%) |
3 |
5 |
Jean-Julien Rojer
Horia Tecău |
7–6(7–4), 3–6, [6–10] |
3–6, 6–3, [10–12] |
|
4−6, 6−7(4−7) |
0–3 |
2–6 (25.0%) |
29–36 (44.6%) |
4 |
8 |
Łukasz Kubot
Robert Lindstedt |
7–6(7–3), 6–3 |
6–4, 3–6, [10–6] |
6−4, 7−6(7−4) |
|
3–0 |
6–1 (85.7%) |
36–29 (55.4%) |
1 |
Group B
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|
Nestor
Zimonjić |
Benneteau
Roger-Vasselin |
Granollers
López |
Dodig
Melo |
RR W–L |
Set W–L |
Game W–L |
Standings |
2 |
Daniel Nestor
Nenad Zimonjić |
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4–6, 7–5, [4–10] |
6–7(5–7), 6–3, [11–9] |
3–6, 5–7 |
1–2 |
3–5 (37.5%) |
32–35 (47.8%) |
3 |
4 |
Julien Benneteau
Édouard Roger-Vasselin |
6–4, 5–7, [10–4] |
|
4–6, 4–6 |
4–6, 6–2, [10–8] |
2–1 |
4–4 (50.0%) |
31–31 (50.0%) |
1 |
6 |
Marcel Granollers
Marc López |
7–6(7–5), 3–6, [9–11] |
6–4, 6–4 |
|
6–7(5–7), 6–7(12–14) |
1–2 |
3–4 (42.9%) |
34–35 (49.3%) |
4 |
7 |
Ivan Dodig
Marcelo Melo |
6–3, 7–5 |
6–4, 2–6, [8–10] |
7–6(7–5), 7–6(14–12) |
|
2–1 |
5–2 (71.4%) |
35–31 (53.0%) |
2 |
Standings are determined by: 1) Number of wins; 2) Number of matches; 3) In two-player ties, head-to-head records; 4) In three-player ties, percentage of sets won, or of games won; 5) Steering Committee decision.
References
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- Rotterdam (S, D)
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- Brisbane (S, D)
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