2006 Stella Artois Championships – Doubles
From Wikipedia, the free encyclopedia
| Doubles | |
|---|---|
| 2006 Queen's Club Championships | |
| Final | |
| Champions | |
| Runners-up | |
| Score | 6–4, 3–6, [10–8] |
| Details | |
| Draw | 24 |
| Events | |
2006 tennis event results
Main article: 2006 Stella Artois Championships
Bob Bryan and Mike Bryan were the defending champions, but lost in the semifinals this year.
Paul Hanley and Kevin Ullyett won in the final 6–4, 3–6, [10–8], against Jonas Björkman and Max Mirnyi.
Seeds
[edit ]All seeds receive a bye into the second round.
- United States Bob Bryan / United States Mike Bryan (semifinals)
- Sweden Jonas Björkman / Belarus Max Mirnyi (final)
- The Bahamas Mark Knowles / Canada Daniel Nestor (semifinals)
- Australia Paul Hanley / Zimbabwe Kevin Ullyett (champions)
- Australia Stephen Huss / South Africa Wesley Moodie (second round)
- Czech Republic František Čermák / Czech Republic Leoš Friedl (quarterfinals)
- Zimbabwe Wayne Black / South Africa Jeff Coetzee (quarterfinals)
- Australia Wayne Arthurs / United States Justin Gimelstob (quarterfinals)
Draw
[edit ]Key
[edit ]- Q = Qualifier
- WC = Wild card
- LL = Lucky loser
- Alt = Alternate
- SE = Special exempt
- PR = Protected ranking
- ITF = ITF entry
- JE = Junior exempt
- w/o = Walkover
- r = Retired
- d = Defaulted
- SR = Special ranking
Finals
[edit ] Semifinals
Final
4
6
3
[10]
2
4
6
[8]
3
77
64
[7]
2
65
77
[10]
Top half
[edit ] First round
Second round
Quarterfinals
Semifinals
1
77
6
6
3
[10]
60
4
3
6
[6]
1
65
78
[10]
4
6
[10]
8
77
66
[7]
1
5
64
78
6
6
5
[6]
66
3
4
6
6
4
77
[11]
7
4
2
6
61
[9]
6r
Bottom half
[edit ] First round
Second round
Quarterfinals
Semifinals
3
6
[9]
3
7
6
6
2
[11]
77
66
[11]
3
64
78
[13]
3
77
64
[7]
2
65
77
[10]
5
4
65
5
65
WC
6
77
WC
7
77
WC
64
2
WC
6
77
WC
77
4
[6]
2
64
6
77