2015 Wimbledon Championships – Women's Doubles Qualifying

Women's Doubles Qualifying
	2015 Wimbledon Championships

Seeds

Oksana Kalashnikova / Laura Siegemund (First round)
Johanna Larsson/ Petra Martić (Qualified)
Chan Chin-wei / Nicole Melichar (Qualifying competition, Lucky losers)
Magda Linette / Mandy Minella (Qualified)
Han Xinyun / Junri Namigata (First round)
Wang Yafan / Zhang Kailin (Qualified)
Barbora Krejčíková / Shahar Pe'er (First round)
Yuliya Beygelzimer / Anna Tatishvili (First round)

Qualifiers

Elizaveta Kulichkova / Evgeniya Rodina
Johanna Larsson / Petra Martić
Wang Yafan / Zhang Kailin
Magda Linette / Mandy Minella

Lucky losers

Qualifying draw

First qualifier

First Round					Qualifying Competition


1	Oksana Kalashnikova Laura Siegemund	3	6	2

	Misaki Doi Stephanie Vogt	6	1	6
						Misaki Doi Stephanie Vogt	6	3	5

						Elizaveta Kulichkova Evgeniya Rodina	4	6	7
	Elizaveta Kulichkova Evgeniya Rodina	6	4	6

5	Han Xinyun Junri Namigata	4	6	3

Second qualifier

First Round					Qualifying Competition


2	Johanna Larsson Petra Martić	5	6	6

WC	Tara Moore Nicola Slater	7	4	2
					2	Johanna Larsson Petra Martić	6	6

						Annika Beck Anna-Lena Friedsam	4	2
	Annika Beck Anna-Lena Friedsam	6	6

8	Yuliya Beygelzimer Anna Tatishvili	3	3

Third qualifier

First Round					Qualifying Competition


3	Chan Chin-wei Nicole Melichar	6	6

	Danka Kovinić Teliana Pereira	4	1
					3	Chan Chin-wei Nicole Melichar	5	4

					6	Wang Yafan Zhang Kailin	7	6
	Nicole Gibbs Eva Hrdinová	4	6	4

6	Wang Yafan Zhang Kailin	6	1	6

Fourth qualifier

First Round				Qualifying Competition


4	Magda Linette Mandy Minella	6	6

WC	Harriet Dart Katy Dunne	1	3
				4	Magda Linette Mandy Minella	2	6	7

					Jana Čepelová Stefanie Vögele	6	2	5
	Jana Čepelová Stefanie Vögele	6	6

7	Barbora Krejčíková Shahar Pe'er	2	3

gollark: For all (values of) f there exists a (value) g such that f (x, y) = (g x) y. In other words, you can convert any function which takes two values as a tuple or something to a curried one. I think.

gollark: I knew it would eventually be useful setting that as my status!

gollark: What part of ∀f ∃g (f (x,y) = (g x) y) did you not understand?

gollark: Also written as ∀, I think.

gollark: It's not for, it's forall, which is a bizarre thing Haskell people like.

External links

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