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2007 Pattaya Women's Open – Doubles

Main article: 2007 Pattaya Women's Open

Li Ting and Sun Tiantian were the defending champions, but Li did not compete that year.

Sun partnering Ji Chunmei lost in the semifinals.

Seeds

  1. TPE Chan Yung-janTPE Chuang Chia-jung (final)
  2. AUS Nicole PrattITA Mara Santangelo (champions)
  3. UKR Yuliana FedakRUS Anastassia Rodionova (quarterfinals)
  4. SVK Jarmila GajdošováCRO Jelena Kostanić Tošić (semifinals)

Draw

Draw

| score-width=20 | font-size=85%

| RD1-team01=TPE Y-j ChanTPE C-j Chuang | RD1-seed01=1 | RD1-score01-1=6 | RD1-score01-2=6 | RD1-score01-3= | RD1-team02=SVK M SucháTHA N Tongsalee | RD1-seed02=WC | RD1-score02-1=1 | RD1-score02-2=1 | RD1-score02-3=

| RD1-team03=TPE S-w HsiehRUS A Kudryavtseva | RD1-seed03= | RD1-score03-1=4 | RD1-score03-2=6 | RD1-score03-3=[10] | RD1-team04=UKR K BondarenkoTHA T Tanasugarn | RD1-seed04= | RD1-score04-1=6 | RD1-score04-2=3 | RD1-score04-3=[12]

| RD1-team05=UKR Y FedakRUS A Rodionova | RD1-seed05=3 | RD1-score05-1=77 | RD1-score05-2=4 | RD1-score05-3=[10] | RD1-team06=TPE C-w ChanUSA C Gullickson | RD1-seed06= | RD1-score06-1=64 | RD1-score06-2=6 | RD1-score06-3=[8]

| RD1-team07=CHN C JiCHN T Sun | RD1-seed07= | RD1-score07-1=6 | RD1-score07-2=3 | RD1-score07-3= | RD1-team08=COL C CastañoCHN S Peng | RD1-seed08= | RD1-score08-1=1 | RD1-score08-2=1r | RD1-score08-3=

| RD1-team09=USA N UberoiIND S Uberoi | RD1-seed09= | RD1-score09-1=4 | RD1-score09-2=4 | RD1-score09-3= | RD1-team10=RUS V BardinaITA R Oprandi | RD1-seed10= | RD1-score10-1=6 | RD1-score10-2=6 | RD1-score10-3=

| RD1-team11=AUT S BammerHUN M Czink | RD1-seed11= | RD1-score11-1=0 | RD1-score11-2=66 | RD1-score11-3= | RD1-team12=SVK J GajdošováCRO J Kostanić Tošić | RD1-seed12=4 | RD1-score12-1=6 | RD1-score12-2=78 | RD1-score12-3=

| RD1-team13=CHN S SunCHN S Zhang | RD1-seed13= | RD1-score13-1=6 | RD1-score13-2=5 | RD1-score13-3=[11] | RD1-team14=UZB A AmanmuradovaSLO A Klepač | RD1-seed14= | RD1-score14-1=4 | RD1-score14-2=7 | RD1-score14-3=[9]

| RD1-team15=AUS A MolikAUS B Stewart | RD1-seed15= | RD1-score15-1=2 | RD1-score15-2=0 | RD1-score15-3= | RD1-team16=AUS N PrattITA M Santangelo | RD1-seed16=2 | RD1-score16-1=6 | RD1-score16-2=6 | RD1-score16-3=

| RD2-team01=TPE Y-j ChanTPE C-j Chuang | RD2-seed01=1 | RD2-score01-1=7 | RD2-score01-2=4 | RD2-score01-3=[10] | RD2-team02=UKR K BondarenkoTHA T Tanasugarn | RD2-seed02= | RD2-score02-1=5 | RD2-score02-2=6 | RD2-score02-3=[7]

| RD2-team03=UKR Y FedakRUS A Rodionova | RD2-seed03=3 | RD2-score03-1=2 | RD2-score03-2=1 | RD2-score03-3= | RD2-team04=CHN C JiCHN T Sun | RD2-seed04= | RD2-score04-1=6 | RD2-score04-2=6 | RD2-score04-3=

| RD2-team05=RUS V BardinaITA R Oprandi | RD2-seed05= | RD2-score05-1=3 | RD2-score05-2=6 | RD2-score05-3=[9] | RD2-team06=SVK J GajdošováCRO J Kostanić Tošić | RD2-seed06=4 | RD2-score06-1=6 | RD2-score06-2=3 | RD2-score06-3=[11]

| RD2-team07=CHN S SunCHN S Zhang | RD2-seed07= | RD2-score07-1=3 | RD2-score07-2=3 | RD2-score07-3= | RD2-team08=AUS N PrattITA M Santangelo | RD2-seed08=2 | RD2-score08-1=6 | RD2-score08-2=6 | RD2-score08-3=

| RD3-team01=TPE Y-j ChanTPE C-j Chuang | RD3-seed01=1 | RD3-score01-1=6 | RD3-score01-2=77 | RD3-score01-3= | RD3-team02=CHN C JiCHN T Sun | RD3-seed02= | RD3-score02-1=2 | RD3-score02-2=65 | RD3-score02-3=

| RD3-team03=SVK J GajdošováCRO J Kostanić Tošić | RD3-seed03=4 | RD3-score03-1=5 | RD3-score03-2=6 | RD3-score03-3=[4] | RD3-team04=AUS N PrattITA M Santangelo | RD3-seed04=2 | RD3-score04-1=7 | RD3-score04-2=3 | RD3-score04-3=[10]

| RD4-team01=TPE Y-j ChanTPE C-j Chuang | RD4-seed01=1 | RD4-score01-1=4 | RD4-score01-2=64 | RD4-score01-3= | RD4-team02=AUS N PrattITA M Santangelo | RD4-seed02=2 | RD4-score02-1=6 | RD4-score02-2=77 | RD4-score02-3=

Notes

  • The winners will receive $6,960 and 115 ranking points.
  • The runners-up will receive $3,740 and 80 ranking points.
  • The last direct acceptance team was Vasilisa Bardina and Romina Oprandi (combined ranking of 390st).
  • The player representative was Aiko Nakamura.
Info: Wikipedia Source

This article was imported from Wikipedia and is available under the Creative Commons Attribution-ShareAlike 4.0 License. Content has been adapted to SurfDoc format. Original contributors can be found on the article history page.

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