Ankita Raina thread
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Re: Ankita Raina thread
At $60K ITF Women,Quanzhou
[R1] Ankita Raina (IND,194) vs. Ayano SHIMIZU (JPN,237)
Late last year, Shimizu had won 2 25K titles in Japan but this year she has not done much this year. Hope Ankita prevails.
Second match on court 3 tomorrow.
[R1] Ankita Raina (IND,194) vs. Ayano SHIMIZU (JPN,237)
Late last year, Shimizu had won 2 25K titles in Japan but this year she has not done much this year. Hope Ankita prevails.
Second match on court 3 tomorrow.
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Re: Ankita Raina thread
Ankita tied at 1 set all against the Japanese in her first round match. She lost the first set 4-6 but just won the second set 6-2.
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Re: Ankita Raina thread
And she gets the win in 3 sets.
At $60K ITF Women,Quanzhou
[R1] Ankita Raina (IND,194) d. Ayano SHIMIZU (JPN,237) 46 62 62
Plays the winner of 5th seed Naomi Broady and 195th ranked Imanishi in second round.
At $60K ITF Women,Quanzhou
[R1] Ankita Raina (IND,194) d. Ayano SHIMIZU (JPN,237) 46 62 62
Plays the winner of 5th seed Naomi Broady and 195th ranked Imanishi in second round.
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Re: Ankita Raina thread
Fantastic run continues for Ankita! Great. With that she reaches a other career high of 193 in the live rankings.
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Re: Ankita Raina thread
Terrific consistency in her performance ... Before singes Ankita is facing Broady in the doubles ... Ankita and Shimizu r facing 2nd seed Broady and Zvonareva.
At $60K ITF Women's Tournament 2018, Quanzhou, China
[R1] Ankita Raina (IND)/ Ayano Shimizu (JPN) vs (2) Naomi Broady (GBR)/ Vera Zvonareva (RUS)
At $60K ITF Women's Tournament 2018, Quanzhou, China
[R1] Ankita Raina (IND)/ Ayano Shimizu (JPN) vs (2) Naomi Broady (GBR)/ Vera Zvonareva (RUS)
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Re: Ankita Raina thread
That is a ridiculously difficult pair to be facing in R1 of doubles. Vera Zvonareva is the former world #2 (in singles) who made the final of both Wimbledon and the US Open in 2010. She is now on a nearly 1-year comeback trail after becoming a mother, and progress has been slow -- but she did win the doubles title at the St Petersberg Open this year (partnering Timea Baczynski). She also made R2 of the singles there, losing a close match to Ostapenko, so she is gradually getting back to reasonable form, although still suffering from a low ranking.
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Re: Ankita Raina thread
Someone on tennisforum pointed out how rigged the draw in these Chinese tournaments are, or appear to be; with a lot of Chinese players especially qualifiers, WC, LLs drawing each other.
(LL)Jia-Qi Kang (CHN) v (JE)Xinyu Wang (CHN)
Fangzhou Liu (CHN) v (Q)Kai-Lin Zhang (CHN)
(wc)Yue Yuan (CHN) v (Q)Xu Liu Sun (CHN)
(6)Saisai Zheng (CHN) v (wc)Hanyu Guo (CHN)
(Q)Wushuang Zheng (CHN) v Xinyun Han (CHN)
(LL)Jia-Qi Kang (CHN) v (JE)Xinyu Wang (CHN)
Fangzhou Liu (CHN) v (Q)Kai-Lin Zhang (CHN)
(wc)Yue Yuan (CHN) v (Q)Xu Liu Sun (CHN)
(6)Saisai Zheng (CHN) v (wc)Hanyu Guo (CHN)
(Q)Wushuang Zheng (CHN) v Xinyun Han (CHN)
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Re: Ankita Raina thread
16/32 players in the draw are Chinese. So 10 to them playing each other doesn't appear like an extreme event. I haven't done any calculations and so my comment is based on gut feeling.
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Re: Ankita Raina thread
Ankita has an off day. She won her doubles match without playing a point thanks to a walkover from the second seeds. Brody won her R1 match and so will play Ankita in R2. Zvonareva also won her R1 match and so I don't get why they didn't play doubles.
[R1] Ankita Raina (IND)/ Ayano Shimizu (JPN) d (2) Naomi Broady (GBR)/ Vera Zvonareva (RUS) w/o
[R1] Ankita Raina (IND)/ Ayano Shimizu (JPN) d (2) Naomi Broady (GBR)/ Vera Zvonareva (RUS) w/o
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Re: Ankita Raina thread
Agree with Suresh. Seems like you could get between 0 Chinese pairs (all 16 Chinese playing someone else) to 8 Chinese pairs (all Chinese playing each other). 5 is near the middle of that range even if it is slightly on the higher side...
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Re: Ankita Raina thread
The actual numbers are slightly lower because of seedings and only 2 of the Chinese being seeded.
For any of the non-seeded Chinese, the likelihood of drawing a Chinese player is 1/3 * 2/8 + 2/3 * 13/23 = 0.0833 + 0.3768 = 0.46; and for the 2 seeds, the likelihood of drawing a Chinese player is 14/24 = 0.58333.
So total expected value is (1.1667 + 0.46*14)/2 = 3.8
I am not inclined to do a montecarlo simulation to determine the standard deviation to check how likely a value of 5 is.
Just out of interest, I checked last week's event. Once again, 16 Chinese with 4 seeds. 5 all Chinese matches. This is right in line with expected value.
Overall, I agree that the claim on tennisforum appears to be unfounded.
For any of the non-seeded Chinese, the likelihood of drawing a Chinese player is 1/3 * 2/8 + 2/3 * 13/23 = 0.0833 + 0.3768 = 0.46; and for the 2 seeds, the likelihood of drawing a Chinese player is 14/24 = 0.58333.
So total expected value is (1.1667 + 0.46*14)/2 = 3.8
I am not inclined to do a montecarlo simulation to determine the standard deviation to check how likely a value of 5 is.
Just out of interest, I checked last week's event. Once again, 16 Chinese with 4 seeds. 5 all Chinese matches. This is right in line with expected value.
Overall, I agree that the claim on tennisforum appears to be unfounded.
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Re: Ankita Raina thread
Ankita off to an excellent start breaking Naomi twice in the first set, currently on serve at 4-0