Ankita Raina thread

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Re: Ankita Raina thread

Post by sameerph » Wed Apr 25, 2018 4:51 am

Broady made a comeback of sorts but Ankita won the first set 6-3. :D

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Re: Ankita Raina thread

Post by sameerph » Wed Apr 25, 2018 5:21 am

Broady comes back strongly and bagles Ankita in the second set. :-(

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Re: Ankita Raina thread

Post by suresh » Wed Apr 25, 2018 6:10 am

Sin Hombre wrote:
Wed Apr 25, 2018 2:49 am
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
Ran a simulation with 1 million draws (around 2-3 times) taking into account that there were two Chinese seeds. I find that the mean number of pairs is 1.48 (mode=1) and the chances of 5 Chinese pairs is around 70-80 per million. So it is a rare event. Didn't get a single simulation where 6/7/8 pairs occurred. For 4 Chinese seeds, I find that the mean goes down to 1.44 but results are similar.

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Re: Ankita Raina thread

Post by suresh » Wed Apr 25, 2018 6:16 am

Ankita loses the 3rd set 2-6. She has her doubles match scheduled later in the day.

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Re: Ankita Raina thread

Post by jayakris » Wed Apr 25, 2018 6:49 am

suresh wrote:
Wed Apr 25, 2018 6:10 am
Ran a simulation with 1 million draws (around 2-3 times) taking into account that there were two Chinese seeds. I find that the mean number of pairs is 1.48 (mode=1) and the chances of 5 Chinese pairs is around 70-80 per million
So you mean there is a chance! (As Jim Carrey would say)

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Re: Ankita Raina thread

Post by suresh » Wed Apr 25, 2018 7:07 am

jayakris wrote:
Wed Apr 25, 2018 6:49 am
suresh wrote:
Wed Apr 25, 2018 6:10 am
Ran a simulation with 1 million draws (around 2-3 times) taking into account that there were two Chinese seeds. I find that the mean number of pairs is 1.48 (mode=1) and the chances of 5 Chinese pairs is around 70-80 per million
So you mean there is a chance! (As Jim Carrey would say)
Yes, especially as this has happened for two weeks in a row.

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Re: Ankita Raina thread

Post by PKBasu » Wed Apr 25, 2018 7:50 am

[R2] Ankita Raina (IND, 195) l. Naomi Broady (GBR, 134) 63 06 26

The second set bagle suggests Ankita has trouble sustaining a lead, or perhaps had stamina issues today. Naomi is a tall player, and probably has a big serve (she is ranked in the top-70 in doubles, and won a WTA doubles title recently).

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Re: Ankita Raina thread

Post by Sin Hombre » Wed Apr 25, 2018 11:59 pm

suresh wrote:
Wed Apr 25, 2018 6:10 am
Sin Hombre wrote:
Wed Apr 25, 2018 2:49 am
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
Ran a simulation with 1 million draws (around 2-3 times) taking into account that there were two Chinese seeds. I find that the mean number of pairs is 1.48 (mode=1) and the chances of 5 Chinese pairs is around 70-80 per million. So it is a rare event. Didn't get a single simulation where 6/7/8 pairs occurred. For 4 Chinese seeds, I find that the mean goes down to 1.44 but results are similar.
I find that mean to be low and would double check the code.

Just the average of 2 Chinese non-seeds drawing each other should be higher than that.

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Re: Ankita Raina thread

Post by suresh » Thu Apr 26, 2018 4:50 am

Sin Hombre wrote:
Wed Apr 25, 2018 11:59 pm

I find that mean to be low and would double check the code.

Just the average of 2 Chinese non-seeds drawing each other should be higher than that.
Yes, will do that!! My low mean surprised me as well but I did not verify your calculation either.

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Re: Ankita Raina thread

Post by suresh » Thu Apr 26, 2018 6:27 am

Sin Hombre wrote:
Wed Apr 25, 2018 11:59 pm

I find that mean to be low and would double check the code.

Just the average of 2 Chinese non-seeds drawing each other should be higher than that.
I found the mistake in my code. For some unknown reason my simulation was for 10 Chinese out of 32 and not 16! Now the mean is 3.8 and the probability of 5 pairs is 0.198 and so it is not an extreme event. Apologies for the mess up.

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Re: Ankita Raina thread

Post by bujilover » Thu Apr 26, 2018 11:26 pm

Ankita will be playing the Anning WTA125 next week. IIRC this will be the first time she has secured a direct entry to an WTA125 main draw without needing a wildcard. Let’s hope she draws motivation from our Indian players this week to continue next week :goodluck:

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Re: Ankita Raina thread

Post by Sin Hombre » Fri Apr 27, 2018 12:35 am

suresh wrote:
Thu Apr 26, 2018 6:27 am
Sin Hombre wrote:
Wed Apr 25, 2018 11:59 pm

I find that mean to be low and would double check the code.

Just the average of 2 Chinese non-seeds drawing each other should be higher than that.
I found the mistake in my code. For some unknown reason my simulation was for 10 Chinese out of 32 and not 16! Now the mean is 3.8 and the probability of 5 pairs is 0.198 and so it is not an extreme event. Apologies for the mess up.
Great, thanks for fixing.

Also good to know that I can still do basic probability :D

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Re: Ankita Raina thread

Post by sameerph » Sat Apr 28, 2018 3:38 pm

Ankita is finally included in TOPS scheme -

Ankita Raina included in TOPS

Some justice prevails at last.

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Re: Ankita Raina thread

Post by PKBasu » Sat Apr 28, 2018 4:32 pm

Justice at last, indeed! The India #1 woman has earned this the hard way. Well done, Ankita!!

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Re: Ankita Raina thread

Post by Omkara » Sat Apr 28, 2018 5:38 pm

Who all are there in the Tops scheme?

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