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I am just curious to know. For acceptance rate calculation, does it consider deleted questions as unaccepted? And what about close questions?

This question is so simple yet the site shows that it doesn't meet our quality standards. (I just typed the last sentence to make it meet the quality standards.)

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    $\begingroup$ See this and this. $\endgroup$ – J. M. is a poor mathematician Jul 25 '11 at 8:34
  • $\begingroup$ @J.M. : I couldn't understand the second link. $\endgroup$ – Rajesh Dachiraju Jul 25 '11 at 8:40
  • $\begingroup$ @Rajesh: It is generally a good idea, when asking for clarification, to try your best to pinpoint what exactly it is that you do not understand. $\endgroup$ – Willie Wong Jul 25 '11 at 23:34
  • $\begingroup$ @Willie Wong : my question is about the second link.....which took me to J.M.'s profile page. The second 'this'. $\endgroup$ – Rajesh Dachiraju Jul 26 '11 at 14:18
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    $\begingroup$ @Willie : Oh! I miss clicked it...sorry for the trouble. $\endgroup$ – Rajesh Dachiraju Jul 26 '11 at 14:20
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    $\begingroup$ Whenever one clicks a link and it takes him to an unexpected page, a mouseover of the link previously clicked certainly helps... :) $\endgroup$ – J. M. is a poor mathematician Jul 27 '11 at 1:58
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From J.M.'s link:

Let $Q$ be the set of all questions that you asked that satisfies the following conditions:

  1. More than 3 days old
  2. Have at least one answer
  3. Is not closed
  4. Is not community wiki.

Let $S$ be the subset $S\subset Q$ satisfying the additional condition

  • You have accepted an answer on that question.

The accept rate is $|S|/|Q|$.

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    $\begingroup$ I can't understand this. :) $\endgroup$ – Jeff Atwood Jul 25 '11 at 23:21
  • $\begingroup$ Don't worry Jeff, I'm sure you understood the blog post. :) Incidentally @Jeff, something just came to mind: when you said that the questions counting for the acceptance count are those which do not have "no answers" (sorry for the convoluted phrasing), is that "no answers" as in "no answer", or "no answers" in the sense of the Community User autobump of unanswered questions? $\endgroup$ – Willie Wong Jul 25 '11 at 23:33
  • $\begingroup$ "no answers" does not mean "unanswered". "no answers" means what it says.. "unanswered" is more of a logical concept. $\endgroup$ – Jeff Atwood Jul 26 '11 at 0:11
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    $\begingroup$ @Jeff: I think that this is one of them places where mathematicians differ from the rest of the world... :-) $\endgroup$ – Asaf Karagila Jul 26 '11 at 8:55
  • $\begingroup$ @Willie Wong : -1; Thank you for the wonderful answer....but just curious ..why do you expect paople using meta to know elementary set theory !!!....isn't it inappropriate ! $\endgroup$ – Rajesh Dachiraju Jul 26 '11 at 14:15
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    $\begingroup$ @Rajesh: since you didn't specify what you didn't understand, one possibility could've been that the colloquial phrasing on the blog was not precise enough for a mathematician! ;p $\endgroup$ – Willie Wong Jul 26 '11 at 15:06
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    $\begingroup$ @Rajesh: I think this answer is far more understandable then the blog post. Also, it doesn't use elementary set theory... The words "set" and "subset" are easily understood by almost all English speakers... $\endgroup$ – Eric Naslund Jul 27 '11 at 22:23
  • $\begingroup$ @Eric : If we start this utterly pointless argument, then I wouldn't be surprised to hear from you that you thought all most English speakers are from US and UK. Any way the point here is that Willie's answer serves as same info in the link but given in a different flavour. $\endgroup$ – Rajesh Dachiraju Jul 28 '11 at 6:18
  • $\begingroup$ @Eric Naslund : What I do not understand is your last comment, 'all most all english speakers', as the set of all english speakers is countable even more is finite. What exactly do you mean by 'almost all'. What do you actually mean I do not get anything in mathematical sense :-). $\endgroup$ – Rajesh Dachiraju Jul 28 '11 at 6:43
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    $\begingroup$ Addendum: The acceptance rate is only defined for $|Q|\geq 4$. In particular, we cannot use it to settle the question of what $0/0$ is. ;.) $\endgroup$ – Michael Greinecker Nov 24 '12 at 12:44

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