User talk:New Thought

Help, please
Thanks for you compliments on the procrastination article. Another editor is trying to build a consensus to remove most of the contents of the article, and I need your help in the discussion to stop him. Please come right away. Go for it! 10:48, 9 December 2005 (UTC)

Thanks for the reply. The danger has subsided for now, but may resurface at anytime. The page is on a certain user's watchlist, and whenever there is a lot of editing activity on the page, he shows up to try to get rid of it. So the best thing you can do is put the procrastination article on your watchlist and monitor it. The article isn't finished yet, and needs some TLC. Several sections need work. Have a look. Go for it! 01:24, 22 December 2005 (UTC)

Notice
The Community Portal was recently reverted to a version that appeared months ago. Therefore, I've called for a vote to restore to the Community Portal the version that had developed there up until that reversion. There are three drafts competing for the privilege, each representing entirely different approaches, including the current revert version. To show your support for which design should be displayed as the Community Portal, VOTE HERE. Sincerely, --Go for it! 18:14, 3 April 2006 (UTC)

binomial distribution
Your "tiny style change" actually substantially altered the meaning of the sentence. Is that not evident? The newly rearranged sentence was incorrect; the old one was correct. Michael Hardy 21:03, 10 May 2006 (UTC)


 * When calculating $$F(k;n,p) = \sum_{j=0}^{\lfloor k\rfloor} {n\choose j}p^j(1-p)^{n-j}$$, please explain how $$k$$ could be anything other than 0 or a positive integer. Thanks! --New Thought 07:45, 11 May 2006 (UTC)

If, for example, k = 3.6, then


 * $$\lfloor k \rfloor = 3,\,$$

i.e., we have


 * $$\lfloor 3.6 \rfloor = 3.\,$$

That's explained in the article after my edit, where it says that


 * $$\lfloor k \rfloor\,$$

is the greatest integer less than or equal to k. If k is negative, then I would take


 * $$\sum_{j=0}^k a_j$$

to be a sum with no terms, and therefore zero. If j &ge; n, then take


 * $${n \choose j}$$

to be zero. Michael Hardy 18:46, 11 May 2006 (UTC)

Historicity of Jesus
Wanted to pop in and say that I understand your frustrations with the page. Just wanted to leave a reminder that focusing on the content is usually more productive than focusing on other editors' intentions. Hope you don't take this the wrong way, because I think you've got a point. It's easier for me to support a particular argument when it focuses on content. Phyesalis (talk) 17:41, 29 December 2007 (UTC)

ArbCom elections are now open!
MediaWiki message delivery (talk) 13:01, 23 November 2015 (UTC)

Live updates
Please do not update match stats (whether a player’s or a club’s or a league table etc.) until the player’s involvement in the match has finished / the match itself has finished. That means (especially, for example, if another goal is scored) that errors can be minimised. Many thanks. GiantSnowman 17:14, 26 August 2023 (UTC)