Anyone who visits this page can judge the merits themselves: there's no argument from authority involved. No one is claiming this form of argument is invalid because it's on LW, or because Yvain wrote it, or because it has a catchy name that's published on a website, or because it now has an easy-to-remember URL. I made a simpler citation, nothing more.
Hahaha, nice.
I was imagining a situation in which someone makes an argument of this type, you say something along the lines of "that's a great example of the 'Worst Argument in the World'," and the person replies "you just made that up..." or "that's just your opinion..."
Providing a pre-existing URL that links to a well-written page created by a third-party is a form of evidence that shifts "Worst Argument in the World" from something that feels like an opinion to the title of a logical fallacy. That can be quite useful in certain circumstances.
I just registered http://worstargumentintheworld.com - it redirects to this post, and should be available shortly. Much easier to mention in conversation when other people use this argument, and don't believe it's a "real thing."
Great piece of work, Yvain - it's now on my list of all-time favorite LW posts.
Wow, that's so simple it could possibly work!
Many thanks - this is most likely what I'll go with. I appreciate your help. :-)
Thanks - the voting system analogy didn't occur to me. Reading up on ranked pairs: http://en.wikipedia.org/wiki/Ranked_pairs
Ah, I see. Instead of updating half the lists, I was updating the 720 sets where C is the #1 preference. Thanks for the clarification.
Very helpful - reading about this now. Starting here: http://en.wikipedia.org/wiki/Ranking
Okay, if A is preferred from { A , [B-G] }, that should add probability mass to [A, [B,...,G] ], where [A, [B,...,G] ] is a ranked set of objects where the first slot is most preferred. That would represent 720 (6!) sets out of 5040.
All other sets (7! - 6! = 4,320) should either stay the same probability or have probability mass removed.
Then, the probability of A being "most preferred" = the sum of the probability mass of all 720 sets that have A as the highest ranked member. Likewise for B through G. Highest total probability mass wins.
Am I understanding that correctly?
Right - thanks for the correction. Posted a correction in the main text.
Update: as it turns out, this is a voting system problem, which is a difficult but well-studied topic. Potential solutions include Ranked Pairs (complicated) and BestThing (simpler). Thanks to everyone for helping me think this through out loud, and for reminding me to kill flies with flyswatters instead of bazookas.
I'm working on a problem that I believe involves Bayes, I'm new to Bayes and a bit rusty on statistics, and I'm having a hard time figuring out where to start. (EDIT: it looks like set theory may also be involved.) Your help would be greatly appreciated.
Here's the problem: assume a set of 7 different objects. Two of these objects are presented at... (read 267 more words →)
Sorry for the downtime - transferred the domain to a new registrar, and thought the forward would be automatically detected and carried over. It wasn't. Should be back up once the record updates.