: Statistical Question
Feb 5th, 2003, 05:25 PM
I am not math whiz, but for the past 5 minutes I have had the last recorded post on all five forum threads. This is no great feat, I am sure, but I was wondering how one might figure out the odds of such an event happening? Given the number of members in the group is one given, but I was wondering how one could figure out the odds of anyone of us having the last post in every thread? Any math Mavens out there in ehMacLand???
Feb 5th, 2003, 05:30 PM
If a comic told a joke, and no one was listening, would it still be funny???
Feb 5th, 2003, 05:51 PM
Does he laugh at his own jokes?
Feb 5th, 2003, 05:52 PM
gg, does a tree fall down in the forest? Now, if no one is there, will the sound of the tree be heard.........especially if it falls on top of the comic???
Feb 5th, 2003, 05:54 PM
Only if he sees the humour in it.
Feb 5th, 2003, 05:55 PM
Visualization is not an auditory perception.
Feb 5th, 2003, 05:56 PM
The falling of the tree will be heard by the comic, with the axiom that the comic is not deaf.
But, will anyone hear his scream as the tree falls onto the comic?
Feb 5th, 2003, 06:06 PM
I don' get it... :rolleyes: tongue.gif
Feb 5th, 2003, 06:20 PM
<BLOCKQUOTE>quote:</font><HR> You see it as half empty, I see it as half full.<HR></BLOCKQUOTE>
If you have all that you need in life, then your glass is always full.
Feb 5th, 2003, 07:05 PM
Forget the philosophical discourse, what about the statistical equation?
Feb 5th, 2003, 07:08 PM
If I remember correctly it's number of desireable outcomes over number of possible outcomes.
The formula is the easy part, getting the outcomes quantified requires a bit more research; I suggest a grant applicaton. After all, you will need a powerful computer, Dr G ;)
used to be jwoodget
Feb 5th, 2003, 07:13 PM
For anyone else, the odds could be astronomical (depending on the popularity of/number of independent contributors to a thread, time of day/night and the wpm typing rate). For Dr. G. on his quest to 1000 posts, I'd say its better than evens smile.gif
Feb 5th, 2003, 07:33 PM
gordguide, as Peter would say " 'taint funny, McGee". No, wait, that was Molly in "Fibber McGee and Molly", a popular radio show back when I was a little boy.
jwoodget, I was actually hoping that someone might have an understanding of the variables and the permutations/combinations. The quest for 1000 has nothing to do with this post. I'll tell you what, when I get a new Mac I shall log in under a new name, such as "He once was Dr.G.". Thus, I shall have a new name and new post number, starting from 0.
Feb 5th, 2003, 08:29 PM
the 'math' behind it all
in order to 'calculate the odds' (said in my best spock voice) one needs to know the distribution of postings vs. members
how often do members post? number of posts per member, etc. stuff like that
the 'odds' you are looking for are directly dependant on the users and usage which cannot be described by a mathematical model, but only approixmated by observation.
it's not a simple question like;
- the odds of a coin coming up heads or tails
- the odds of picking a blue ball out of a bag that has 3 blue balls and 2 red balls
- the odds of rolling a 'seven' using 2 dice
that's the easy stuff
your request would require observation and analysis for some time
maybe a project one of you grad. students could undertake or perhaps something you could suggest to the Stats dept. over at MUN
Feb 5th, 2003, 08:38 PM
I guess I shall have to go that route, Macspectrum.
Feb 5th, 2003, 09:38 PM
Well, I am not maven of math but from my little non-understanding of the abstract concepts involved you would have to factor in a number of things.
-Time of day, as some people (such as myself) post quite a bit late at night when no one else is up and are therefore quite a bit more likely to be able to pull off such a feat (I have had the most recent post on a number of threads at the same time numerous times, but I am posting after midnight my time which is after 3.00 am eastern)
-Individual Poster's likelyness to actually post, which if you wanted to do it the simplest way I can think of (which is inacurate) would be to divide the number of posts they have by the number of days they have been here and get an average. Then with that average you could get a basic understanding as to whether the poster would be likely to post in all of them or not by taking their average and subtracting the number of available active threads.
-Divide the whle thing by PI, just for fun, because posting seems to go in circles, as in the person who starts a thread is fairly likely to post back in the thread than into a thread they did not create.
If you ignore the fact that I don't really know what I am talking about, some of that actually makes sense.
I suppose if enough information was collected I could make an excel spreadsheet to chart it all, but there is no way I am going to be bored enough to collect all the information on every post of every frequent poster on the board.
It would be interesting though.
Feb 6th, 2003, 04:41 PM
PosterBoy, everything went well until we factored in Pi. Without it, the odds are 7,425,581 to one from this taking place for more that 10 minutes (I was on for 7 minutes with the last post in each of the five fourm threads). I also figured out the odds of how long I would continue to be allowed to stay on this forum if I delayed my decision to buy a Mac before MacWorld...in Boston! Let's just say that the odds were equal to my winning (in the same week) the Lotto 649, The US Powerball, and being elected as the next Pope.
Feb 6th, 2003, 11:13 PM
<BLOCKQUOTE>quote:</font><HR>Originally posted by Dr.G.:
Let's just say that the odds were equal to my winning (in the same week) the Lotto 649, The US Powerball, and being elected as the next Pope.<HR></BLOCKQUOTE>
So pretty good then!
Feb 7th, 2003, 12:27 PM
The formula is (t/m)x(12+h)-(c-d)x(<pi>-1)-?where pi are round, cake are square and all values must be primary numbers and cannot equal 0. Also, no doxies can be consulted or experimented on to reach the answer. You will only need an LCII (or Omega) to compute the answer.