home *** CD-ROM | disk | FTP | other *** search
- Path: sparky!uunet!usc!sol.ctr.columbia.edu!spool.mu.edu!uwm.edu!biosci!ucselx!crash!nosc!pages!bwebster
- From: bwebster@pages.com (Bruce F. Webster)
- Newsgroups: sci.skeptic
- Subject: Re: Innumeracy, humorous ... maybe.
- Message-ID: <1992Nov17.025722.6032@pages.com>
- Date: 17 Nov 92 02:57:22 GMT
- References: <1992Nov16.045407.29782@udel.edu>
- Sender: bwebster@pages.com
- Reply-To: bwebster@pages.com
- Organization: Banzai Research Institute
- Lines: 60
-
- > In article <BxoJK2.KzI@ccu.umanitoba.ca>
- > vnelson@ccu.umanitoba.ca (Gerald Vernon Nelson) writes:
- > >
- > >Speaking of inummeracy...
- > >
- > >There was an article in our local paper about a year back on the
- > >subject. The author went to lengths to point out how little people
- > >understand numbers. In his example he used the 6/49 lottery.
- > >You know, pick 6 different numbers from 49. He stated that people
- > >just didn't understand that the odds of winning that lottery were
- > >about 10 billion to 1 (49 * 48 * 47 *46 * 45 * 44).
- > >
- > >He was of course, inumerate himself, as the odds of winning are
- > >actually about 14 million to 1.
- > >
-
- Well, in a society where high school graduates can't make correct change, I'd
- be a bit hesitant to call someone who doesn't correctly apply statistics
- "innumerate"--but given the tenor of his article, it might be just. :-)
-
- In article <1992Nov16.045407.29782@udel.edu> mccoy@pecan.cns.udel.edu (Don
- McCoy) writes:
- >
- > Uh, I could be wrong here, but I believe 10 billion to 1 is right.
- > How did you arrive at the 14 million to 1 figure???
- >
-
- The fallacy of the author's calculations is an easy one to fall into, but
- here's a quick thought experiment which will illustrate it: if the lottery
- required that you correctly picked 48 out of 49 numbers, would your odds of
- winning be (49*48*47...*4*3*2) to 1? Obviously not; they would be merely 49 to
- 1, the same as correctly picking 1 out of 49 numbers. So the calculations have
- to be symmetric and work appropriately for any selection of numbers.
-
- The correct caculation for the number of possible ways of taking n objects r at
- a time is n!/[r! * (n-r)!]. For picking 6 out of 49 numbers, that leaves you
- (after factoring out 43! from 49!):
- (49*48*47*46*45*44)/(6*5*4*3*2*1) = 13,983,816 possible combinations.
-
- Thus, for any random combination, the odds of having chosen that one is 14
- million to one.
-
- If you're still stuck back on "but the odds should be based on how many numbers
- are left", here's the fallacy the author of the news article made: the odds of
- the first lottery ball chosen being correct are not 1/49, they are 6/49. You've
- picked six numbers ahead of time, remember--any one of those six can match the
- first lottery ball. If one does, then the odds of any of the second ball
- matching any of the remaining five numbers is 5/48; for the third ball, 4/47;
- and so on. Multiplying these odds together, you get a familiar value:
- (6*5*4*3*2*1)/(49*48*47*46*45*44) or 1/13,983,816.
-
- Q.E.D. ..bruce..
-
- -------------------------------------------------------------------------------
- Bruce F. Webster | Am a flamer, goateed, pallid, overweight,
- Chief Technical Officer | Willing to pull two shifts, then (hell) a third,
- Pages Software Inc | To save a session from a deadlocked state;
- bwebster@pages.com | At times, (to put it mildly) unrestrained--
- #import <pages/disclaimer.h> | Almost, at times, a nerd. -- Jeff Duntemann
- -------------------------------------------------------------------------------
-