From: bwebster@pages.com (Bruce F. Webster)
Newsgroups: sci.skeptic
Subject: Re: Innumeracy, humorous ... maybe.
Date: 17 Nov 92 02:57:22 GMT
Reply-To: bwebster@pages.com
Organization: Banzai Research Institute

> 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..

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Bruce F. Webster             | Am a flamer, goateed, pallid, overweight,
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#import <pages/disclaimer.h> | Almost, at times, a nerd.      -- Jeff Duntemann
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