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- From: dsg@impmh.uucp (Dave Gordon)
- Newsgroups: sci.math,alt.politics.kibo
- Subject: Re: three sided coin
- Message-ID: <1992Nov9.181231.9473@impmh.uucp>
- Date: 9 Nov 92 18:12:31 GMT
- References: <MJD.92Oct29173230@saul.cis.upenn.edu> <1992Oct30.092113.24403@jarvis.csri.toronto.edu> <1992Nov3.094443.10479@impmh.uucp> <1d6ti8INN976@CAUCHY.MATH.YALE.EDU>
- Organization: Integrated Micro Products Ltd
- Lines: 87
-
- In <1d6ti8INN976@CAUCHY.MATH.YALE.EDU> ferenbaugh-charles@math.yale.edu (Charles Ferenbaugh) writes:
- >
- >In article <1992Nov3.094443.10479@impmh.uucp> dsg@impmh.uucp (Dave Gordon) writes:
- >> [original solution deleted]
- >>
- >>The original poster mentioned adjusting the weights by allocating smaller
- >>ranges to some candidates and giving the 'excess' to Kibo. This solves
- >>the problem! By allocating the ranges thus:
- >> [0,1/4) to Bush
- >> [1/4,3/8) to Kibo
- >> [3/8,5/8) to Clinton
- >> [5/8,3/4) to Kibo
- >> [3/4,1) to Perot
- >>(this is the equally-weighted version; note that each candidate has
- >>1/4 of the total range).
- >>
- >>Now, we can guarantee that in only three tosses, two of the candidates
- >>will be eliminated, leaving a choice between Kibo and the other one ...
- >>no contest!
- >>
- >>All hail Kibo, champion of democracy and stochastic processes!
- >>--
- >
- >Sorry, but this won't work. (Or should I say, "no dice"? :-)
- >By altering the process so that it terminates after 3 steps,
- >you've changed the distribution of the outcome: it's no longer
- >uniformly distributed over [0,1), but instead takes on a
- >discrete set of values each with probability 1/8:
- >
- > 0 to Bush -> Bush
- > 1/8 to Bush -> Bush
- > 1/4 to Bush/Kibo -> Bush
- > 3/8 to Kibo/Clinton -> Clinton
- > 1/2 to Clinton -> Clinton
- > 5/8 to Clinton/Kibo -> Clinton
- > 3/4 to Kibo/Perot -> Perot
- > 7/8 to Perot -> Perot
- >
- >so you've shortchanged Perot. (Now this may be a good thing, but
- >discussing that would be another matter entirely.)
- >
- >In fact, it's not hard to show that you can't make a fair three-way
- >choice with any procedure guaranteed to finish in n tosses
- >(for any n). Why? Well, there are 2^n possible outcomes, and
- >for equal distribution you would want each candidate to get
- >2^n/3 of those - but that's not an integer. So if you really
- >want to use this method, you have to allow some small probability
- >of non-termination - it's unavoidable.
- >
- >- Charles Ferenbaugh
- > crf@math.yale.edu
- >
- >PS. Pardon my ignorance - who or what is Kibo, anyway?
-
- Aha, the PS explains all. I was going to flame you for getting
- the mapping above wrong, but perhaps ignorance is an excuse ...
- ... so I'll merely point out politely where you've gone wrong.
-
- Actually, there are two things wrong with it ... the non-mathematical
- one is that you've assumed that any split such as "Bush/Kibo" goes
- to the non-Kibo! If you look at what I wrote originally
- >"a choice between Kibo and the other one ... no contest!"
- this was supposed to imply that any such split goes to Kibo!
-
- The mathematical error is to suppose that there are any such splits;
- with the range [0,1) divided into eight, three tosses identifies
- *precisely* one candidate. In fact, only two tosses are strictly
- necessary (as there are only four names) - I just put the extra one
- in to make it more interesting. So, to simplify, we could enumerate
- the results thus:
-
- T T Bush
- T H Clinton
- H T Kibo
- H H Perot
-
- For anyone who objects that Kibo isn't on their voting paper, why not
- just write him in? Can't do that? Not much of a democracy, is it?
-
- Binarily,
- Dave
- (writing from another non-democratic country)
- --
-
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- Dave | dsg@imp.co.uk | +44 753 516599 | "Ceci n'est pas une .signature"
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