From: dsg@impmh.uucp (Dave Gordon)
Newsgroups: sci.math,alt.politics.kibo
Subject: Re: three sided coin
Date: 9 Nov 92 18:12:31 GMT
Organization: Integrated Micro Products Ltd

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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