From: 2893684@QUCDN.QueensU.CA (andy)
Newsgroups: can.general,can.politics,soc.culture.canada
Subject: Re: Voting Systems (was Re: Proportional representation)
Date: 9 Nov 92 01:56:39 GMT
 <1dk6adINN5jb@iskut.ucs.ubc.ca> <1992Nov9.022817.20512@ee.ubc.ca>
Organization: Queen's University at Kingston

In article <1992Nov9.022817.20512@ee.ubc.ca>, jmorriso@ee.ubc.ca (John Paul
Morrison) says:
>
>In article <1dk6adINN5jb@iskut.ucs.ubc.ca> ramsay@unixg.ubc.ca (Keith Ramsay)
>writes:
>>In article <1992Nov8.074732.7435@ee.ubc.ca> jmorriso@ee.ubc.ca
>>(John Paul Morrison) writes:
>>[Regarding the Arrow impossibility theorem:]
>>>I would like to see how this following system is affected. I believe
>>>that the theorem you mention can't apply in this case.
>>[...]
>>>This is similar to the way shareholders elect a board (but not really,
>>>so don't bother picking on this point).
>>>
>>>1. A voter can only vote for one person, no prioritizing
>>>2. The candidate is elected if he gets at least one vote.
>>>3. The candidate is actually a delegate, so in parliament, he or she
>>>   represents as many votes as they received.
>>
>>There's no way to express the preference, for instance, of those who
>>might mainly want some kind of coherence and consensus of purpose in
>>the parliament, for efficiency, and aren't as concerned about getting
>>their own first choice of representative into it.

Is this done on a riding basis or a national basis



>
>You don't read very closely do you? Any candidate who gets a vote, takes
>that vote with him to the parliament. The voter is saying: here I trust
>you, I transfer my vote to you. I am delegating my responsibility to you.
>The more votes (ie the more people who trust that delegate) the delegate
>gets, the more influence he has. (Since a delegate who represents 2,000
>people has more voting power than a delegate who has onlt 500 people
>supporting him).
>
>What you don't understand, is that in this hypothetical system, the votes
>of each delegate are not equal
>>
>>It also fails another condition of the theorem, in that it is possible
>>for a person to cause an option A to be selected instead of an option
>>B, by deciding to try to get C instead of D. Suppose there are a
>>number of us who'd like to vote for Smith, but for whatever reason
>>we'd rather she not have too much power. If, however, she already has
>>sufficient votes, then we'd rather vote for Jones.
>
>This isn't the case. Your logic system requires that out of the set
>{A,B,C,D}, exactly one item is selected. In the system I am suggesting
>(to demonstrate that this line of reasoning does not apply) you actually
>have a "fuzzy" set where ALL of the set {A,B,C,D} is chosen. But the
>weights are different:  they might be 700A, 200B, 50C, 50D (sums to
>1000. or 0.7, 0.2, 0.05, .05 of the local vote).
>Thus each delegate represents exactly the number of voters who
>voted for him. Put A,B,C,D in a larger parliament with AA, BB, CC, DD etc
>and A still represents 700 votes. IF AA..DD get 500, 300, 100, and 100,
>(sums to 1000) the total parliament would represent 2000 votes. But
>A still represents 35% of the popular vote.
>>
>>Let A, B, C, and D be possible compositions of the legislature, where
>>A differs from B only by Smith getting some of the votes Jones got in
>>case B, and likewise for C and D, where A and B are alternatives in
>>which Smith gets enormous power, and C and D are alternatives in which
>>Smith gets either a modest amount or none at all. The condition
>>described by the theorem, requiring that changing preferences between
>>A and B will not affect whether C or D is chosen, is violated by this
>>voting system. By trying to ensure Smith gets a modicum of influence,
>>we could end up giving her more than we had wanted, or conversely.
>
>It is completely independent! You haven't proven anything.
>
>If there are 100 voters:
>
>suppose delegate A gets 50 votes, B gets 30, c gets 15 and d gets 5
>but then some voters take away their vote from And give it to B,
>giving    A = 40, B = 40, C = 15 and D = 5
>C and D are NOT affected. In both cases, all 4 delegats go to parliament,
>however A's vote will only be worth 40 instead of 50. D still has 5 votes
>to cast in the parliament.
>

How are these people paid. Do all delegates get the same wages, do they all get
 support staff, offices, constituency offices etc. Your system has the
potential for dramatically increasing the number of reps we have and the
related expenses. Also, you ignore that there is more to being an MP than
voting, there is also lobbying on your constituents' behalf and helping them
with their problems and bureaucracy. Obviously a riding with 4 MPs has an
advantage, even if their delegates votes have the same weight as a riding
with one delegate, they are still better represented and have better service
and better access to the government and bureaucracy.

>
>>
>>The conditions about "independence of preferences" are in some sense
>>the stiffest constraints placed by the theorem, and are often
>>violated. It is often possible for it to be the case that if you
>>prefer A to B to C, you are well advised to vote as if you prefer B
>>most of all, so as to make sure that B rather than C occurs.  This
>>"sophistical voting" is what the Arrow conditions are formulated to
>>preclude, and the upshot is that there always is a price paid for
>>avoiding it.
>
>
>the Arrow theorem assumes that only one candidate is elected. The example
>posted does not follow Arrows assumptions, so it doesn't apply.
>
>IF you prefer A to B to C, voting for A will always help A and never C,
>because in this system, a delegate's power depends only on the votes
>he receives. IF A has 100 votes, B has 100 and C has 100 votes, and
>you vote for your first choice A, A will  get 101 votes, when ALL of
>A, B and C go to parliament, A will still have 101 votes, while C
>will always have 100 votes.
>
>
>This does not fall into range where Arrow's theorem applies, because it
>Assumes only one of ABC is selected, but this system selects all of ABC,
>and A,B, and C have different weights. Arrows's reasoning requires
>that if A is selected, A has weight 1.0 and B and C get 0.0 (even
>if B and C get voted for), but in this example, A would get a weight of
>a, B gets b, and C gets c, where a+b+c = 1.0.

It's a very complicated system whose goals could be better achieved through
other methods, like the German model.

>
>
>
>hope this cleats it up. it seems that Arrow's theorem doesn't apply if
>you have a "fuzzy" (ie statistically weighted) system
>
>>
>>Keith Ramsay
>>ramsay@unixg.ubc.ca
>
>
>--
>__________________________________________________________________________
> John Paul Morrison                     |
> University of British Columbia, Canada |
> Electrical Engineering                 |   .sig file without a cause
> jmorriso@ee.ubc.ca              VE7JPM |
>________________________________________|_________________________________