[EM] Voting paradoxes article
David Catchpole
s349436 at student.uq.edu.au
Sun Sep 19 17:00:00 PDT 1999
On Fri, 17 Sep 1999, Markus Schulze wrote:
> Example 1:
>
> There are 120 voters and 4 candidates for 2 seats.
> 8 voters vote A > C > B > D.
> 8 voters vote A > C > D > B.
> 8 voters vote B > C > A > D.
> 8 voters vote B > C > D > A.
> 8 voters vote C > A > B > D.
> 8 voters vote C > A > D > B.
> 8 voters vote D > A > B > C.
> 8 voters vote D > A > C > B.
> 7 voters vote A > D > B > C.
> 7 voters vote A > D > C > B.
> 7 voters vote B > D > A > C.
> 7 voters vote B > D > C > A.
> 7 voters vote C > B > A > D.
> 7 voters vote C > B > D > A.
> 7 voters vote D > B > A > C.
> 7 voters vote D > B > C > A.
Let's start with A vs. B vs. C .
> 8 voters vote A > C > B
> 8 voters vote A > C > B
> 8 voters vote B > C > A
> 8 voters vote B > C > A
> 8 voters vote C > A > B
> 8 voters vote C > A > B
> 8 voters vote A > B > C.
> 8 voters vote A > C > B.
> 7 voters vote A > B > C.
> 7 voters vote A > C > B.
> 7 voters vote B > A > C.
> 7 voters vote B > C > A.
> 7 voters vote C > B > A
> 7 voters vote C > B > A.
> 7 voters vote B > A > C.
> 7 voters vote B > C > A.
I'm using the Droop quota for my n-out-of-n+1 system so-
120/3=40
A is selected, B is selected.
Try A vs. B vs. D .
> 8 voters vote A > B > D.
> 8 voters vote A > D > B.
> 8 voters vote B > A > D.
> 8 voters vote B > D > A.
> 8 voters vote A > B > D.
> 8 voters vote A > D > B.
> 8 voters vote D > A > B
> 8 voters vote D > A > B.
> 7 voters vote A > D > B
> 7 voters vote A > D > B.
> 7 voters vote B > D > A
> 7 voters vote B > D > A.
> 7 voters vote B > A > D.
> 7 voters vote B > D > A.
> 7 voters vote D > B > A
> 7 voters vote D > B > A.
A is selected, B is selected. So A and B is at least one of the
"Condorcet solutions" using Droop.
The following solutions may exist-
AB OK
AC eliminated
AD eliminated
BC eliminated
BD eliminated
CD not eliminated.
So try A vs. C vs. D
> 8 voters vote A > C > D.
> 8 voters vote A > C > D
> 8 voters vote C > A > D.
> 8 voters vote C > D > A.
> 8 voters vote C > A > D.
> 8 voters vote C > A > D
> 8 voters vote D > A > C.
> 8 voters vote D > A > C
> 7 voters vote A > D > C.
> 7 voters vote A > D > C
> 7 voters vote D > A > C.
> 7 voters vote D > C > A.
> 7 voters vote C > A > D.
> 7 voters vote C > D > A.
> 7 voters vote D > A > C.
> 7 voters vote D > C > A.
C and D are selected- Aaargh! Try B vs. C vs. D
> 8 voters vote C > B > D.
> 8 voters vote C > D > B.
> 8 voters vote B > C > D.
> 8 voters vote B > C > D
> 8 voters vote C > B > D.
> 8 voters vote C > D > B.
> 8 voters vote D > B > C.
> 8 voters vote D > C > B.
> 7 voters vote D > B > C.
> 7 voters vote D > C > B.
> 7 voters vote B > D > C.
> 7 voters vote B > D > C
> 7 voters vote C > B > D.
> 7 voters vote C > B > D
> 7 voters vote D > B > C.
> 7 voters vote D > B > C
So yes, damnit, it's one of the situations I don't like. The choice then
is of a resolution procedure between A,B,C, and D (remember, the choice of both
n-out-of-n+1 and resolution system (OK later on I admit this is wrong) is
arbitrary... so long as they're used the right way the system is
independent of the removal of irrelevant alternatives (I thought- anyway
it works most of the time). The imposition of norms such as the ones you
have below might shorten the preferred list, though). Yeeargh! You
b******, Marcus, you just had to give them equal votes! What am I going to
do with you? ;)
>
> If I understand you correctly, then either
> (A and B) or (C and D) must be elected in
> Example 1. Because of the Anonymity Criterion
> and the Neutrality Criterion, you can presume
> without loss of generality that C and D are
> elected.
What are the anonymity and neutrality criteria? I assume that the
anonymity principle is the same as that of Arrow- swapping between
candidates does not change the result (excepting any inversions between
winners and losers). Not sure though. Why do they impute that C and D are
elected?
>
> ******
>
> Example 2:
> Now, candidate E is added so that there are
> 120 voters and 5 candidates for 2 seats.
> 8 voters vote A > E > C > B > D.
> 8 voters vote A > E > C > D > B.
> 8 voters vote B > C > A > E > D.
> 8 voters vote B > E > C > D > A.
> 8 voters vote C > A > B > E > D.
> 8 voters vote C > A > D > B > E.
> 8 voters vote D > A > B > E > C.
> 8 voters vote D > A > C > B > E.
> 7 voters vote A > D > B > E > C.
> 7 voters vote A > E > D > C > B.
> 7 voters vote B > E > D > A > C.
> 7 voters vote B > E > D > C > A.
> 7 voters vote C > B > A > E > D.
> 7 voters vote C > B > D > A > E.
> 7 voters vote D > B > A > E > C.
> 7 voters vote D > B > C > A > E.
>
> What does this do?
Well, let's try A vs. B vs. E and C vs. D vs. E to see if they remain
Condorcet favourites.
> 8 voters vote A > E > B
> 8 voters vote A > E > B.
> 8 voters vote B > A > E.
> 8 voters vote B > E > A.
> 8 voters vote A > B > E
> 8 voters vote A > B > E.
> 8 voters vote A > B > E
> 8 voters vote A > B > E.
> 7 voters vote A > B > E
> 7 voters vote A > E > B.
> 7 voters vote B > E > A
> 7 voters vote B > E > A.
> 7 voters vote B > A > E
> 7 voters vote B > A > E.
> 7 voters vote B > A > E
> 7 voters vote B > A > E.
A and B are alright.
> 8 voters vote E > C > D.
> 8 voters vote E > C > D
> 8 voters vote C > E > D.
> 8 voters vote E > C > D
> 8 voters vote C > E > D.
> 8 voters vote C > D > E.
> 8 voters vote D > E > C.
> 8 voters vote D > C > E.
> 7 voters vote D > E > C.
> 7 voters vote E > D > C
> 7 voters vote E > D > C.
> 7 voters vote E > D > C
> 7 voters vote C > E > D.
> 7 voters vote C > D > E.
> 7 voters vote D > E > C.
> 7 voters vote D > C > E.
Aargh! Say C and E win. Let's try A vs. C vs. E .
> 8 voters vote A > E > C
> 8 voters vote A > E > C
> 8 voters vote C > A > E
> 8 voters vote E > C > A.
> 8 voters vote C > A > E
> 8 voters vote C > A > E.
> 8 voters vote A > E > C.
> 8 voters vote A > C > E.
> 7 voters vote A > E > C.
> 7 voters vote A > E > C
> 7 voters vote E > A > C.
> 7 voters vote E > C > A.
> 7 voters vote C > A > E
> 7 voters vote C > A > E.
> 7 voters vote A > E > C.
> 7 voters vote C > A > E.
A and C are selected.
AB OK
AC eliminated
AD eliminated
AE eliminated
BC eliminated
BD eliminated
BE eliminated
CD Whoa. Bummer.
CE eliminated
DE eliminated
So the addition of E may well bring the result over to AB where before it
was CD's... Uh oh. I guess-
- either further conditions must be placed on the resolution procedure to
ensure that the addition of any additional candidate would still give the
same result; or
- I give up and say "OK- where multiple Condorcet winners exist, the
possibility exists of adding extra alternatives such that one loses and so
the other takes over, causing a violation of IIA."
Where single CWs exist the system still works for removal... as long as
none of the sub-systems have a "sleeping multiple." I'll have to go back
to the drawing board and figure out if anything can be done about
"sleeping multiples." Any ideas?
What a silly person I am! When I was thinking up this, I considered
only the removal of candidates who had no impact on whether or not a
group of candidates was a CW.
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