On Fri, 29 Oct 1999, Craig Carey wrote:
> At 19:52 28.10.99 , David Catchpole wrote:
> >On Thu, 28 Oct 1999, Craig Carey wrote:
> ...
> >couldn't work it out from the plain English, I'll have a go at putting it
> >in a formal structure of proof. Later, though...
>
> Proof of what?.
Hopefully next week some time, I'll be sending a formal version of
the proof all of this started with (you know- Neutrality to voters +
Majority rules + Monotonicity -> Condorcet) in microsoft equation form
(i.e. as a word document) to this list. At the same time, I'll attempt to
use the same method to establish ramifications for (P1).
> >> FPTP is a monotonic preferential voting method, and it isn't Condorcet.
> >> So therefore the and rule is finally rejected.
> >
> >It's not! Consider the following example I gave to Markus Schultze-
>
> FPTP is monotonic method. There need not be misunderstanding over that.
> Mr Catchpole is redefining the word monotonic. I suggest it be called
> David-Catchpole-Monotonicity-version-0.1.
Finding Arrow...(more further down)
>
> >Say two voters have preferences A>C>B in an election where A has 5
> >votes (including that of this voter), B has 4 votes and C has 1. A
> >wins. Say that these two voters change their preferences to C>A>B. B now
> >wins, even though nobody changed their preferences between A and B. While
> >some definitions of monotonicity fail to rule on such a scenario, the one
> >I gave in my message does- and rules FPTP out. (more further down)
>
>
> First example: (1)
> 3 A
> 2 AC A wins
> 4 B
> 1 C
> FPTP,STV,IFPP: A wins
>
> The Second example: (2)
> 3 A
> 4 B
> 2 CA B wins
> 1 C
> FPTP,IFPP: B wins
> STV : draw between A and B
>
> Mr Catchpole probably wants A to win election (2)
> So A wins the first if and only if A wins this: (3)
> 5 A
> 4 B
> 1 C
It's not whether A _has_ to win the election (remember, A could lose the
first, and that too would eliminate the problem for this particular case)!
It's whether voters C>A>B will vote insincerely- you can plainly see that
with FPTP it benefits them to do so.
> 0.9 A
> 2.1 AC
> 4 B
> 2 CA
> 1 C
I II III IV | I II III IV
A>B 5 5 5 5 |C>A 1 3 1 1
B>A 4 4 4 4 |B>C 4 4 4 4
A>C 5 3 5 5 |C>B 3 3 1 1
VI VI
A>B 5 |C>A 3
B>A 4 |B>C 4
A>C 3 |C>B 5.1
The possible winners are therefore-
I II |I III |II III
A C |C B |C A
A A |A A |C B
B B |B B |A A
C C |C C |B B
| |C C
-----------------------------------------
I IV |II IV |III IV
C B |C A |A A
A A |C B |B B
B B |A A |C C
C C |B B |
|C C |
---------------------------------------------------------
I VI |II VI |III VI |IV VI
A C |B C |A C |A C
B C |A A |B C |B C
A A |B B |A A |A A
B B |C C |B B |B B
C C | |C C |C C
Logically, the possible winners are syphoned down to-
I II III IV VI
A A A A A
A C A A C
B B B B B
B B B B C
C C B B C
C C C C C
So there you go, these are the options which are not ruled out outright
for these 5 schema. REMEMBER- I can prove that monotonicity may imply _no_
answer! (This was discussed in the dim distant past with Blake Cretney)
However, where we expect monotonicity to apply for _most_ transitions,
this case-by-case method demonstrates some not-so-noxious options.
(more further down)
