[EM] 09/20/02 - The Manipulation Test:

[Elisabeth Varin/Stephane Rouillon]

Alex Small a écrit :

> Elisabeth Varin/Stephane Rouillon said:
> > However, do not dig your own grave. Yes,
> > in a Condorcet method lower ranks can help defeat
> > our own first choice. I admit it.
>
> I don't admit that.  In each pairwise contest involving your first choice,
> your ballot always favors your first choice over the others.  Your ballot
> also helps your second choice defeat the 3rd, 4th, etc.
>
> If there is no Condorcet winner, whether we use margins or winning votes,
> your ballot will help improve your favorite's strength of victory over
> other candidates, to lessen the chance of some other candidate's defeat
> being nullified and letting him win instead of your favorite.  Also, your
> ballot helps minimize your favorite's margin of defeat, improving the
> chance that his defeat will be nullified and your favorite will win.
>
> Also, consider the situation of a 3-way cycle where your second choice has
> the weakest defeat, and your favorite has the next weakest defeat.  Your
> second choice wins if we use margins or wv.  With winning votes the
> weakness of that defeat was only due to the fact that very few people
> preferred a particular candidate to your second favorite.  You did nothing
> to help your second choice win.
>
> So, I'm still trying to see where in Condorcet your vote for your second
> choice works against your first choice.
>
> Alex

Alex,

I agree with you that in general producing all your sincere
preferences would help enhance any voters result from its
own personal point of view. I am the kind of people that
would give my exact ranking even if the polls show I am
shooting in my foot at 99% odds. This is what I have been
saying from the start, the mean (average) gain of truncature
is not a gain, it is a loss (for margin, relative margins and
winning votes all the same). However there are special cases
where your 2nd and following preferences could make your
first pick loose. This is why I said:
"in a Condorcet method lower ranks CAN help defeat
 our own first choice." They usually don't.
But there are cases where it does.

A family of examples with (11+2x) (x>=0) voters and 3 candidates is:
2+x: A
2: A > B > C
2: B > A > C
1+x: B > C > A
4: C
Ranked pairs with winning votes produces:
A (6+x) > C (5+x) , B (5+x) > C (4) and A (4+x) > B (3+x).
A is the Condorcet winner and wins.
Margins and relative margins produce of course the same result.
If I am one of the two B > A > C voter, my 2nd (A)
choice harms my favorite 1st choice (B).
The proof is, if I and my co-thinker vote B only:
2+x: A
2: A > B > C
2: B   (truncated !)
1+x: B > C > A
4: C
Ranked pairs with winning votes produces:
B (5+x) > C (4), C (5+x) > A (4+x) and A (4+x) > B (3+x) can't lock.
B wins now.

By the way,
with x=0, margins finds a triple equality.
With x>0 it locks B>C, then gets stuck between the two last pairwise
comparison.
Winning votes used as tie-breaker would give C>A. Thus B wins.
Relative margins used as tie-breaker would give A>B. Thus A wins.

With relative margins, B>C ((1+x)/(9+x)), A>B (1/(7+2x)) and C>A (1/(9+2x)).
With x=0, it locks A>B, then gets stuck between the two last pairwise
comparison.
Neither winning votes, nor margins can be used as tie-breaker.
With x>0, A still wins.

Any publisher in the room ?
Just kidding, but advise the winning votes advocates...
Winning vote cannot garantee always protecting
a Condorcet winner. And if it can when every ballot
has no truncation from start, so can margins and relative margins
because with fully ranked ballots, all three methods
are equivalent...

Steph.
PS: Sorry Adam, thus the 2nd counter-example should have been:
>3: A
>2: A >B >C
>2: B >A >C
>2: B >C >A
>4: C                    =======> (not 3!!!) I messed again.

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