[EM] More results about computer simulations of elections pt 2
Kevin Venzke
stepjak at yahoo.fr
Tue Dec 6 14:36:22 PST 2005
Warren,
--- Warren Smith <wds at math.temple.edu> a écrit :
> But in this post we shall consider some rank-ballot voting methods that
> allow rank-equalities.
>
> Thm 7.
> Suppose we are using a Condorcet method but this time allowing
> rank-equalities.
> Suppose there are two random but pre-fixed candidates A and B.
> Suppose the voters behave thusly:
> Each gives either A or B (whichever they consider better - call him T)
> top rank.
> Then they rank everybody else honestly (generically with no equalities)
> except
> that anybody who would have been ranked above T is instead ranked
> coequal-top with T.
>
> Then: either A or B will, with 100% probability asymptotically for
> V=large,
> win the election, i.e. the same winner results as with strategic
> plurality voting.
>
> Proof sketch:
> Because that winner X will be Condorcet winner.
> For each Y not in {A,B}, X is ranked above Y one-half of the time,
One-half isn't a reasonable guess for this. You're not considering the
equal ranking at all.
> is ranked equal with Y at top 1/6 of the time, and is ranked below Y
> one-third of the time (all almost-surely in the V=large limit by
> the law of law numbers...).
Not sure how you get these numbers.
> So X pairwise-defeats all Y (whether we use
> margins or winning-votes, does not matter). Q.E.D.
Suppose 51% of the voters rank X=Z>Y and 49% vote Y>Z>X? Then Z wins even
though Z can't (by these ballots) be either A or B.
> It is very interesting, however, that if voters adopt the following
> alternate strategy for Condorcet:
> Give either A or B (whichever they consider better - call him T) top
> rank.
> Give the other bottom-rank (call him Z).
> Then rank everybody else honestly (generically with no equalities)
> except that anybody who would have been ranked above T is instead ranked
> coequal-top with T, and anybody who would have been below Z is instead
ranked
> coequal-bottom with Z.
> Then:
> it is unclear to me who will win, and I do not see any reason it
> generally must be A or B.
I agree.
> [If you try to redo the above proof, then
> X is ranked above Y one-third of the time, below one-third, coequal
> bottom 1/6, coequal top 1/6,
How do you get 1/3 and 1/6 figures?
> up to additive noise of order o(1) in the V=large limit, and so we cannot
> draw a conclusion that
> anybody is going to be a Condorcet winner. In fact in general I would
> expect no Condorcet winner will exist if C is reasonably large.]
>
> In other words: claims have been made that my notion that voters would
> "bury" their least-liked among {A,B} to bottom rank, was silly,
> that it was stupid for Condorcet voters to behave that way, etc.
>
> But those claims now seem wrong - because as we now see, if voters behave
> in this burial-manner, then in this randomized election model, they will
be able
> to get a winner better
> than the strategic-plurality-winner, but without the burial (as in thm 7)
> they will just get the same winner as with strategic-plurality voting.
Two problems:
1. You didn't show burial here. Burying involves order reversal.
2. How did you show that the second scenario is superior?
> Thm 9.
> Suppose we are using an IRV (or Condorcet)
> method but this time allowing rank-equalities, where
> all equalities X=Y in votes are replaced by X>Y and Y>X etc in all
> possible ways
> (i.e. C! near-copies are made of each vote to allow this) which is the
> usual
> very nasty-to-deal-with proposal for allowing IRV to handle
> rank-equalities in votes.
>
> Suppose there are two random but pre-fixed candidates A and B.
> Suppose the voters behave thusly:
> Each gives either A or B (whichever they consider better - call him T)
> top rank.
> Then they rank everybody else honestly (generically with no equalities)
> except
> that anybody who would have been ranked above T is instead ranked
> coequal-top with T.
>
> Then: either A or B will, with 100% probability asymptotically for
> V=large,
> win the election, i.e. the same winner results as with strategic
> plurality voting.
>
>
> Proof sketch:
> For Condorcet:
> For each Y not in {A,B}, but with X in {A,B} is ranked above Y 7/12 of
> the time,
> and is ranked below Y 5/12 of the time (all almost-surely in the V=large
> limit by
> the law of law numbers...). So X pairwise-defeats all Y (whether we use
> margins
> or winning-votes, does not matter, in fact the two are the same with this
> method of handling
> X=Y votes). So an X in {A,B} will be a Condorcet winner.
Isn't this identical to Thm 7?
> Now for IRV:
> X is ranked top by V/2+o(V) voters before vote-alteration
> and after alteration is
> top-ranked by 3V/(2C) voters. Meanwhile Y is top-ranked 1/3+o(1)
> of the time
Where does this come from? Y could be top-ranked on *all* the ballots.
> before vote-alteration and after alteration is
> top-ranked by V/C voters. So each Y will get eliminated (and since the
> vote-transfers
> are independent random, this will happen in subsequent IRV rounds also)
> Q.E.D.
It looks like you forgot that the votes have to be split up symmetrically.
Counter-example:
51 X=Z
49 Y=Z
Z wins even if we say that A and B are X and Y.
> So put those theorems in your pipe, and smoke them.
Ok. It looks to me that simulations could do a much better job of
estimating
realistic scenarios than your (apparent) assumption here that the voters
will appear in equal numbers or something.
In my simulations, voters were grouped into like-minded factions, rather
than each voter being a distinct faction.
Kevin Venzke
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