[EM] Why Clone Independence?

Forest Simmons forest.simmons21 at gmail.com
Wed Jan 25 19:01:20 PST 2023


The clone dependence of Borda and Kemeny-Young are symptoms of the same
root malady ... the clone dependence distortion of the Kendall-tau metric.

You know that Kemeny-Young elects the head of the candidate ranking that
minimizes the total Kendall-tau distance from it to the ballot rankings.

The contribution of one ballot B to the Borda count of candidate X is the
distance from the ballot B ranking to the ranking with X moved to the
bottom of B ... that is the number of swaps it takes to lower X to the
bottom of B.

Clones distort the Kendall-tau metric like the rear view mirror that says
"objects may be closer than they appear."

Let's look at the example ...

60 A>B>C>D>E
40 B>C>D>E>A

The Kendall tau swap cost of moving A to the bottom of one first faction
ballot is 4 swaps per ballot ... a total Kendall-tau distance of 4*60=240

The second faction takes zero swaps to get A to the bottom, so the Borda
total for A is 240.

Similarly, the Borda total for B is
60*3+40*4=340 total swaps.

The majority candidate A gets a lower Borda score than B!

Without the clones it would take 60 swaps to get A to the bottom of all
ballots, but only 40 swaps to move B to its faction bottom.

Note that the candidates get credit for moving through the crowd of clones.
This gives most advantage to the clone ranked bighest among its fellow

If Kendall-tau is decloned by weighting each swap with the product of the
first place scores of the candidates being swapped, then swapping A and B
yields a cost of 60*40 per swap ... while all other swaps yield zero.

So the total weighted swap cost of moving A to the bottom is 60*(60*40) ...
much greater than the total cost of moving B to the bottom ... 40*(60*40).

The ratio of the two costs is 60/40 ... the same as it was before the
clones were introduced.

Decloning Kendall-tau rectifies the distance distortion at the root of
Kemeny-Young crowding in a similar manner.

We have used the random ballot favorite lottery probabilities to declone
Kendall-tau. The probability distribution of any proportional lottery could
be used.

For example we could use the random implicit approval ballot lottery

This can be done in three ways ... counting approvals fractionally ... or
repeated drawings to narrow down to a winner ... or the Martin Harper
trick: all of B's probability (1 over the number of ballots) goes to the
candidate approved by B with the greatest approval among such candidates.

Here's another way to get approval cutoffs automatically ... on each ballot
B approve every candidate not outranked by any Smith candidate.

Then use one of the three methods in the previous paragraph to extract
proportional lottery probabilities from the resulting automatically
generated approvals.

That's enough for now!


On Tue, Jan 24, 2023, 4:14 PM Kristofer Munsterhjelm <km_elmet at t-online.de>

> On 1/24/23 22:32, Kristofer Munsterhjelm wrote:
> > I would say that it's not so much that clone winner is linked to
> > compromising and clone loser to burial, as that they're linked to
> > nomination incentive. For instance, with enough candidates in impartial
> > culture, Ranked Pairs and Schulze are plenty susceptible to burial, even
> > though they're cloneproof.
> >
> > (Though perhaps there is a more clear relation in say, a spatial model.
> > I don't know as I haven't checked.)
> A thought occurred to me: it might be that the reverse implication is
> true: that we can't have vote splitting clone failure without
> compromising incentive, and we can't have teaming without burial incentive.
> This seems intuitively right for Plurality and Borda: suppose for
> Plurality that A loses after being cloned. Then if everybody decides to
> rank A1>A2>A3, then that will make A1 win again; this is a compromising
> strategy for the A-voters. Conversely, in Borda, suppose that after
> cloning A, A1 wins; then in at least some elections, the B>A voters
> moving every A clone except A1 to equal last should make A lose again,
> which is a burial strategy. These countermeasures only work if the A
> voters or the not-A voters (respectively) hold a large enough share of
> the votes.
> But generalizing it to *every* method would be much harder.
> And there's the obvious question: if there are implications for
> vote-splitting and teaming, then what's the implication for crowding?
> You'd think nonmonotonicity (due to the chaos), but nope - Kemeny has
> crowding and is monotone.
> -km
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