[EM] A set of clone independence criteria more applicable to multiwinner
Kristofer Munsterhjelm
km-elmet at munsterhjelm.no
Tue Oct 7 06:35:39 PDT 2025
Here's a way to split up clone independence so that it should be
applicable to multiwinner. Define three criteria:
Clone-add: Replacing some candidate X in an election with a number of
new candidates X1..Xn, where every voter ranks them consecutive in order
X1>X2>...>Xn where they used to rank X, should not change the win
probability for any candidate outside X. (This captures some teaming,
e.g. in Borda)
Now, let S be a clone set: a set of candidates with the property that
nobody ranks a candidate not in S between candidates in S. Then
rearranging the order that the candidates in S are ranked on one or more
voter's ballots should not:
1. Change the probability that a winner comes from S ("Clone-reorder")
2. Change the probability that a particular candidate X is elected
instead of Y, with both X and Y being outside S ("Clone-no-crowding").
For simplicity's sake, you could collapse both to:
(unnamed) ... "should not change the probability that X is elected, for
any X outside of S",
but I think it's more useful to keep these criteria separate because
they allow us to distinguish vote-splitting and the remaining teaming
incentive from crowding.
Clone-add is incompatible with most multiwinner notions. Consider
something like:
100: A>B>C
two seats. A and B are elected. Then clone A:
100: A1>A2>A3>A4>B>C
Now A1 and A2 are elected.
But suppose we start with:
100: A1>A2>A3>A4>B>C
The largest clone set S is {A1, A2, A3, A4}. Rearranging them doesn't
change the probability that the winner comes from S (which is 100%). Nor
does it change the relative win probability of B and C (they're both
zero and so strictly speaking undefined). The same holds (I think) for
all other clone sets, e.g. suppose we let S = {A2, A3}, then rearranging
these doesn't change the probability that the winner comes from S (50%),
nor does it alter the probability for anyone outside (100% for A1, 0 for
the rest).
So clone-reorder and clone-no-crowding might just be compatible with the
DPC.
The intuition is that it's adding candidates that makes clone
independence troublesome for multiwinner, because some candidate might
be exceeding the quota by enough that he could have two seats if he were
to split himself in two. But that's all clone-add. On the other hand,
vote-splitting happens when clone orders are too evenly distributed, so
that each clone gets too few points; that's covered by clone-reorder.
The weak point of this rearrangement is that clone-add is not entirely
irrelevant to what we're trying to prevent with clone independence. E.g.
single-winner Borda:
2 1 <- points
1: A>B
1: B>A
is a tie, then
3 2 1 <- points
1: A1>A2>B
1: B>A1>A2
A1 gets 5 points, A2 gets 3 points, B gets 4 points, so A1 wins.
However, this is detected by clone-reorder, so not all is lost. S = {A1,
A2} so suppose we reorder A1 and A2 on the second ballot:
3 2 1 <- points
1: A1>A2>B
1: B>A2>A1
Everybody gets four points and we have a three-way tie, so this changed
the probability that the winner comes from the clone set {A1, A2} from
100% to 67%.
67% is higher than 50%, so there's still some teaming effect merely by
adding candidates, and by not applying clone-add to multiwinner, we
appear to lose some teaming detection power.
I don't think this is an artifact: "ordinary" clone dependence is a
proxy for exit and entry incentive, and in the unanimous A>B>C
multiwinner situation, there *is* a real incentive for candidates allied
with A to enter. The only way to remove that is to give winners weights,
party-list style; in which case A would get a weight of one and B and C
would get zero, and then after cloning A1 would still have a weight of
one and everybody else zero.
But for multiwinner methods without weights, we may want to keep this
"true, unavoidable incentive" from obscuring method-specific failures.
The classification above should do that.
(And who knows, maybe whenever there is a clone-add failure for Borda,
there's also a clone-reorder failure? If so, we may be losing less
distinguishing power than we think!)
-km
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