# [EM] A Condorcet-like method that satisfies FBC (I believe)

Kevin Venzke stepjak at yahoo.fr
Wed May 18 15:14:24 PDT 2005

```Hello,

The method I'm going to describe here is a three-slot method.
You wouldn't have to use three slots, but I suggest to for two
reasons: First, each slot will have a specific meaning. Second,
I note that even ranked ballot methods which permit strict
ranking of all the candidates tend to require the voter to make
approval-like decisions at times: Whether to rank top candidates
equally, whether to rank another candidate at all, etc. So I
don't see that the limitations resulting from the use of three
slots are necessarily alleviated by using more slots. This just
allows the voter to falsely believe that he can rank freely
and has no need for strategy.

So we have a three-slot ballot. I asked myself first, how would
I want my ballot to be interpreted? If I vote A>B>C, it means
that I want to vote A>B and A>C pairwise *if* A "can win." I
don't want my ballot to create an A>B vote which causes B to
lose without making A win. In that case I would rather just vote
AB>>C, approval-style; when everyone does this, it's just approval,
so that will be the second of two stages: The method I'm defining
is Something//Approval.

What can we use for the first stage? That is, what does it mean
if A "can win"? An easy answer is Condorcet: If there is a CW, he
wins; else we go to the second stage with all the candidates.
Condorcet is particularly a nice choice because, when A is the CW,
it means that the second-stage compression of voters preferring X
to A could not cause A to not be the CW. (That is, changing a
ballot from B>C>A to BC>>A can't reverse any of A's pairwise wins.)

This method, three-slot Condorcet//Approval, already makes a lot
of sense. If we imagine there to be two rounds of voting, then first
you use three slots, hoping that one of your top-slot candidates
will win outright. If not, you get to revise your vote: Believing
your top-slot candidates to be not as viable as hoped, you compress
the top and middle.

(I want to note that something which would *not* make sense in this
way would be for there to be four slots, with the fourth slot
compressed with the third slot in the second stage: Compressing your
bottom two slots can only turn a bottom-slot candidate into the CW,
and so is not a reasonable reaction to the first stage results.)

We can do one more thing to ensure that our vote is interpreted
well. Suppose we want to vote ABC>DEF>G. Suppose there is no
Condorcet winner, and D or G wins in the second round. Suppose
that B would have been the Condorcet winner, except that he was
beaten pairwise by A, so that if you had put A in the middle slot,
B would have been the CW. That would be an annoying thing for you
to have to consider when voting.

This can be fixed. Rather than, in the first stage, electing a
candidate who beats all others pairwise, narrow the field down
to all candidates who do not have any pairwise *losses* (if such
candidates exist). Then, consider it to be a ("special") pairwise
*tie* between some X and Y if the number of voters ranking X and Y
equally in the top slot, plus the number of voters for the losing
side in the X:Y contest, would exceed the number of voters for the
winning side in this contest.

What's the reasoning for this? The voters ranking X and Y
equally in the top slot would gladly vote however necessary in
order to create a pairwise tie between the two. If these voters
plus the supporters of the pairwise loser outnumber (or equal)
the supporters of the pairwise winner, then there are enough of
these voters to create a pairwise tie, if only they had had

It's this rule which allows FBC to be satisfied, I believe. To
prove it, I think it's *almost* adequate to show that raising
some candidate A to the top slot (due to being the voter's real
favorite) cannot turn some other top-slot candidate (on this ballot)
into a first-stage loser, and cannot turn any candidate below the
top slot into a first-stage winner.

Let v[a,b] signify the number of votes preferring A to B.
Let t[ab] signify the number of voters ranking A and B equally in the top
slot.

1. Raising A to the top slot can't turn B, also in the top slot,
into a first-stage loser:
Note the only affected contest involving B is the A:B contest. B
can be made into a first-stage loser, then, only if it becomes the
case that B has a pairwise loss to A. If B doesn't have such a loss
to begin with, that means that v[b,a]+t[ab]>=v[a,b]. Raising A equal
to B decreases v[b,a], but it also increases t[ab] by the same amount,
so that the condition can't become false.

2. Raising A to the top slot can't turn C, below the top slot,
into a first-stage winner:
Since the only affected contest involving C is A:C, it would have
to be the case that raising A causes C to no longer lose pairwise
to A. But raising A can do nothing but increase v[a,c] and decrease
v[c,a].

3. It's possible that the vote B>AC>D results in C winning in
the first stage, while AB>C>D results in D winning in the second
stage. This could happen if with the latter vote, C has a pairwise
loss to A (and D is the approval winner), while in the former case,
C has no losses. However, the voter could simply raise C and vote
ABC>>D to avoid this, meaning that this is not an FBC failure.

And of course, Approval satisfies FBC, so I think there is no need
to consider the second stage.

I'm interested to hear if anyone thinks there is an FBC failure
in here that I haven't noticed. If I'm right, that this satisfies
FBC, then surely this is the most complicated method yet to do so.

For clarity, here is a brief definition of the method I'm suggesting:

The voter places each candidate into one of three slots.
v[x,y] is the number of voters voting X over Y.
t[xy] is the number of voters ranking X and Y together in the top slot.
Define a set S containing every candidate Z for whom there is no other
candidate W such that v[z,w]+t[zw]<v[w,z].
If S is empty, then S contains all the candidates.
Elect the member of S who is in the first or second slot on the most
ballots.

(I admit that it's weird to have everyone compress in the second
stage in the same way, regardless of who is in S. In ordinary
Condorcet//Approval, the second stage never occurs unless S contains
all the candidates. But I don't see any other neat way of picking a
winner from S, especially if S contains more than two candidates.)

Kevin Venzke

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