## "Winning votes" versus "margins" – two kinds of Condorcet voting

### What they are

In "Condorcet voting methods" in which "equal rankings are permitted in votes," there are two natural ways to reckon the "strength" of a "pairwise victory": margins and winning votes.

Margins: The "margin" of A over B is the number of votes saying A>B, minus the number saying B>A.

Winning votes: If more votes say A>B than B>A, then A wins that pairwise contest, and the number of "winning votes" is precisely the number of votes saying A>B. Otherwise it is zero.

The distinction between these two concepts may sound minor, but it can have some profound consequences.

### Effect on a "favorite betrayal" scenario

Original scenario (honest votes)
#voters Their Vote
9 B>C>A
6 C>A>B
5 A>B>C

Here is a typical election scenario involving a Condorcet cycle. In this situation almost every election method declares B the winner. (Every candidate is defeated pairwise by one other, e.g. B is defeated by A. But since this is the weakest such defeat, B has the strongest claim to win.)

But the six C>A>B voters are strategically motivated to change their vote to the dishonest A>C>B, "betraying" their true favorite C in order to make their "lesser evil" A win. They could also try changing it to the less-dishonest A=C>B, which still leaves C ranked co-equal top and hence does not really betray C, but which still aims to boost support for their lesser evil A to make A win. Let us consider the consequences of these two strategic moves.

Strategic vote change fully-betraying C
#voters Their Vote
9 B>C>A
6 A>C>B
5 A>B>C

In this situation, the tallies are

Pairwise winning votes margins
A>B1111-9=2
B>C1414-6=8
A>C1111-9=2

So that A indeed now wins using either margins or winning votes – the betrayal worked. With the other strategic choice,

Strategic vote change partly-betraying C
#voters Their Vote
9 B>C>A
6 C=A>B
5 A>B>C

The tallies are

Strategic vote change partly-betraying C
Pairwise winning votes margins
A>B1111-9=3
B>C1414-6=8
C>A99-5=4

so that there is still a Condorcet cycle. In this situation reckoned by winning votes C>A is the weakest defeat, so now A wins. But reckoned by margins A>B is the weakest defeat, so that B remains the winner.

In other words, this less-dishonest form of strategic voting, not fully-betraying C, worked to elect the lesser evil A with winning-votes, but failed to change the winner with margins.

Full betrayal works works
Partial betrayal works fails

What is the moral? One moral is that with margins-based Condorcet voting, voters are motivated to fully-betray their favorites that are "third party candidates" with little chance to win. That could lead ultimately to two-party domination caused by third parties dying out. But with winning-votes, voters can often get away with a mere partial betrayal, which causes the third party candidate C still to have a significant number of top-rank votes, which hopefully causes third parties not to die out, avoiding degeneration into locked-in two-party domination.

However, winning-votes does not succeed in fully avoiding situations in which full-strength betrayal is the only strategic move that works. (Example.) It just makes them rarer. Specifically, in the random elections model in 3-candidate scenarios, i.e. where all 12 honest-vote types A>B>C, A>C>B, B>A>C, B>C>A, C>A>B, C>B>A, A=B>C, A=C>B, B>A=C, B=C>A, C>A=B, C>B=A are equally likely, situations where full-favorite-betrayal is the only strategic move that works, arise 74% of the time with margins, which is reduced to 67% with winning votes, in both cases when (and only when) there is a Condorcet cycle. And a Condorcet cycle happens in about one election in 19 in this model. (See puzzles 7 and 8.)

Therefore, it would appear that winning votes has a slight advantage over margins as far as avoiding favorite betrayal is concerned.

### But winning votes has some disadvantages

But before you rush to abandon "margins" and adopt the more subtle "winning votes" (wv) approach, you should take into account that wv also has some disadvantages.

#voters Their Vote
1000 A>B
1000 C>D

Consider this 2000-voter election by Juho Laatu (note, the candidates unmentioned in the votes are regarded as ranked coequal last, that is "A>B" is really "A>B>C=D"). There are two parties whose supporters rank only their own candidates (and leave others into to the default shared-last position). Winning votes seems to think that all candidates are about equal since one additional vote can make any one of the four candidates win (e.g. one additional "D>B" vote makes D the winner). Margins seems to think that A and C are the strongest candidates since some 500 additional votes are needed to make B or D win.

Laatu justifiably considers winning votes to act in a rather dubious way here!

Remark: It has been pointed out that "margins" is equivalent to count each "A=B" vote as equivalent to half A>B and half B>A. In other words, with winning-votes Condorcet, voters could get the effect of having margins Condorcet by flipping a coin whenever they planned to express an equality.

### Effect on the DH3 scenario

A simple DH3 scenario (honest votes). The notation "X,Y" means a mixture of X>Y and Y>X votes whose composition is not immediately known.
#voters Their Vote
37 C>A,B>D
32 A>B,C>D
31 B>A,C>D

In the 100-voter election above, there are three main rivals A, B, and C, as well as a "dark horse" D who is unanimously agreed to be the worst candidate. (There could also be additional dark horses, but for simplicity we consider the case where there is only one.) In this situation, we shall suppose it is not clear to the voters which is larger, 32 or 31, and it also is not completely clear which (X>Y or Y>X) is larger in the "X,Y" mixtures.

Now it depends on the mixtures, but if (e.g.) they are reasonably near equal, then C is going to be the Condorcet winner here (and C is definitely the plurality winner).

Now. What do the A- and B-voters think? They think: "C is probably going to win. We don't like that. But suppose we artificially score D above C..."

DH3 scenario with strategic votes by the A- and B-voters.
#voters Their Vote
37 C>A,B>D
32 A>D>B,C
31 B>D>A,C

Then the pairwise tallies are going to be:

DefinitelyA,B > D > C
ProbablyC > A,B

In which case we (probably) have a Condorcet cycle scenario. (It is actually two 3-cycles which share the common DC arc.) The weakest defeats in these cycles are C>A,B which means, under both all the most-commonly-used Condorcet rules (since they all are equivalent in the 3-cycle case) and Borda, that one of {A,B} is going to be the winner.

I verified that A wins in the 50-50 mixture case under Tideman ranked pairs, Schulze beatpaths, and basic Condorcet by using Eric Gorr's Condorcet calculator using this input

 ```37:C>A>B>D 37:C>B>A>D 32:A>D>B>C 32:A>D>C>B 31:B>D>A>C 31:B>D>C>A ```

In other words, this strategic move worked; in the original situation C won, but by this strategic burial-move, the A- and B- voters converted the situation to one in which either A or B would win. (But note under IRV and of course plurality voting, it didn't work – C still wins.)

The trouble is, the C-voters here are not going to be willing to sit still and take that. They've been deprived of their deserved victory. They feel the need to fight back with the same exaggerating method:

DH3 scenario with strategic votes by all voters.
#voters Their Vote
37 C>D>A,B
32 A>D>B,C
31 B>D>A,C

Then the pairwise tallies are going to be:

DefinitelyD > C,A,B

and D – the candidate unanimously agreed to be worst – now is the clear Condorcet and Borda winner – a disaster for everybody! (But note that with plurality, IRV, range, or approval voting, C still wins. Those voting systems are immune to the DH3 pathology.)

The DH3 pathology is a tremendous problem for all Condorcet methods and the Borda method. But now let us consider what happens if we allow equalities in the Condorcet votes. Then the A- and B-voters have the option of trying a less-dishonest form of strategic voting:

DH3 scenario with strategic vote "truncation" by the A- and B-voters.
#voters Their Vote
37 C>A,B>D
32 A>D=B=C
31 B>D=A=C

Then the pairwise tallies are going to be:

DefinitelyC > A,B > D

And C still wins. (And note: C wins no matter what the "A,B" mixture of A>B, A=B, and B>A votes is, and no matter what Condorcet variant you use: winning votes, margins; Schulze, Tideman ranked pairs, or basic.) So what is the moral? The moral is that this lesser form of strategic dishonesty did not work for the A- and B-voters. Those voters will therefore still go for full dishonesty, pretending D>C and not merely that D=C, because that works. In other words, neither margins, nor winning votes, cures the DH3 pathology. The DH3 pathology hurts every form of Condorcet voting.

### Conclusion

"Margins" and "winning votes" are two ways to handle Condorcet voting methods in which equalities in votes are allowed. Neither succeeds in avoiding the horrific DH3 pathology. Although winning votes is better than margins for avoiding the favorite betrayal pathology, it does not completely succeed, and winning votes causes some strange and dubious voting-system behavior which margins avoids.