Condorcet.orgA method called Borda is defined as follows. Each last place vote is worth no points. Second to last is worth 1 point, third to last 2 points, and so on, up to the candidate who is ranked first, who is given 1 fewer points than there are candidates. In short, each ballot gives each candidate one point for every candidate ranked below it.
The basis for Borda is similar to the basis for Average Ratings. That is, we are attempting to maximize voter satisfaction. Obviously, voter satisfaction with the outcome declines as one goes down the ballot. So, the number of points given for each rank declines as well. It is assumed to be a benign simplification to make the drop in satisfaction equal between each rank. That is, a person may have a greater preference for first over second than third over forth, but since only ranks are provided, Borda assumes the drop of satisfaction to be equal.
This assumption may seem harmless, but it actually has a surprising negative effect on the method. Consider the following example
60 A B
40 B A
Because there are only two candidate, the method proceeds like plurality (or pretty much any method). A wins with 60 points to B's 40. The total point score for a candidate, incidentally, is called the Borda Count.
Let's imagine however, that a candidate very similar to B runs as well. This candidate will be called C.
50 A B C
10 A C B
35 B C A
5 C B A
Now the Borda Counts are
A:120 B:125 C:55
So B wins. Notice that there was no more support for the basic ideological perspective that B represents. The A voters still all voted A first. The lesson is that Borda will often give the election to a group or political party because it runs more candidates. In contrast, Ranked Pairs chooses A in both examples.
This result is caused because Borda assumes that the ABC voters prefer B to C exactly as strongly as they prefer A to B. In the above example, this is probably not true. In fact, they may see B and C as almost identical, but because of this assumption they are tricked into giving support to B.
If we want to try to deduce relative preference from a ranked ballot,
we have to make some assumptions. Ranked Pairs avoids this by
not attempting to deduce levels of preference. To Ranked Pairs,
it is only important how many people prefer A to B, not how strongly
each person feels this way. For a thourough defense of this position,
see the Ratings Method.
60 A B
40 B A
A wins here, but what if a new candidate runs with very similar views to A.
35 A C B
25 C A B
30 B A C
10 B C A
Here, C and A are splitting the vote. B wins. In contrast, Ranked Pairs chooses A in both examples.
Here, the assumption being made is that people only care about their
first choice. This would justify giving 1 point to the first rank
and 0 to all others. However, in the above example, voters view A
and C as very similar, so this assumption proves unjustified. So
much has been written about the inadequacies of plurality that I won't
bother to repeat all its negative characteristics here.
Consider the following votes:
3 A B C
3 B C A
3 C A B
If we look at the majority preferences, we have A>B, B>C, and C>A.
The conclusion reached by some supporters of Borda is that these are crazy,
mixed-up, garbage votes. They don't make any sense, and therefore
a sensible method will discard them. Since the Borda Count from these
votes is
A: 9
B: 9
C: 9
they will cancel each other out and not have any effect on the outcome.
In Ranked Pairs, on the other hand they very well could affect the
outcome. But consider the following. There is no reason to
believe that these voters are insane or that these votes are random.
What we do know, is that some of these voters are wrong. However
this is usually the case. As well, we know that a majority is wrong
about one of the statements A>B, B>C, and C>A. However, majorities
are often wrong, so this should not be too upsetting. So, the question
is, is it possible that these votes could be used as evidence of the correct
winner, when part of a larger election.
Let me give an example to consider when thinking about this. Imagine that I place three cards face down on a table. One is a Queen, the other two are Aces. Now, I tell you that for each card, you can ask whether or not it is the Queen, but there is a catch. I lie exactly 10% of the time. Now, lets say that I claim that the left and rightmost cards are Queens. This is contradictory since there is only one Queen. However, contradictory information is not necessarily useless garbage information. You have increased your knowledge about where the Queen is most likely to be.
Now, consider the above example. We have evidence in favour of each of the three statements A>B, B>C, and C>A. It is quite true that this evidence is contradictory. It is even true that by itself it doesn't help in determining a winner. But it isn't true that we have learned nothing. It is reasonable to conclude, for example that A>B is more likely than B>A. That fact could be useful if these votes were combined with more information from other votes.