Election Reform
One major disadvantage of our present "Winner Take All" (also called Plurality and First Past The Post) method of voting was the spoiler effect illustrated in the 2000 Presidential elections. Ralph Nader's entry into the race "spoiled" it for Al Gore - Nader has no chance of winning, but his appearance on the ballot meant that an election Gore would probably have won instead went to George W. Bush. (Side note: I am not comparing the relative merits of the three candidates - I voted Libertarian <grin> - I'm just comparing their relative strength in the electoral college, an antiquated system we should probably abolish). The other side of the "spoiler" effect is the "lesser of two evils" effect: if I really like Nader, think Gore's a bore, and believe George W. Bush runs around in brimstone-scented red leotards poking sinners with a pitchfork, I would be forced to vote for Gore just to keep "W" out of the White House.
In addition, when you have three or more candidates in a race, it is perfectly possible to have a person who is hated by a majority of the people end up as the winner. Let's say we have three candidates: Mussolini, Eisenhower, and Stalin. Let's also assume that Mussolini and Stalin are far apart on the political spectrum (far right and far left, respectively), and that Mussolini is slightly closer to Eisenhower than Stalin. In this hypothetical election, Mussolini gets 30% of the vote, Eisenhower also gets 30%, and Stalin receives 40%. 100% of Mussolini voters prefer Eisenhower to Stalin, and 100% of Stalin voters prefer Eisenhower to Mussolini. 100 percent of Eisenhower voters, since they are farther away from Stalin than Mussolini, choose the latter over the former. Stalin wins under plurality, even though 60% of the voters rank him dead last. Most people would agree that Eisenhower should win, since he would win any 1 on 1 election with either Mussolini (70% to 30%) or Stalin (60% to 40%).
What properties should we look for in a voting system? Kenneth J. Arrow came up with the following reasonable criteria:
Unfortunately for us, Arrow then goes on to prove that no voting method can follow all criteria, a proof which won for him a Nobel Prize in Economics in 1972. We could throw up our hands and say that, since no voting method is perfect, we might as well keep plurality voting. Fortunately for us, while Arrow's Proof showed that no voting method can fulfill all of his criteria, it did not prove that all methods are equally bad (or good).
There are several dozen methods that have been proposed in the last two hundred years, ever since Jean-Charles de Borda and Marquis de Condorcet argued about the best method at the French Academy of Sciences in the 1790's. In fact, two of the best methods are named for these early political scientists (Borda was actually a Physicist and Condorcet a Mathematician).
In alphabetical order, the most-studied methods are:
As mentioned before, there are literally dozens of different methods. Many use Condorcet as the core, and use different tiebreakers in case there is a circular tie. Others use Borda, but with different weights on the rankings. Still others choose losers to throw out (least number of first place votes, greatest number of last place votes, Condorcet or Borda losers, or even stranger criteria), and keep recalculating until they only have one candidate remaining. Strangely enough, very few people have tried combining methods, other than for Condorcet tiebreakers. If you have method A and method B, and neither method is perfect, why not calculate using both methods? Assuming both methods are pretty good, if they both choose the same candidate, that candidate is probably the right choice. If they choose different candidates, a simple plurality comparison would be enough to choose between them, and you would (more than likely) have chosen the right candidate.
To show this mathematically, suppose we have two distinctly different methods of calculating the winner of an election, A and B. Method A is 90% accurate at choosing the winner of an election. Method B, likewise, is 90% effective. In both cases, the error is spread randomly. We have candidates X, Y, and Z. By some mystical method, we determine that X should be declared the winner (this is so we have something to compare our results to). Method A and B choose X 90% of the time, and choose Y and Z 5% apiece. Using both methods together, we get A and B choosing X 81% of the time, wrongly choosing Y 0.25% of the time, and wrongly choosing Z 0.25% of the time. Of the remainder, 18% of the time we have one method choosing the correct answer, and 0.5% neither method choosing the right answer. Combining the two methods, we are almost certain - 81% to 0.5% - that when both methods agree, we have chosen the correct winner. Randomly choosing between choices when they disagree brings us back to the 90% range (81% + (18/2)%) - seemingly no gain for a lot more calculation. The improvement comes when we consider that even plurality chooses the correct winner more often than random chance, and all good voting methods simplify to plurality when considering two candidates. We take the two possible winners; if we have a 90% chance of choosing correctly between them, then 0.9x0.18=16.2%. Therefore our combined method could very well have a 97.2% chance of correctly choosing the right winner.
If someone were interested in voting methods and had good programming skills, one interesting "computational experiment" would be to test combined election methods, with a plurality tiebreaker if necessary, and then look at all the instances where the sets disagreed. For example, you could take 12 candidates and 5040 voters (both numbers are highly composite, which would allow certain combinations to simplify). Or if you wanted one more representative of a national election, 120 candidates and 73,513,440 voters is another set of highly composite numbers, although in that case you would have to run so-called "Monte Carlo" trials or a quantum computer to figure it out. The best voting "set" would be one where the two combined methods (with plurality tiebreaker) consistently choose the "best" candidate. Analyzing where groups disagreed would quickly show which methods fit together best.
Unfortunately, I don't have the programming skills to run such tests. However, it is easy enough to figure out what kind of ballot would be needed to cover
every possible voting method. In fact, such a ballot is easier to define than it is to vote on or count! If you construct a range voting ballot and add a cutoff line
between acceptable and unacceptable candidates, such a ballot would suffice for every voting method. You also allow ties and truncation. For plurality, just take
the top-scoring candidate (if more that one shares the top spot on a ballot, either consider the ballot "spoiled" or share the vote between the candidates). For
approval, just take all the candidates above the cutoff line. For Condorcet and IRV (and variants), just order the candidates from highest to lowest. For Borda,
order from highest to lowest, and if two or more candidates share a ranking, average the score given them (for example, if the ranking was A, (B,C), D, then A
would receive 3 points, B and C would each receive (2+1)/2 or 1.5 points, and D would receive 0). If you are using range voting, just add the scores.
The CAB Voting Method
The ballots have all been cast, and now we have to choose a way to tally the votes. As far as I know, I am the first person to come up with the CAB method of voting, which combines my three favorite methods. For a single winner election using the CAB method (it would differ slightly if there were two or more seats to fill), we would first check to see if there was a Condorcet winner (the C in CAB). If so, that candidate is declared the winner. If there were no Condorcet winner, then we would take Smith set (the smallest set of candidates where every member of the set beats every member outside the set, beats at least one other member of the set, and loses only to other members of the set). We then use Approval and Borda (the A and B in CAB - I really wanted to call it the ABC method, but I use them in a different order) methods on the members of the Smith set, generating rankings from first to last place using both. If Approval and Borda agree on the winner, then that candidate is the winner. If they disagree, then the two candidates are compared against each other using Plurality, and the winner is chosen as the overall winner. Another way of breaking ties is to drop down one place at a time on both the Approval and Borda lists. The first candidate to show up on both lists is the winner - and if there are two then run the Plurality test. (There are still other methods of breaking ties, but they are all more complicated, and I would not use them unless computer simulations or real-life elections indicated they would be more accurate. My guess is that most would not be more accurate, just more complicated.)
In order to add Approval and Borda to the mix, we need to make a couple of changes. All candidates that are not marked (a truncated ballot) are considered below the cutoff line and tied with each other. If the entire Smith set is tied on a ballot, either above or below the approval line, it is not counted. If a voter does not have any member of the final Smith set above the cutoff, move the voter's top-rated Smith set candidate(s) above the cutoff. If all of the members of the Smith set are approved on a voter's ballot, move the lowest-rated Smith set candidate(s) below the cutoff line. The ballots are then counted for Approval. For the Borda count, if candidates are tied - either marked, or via truncation - then they share the average of the appropriate scores remaining. For example, I have candidates A, B, C, D, E, F, G, and H. I rank A highest, B and C are tied, D is ranked lower, and E-H are unranked. In this case, A would get 7 points, B and C 5.5 points apiece, D 4 points, and E, F, G, and H would each get 1.5 points apiece. Since I would only Borda Count the members of the Smith set, there probably will be far fewer than eight candidates to worry about.
This method is admittedly complex, but it can be easily implemented, and the vast majority of winners chosen by the method would agree with common sense. To simplify the explanation, run an election with the appropriate ballots. If there is a Condorcet winner, great - that person is elected. If not, take the Smith set of the candidates, and find the Approval and Borda winner. If they agree, great - that person is elected. If they disagree, then a simple plurality test between the two winners will determine the overall winner. Most elections will probably have a Condorcet winner, and the only times things will be complicated is when the election is close - which is the point where the method of determining the winner needs to become more precise. The winner chosen by this method will be arguably the best pick of all the candidates in any reasonable election.
Now that we have a method for single-winner elections, what about elections for multiple candidates (like some elections for judges, and how senators should be chosen)? We can modify the existing CAB method. First, we have to define two particular Condorcet sets to choose the winners from. Both sets, like the Smith set, will be winner sets - every candidate in each set beats every candidate outside the set. The inner set will be the largest winner set that does not have more candidates than the number of positions being filled (C<=N, where C is the number of candidates, and N is the number of positions). The outer set will be the smallest winner set that has at least as many candidates as the number of positions being filled (C>=N). If the inner and outer set are the same, then those candidates are the ones chosen (it's possible that the inner set will be a null set, but that just means the outer set is the only one determining the winners, it does not mean the method "blows up").
If the inner and outer set are not the same, then things get interesting. Every candidate in the inner set is guaranteed a position - they are an unbeaten group smaller than the number of positions available, and they beat every candidate in the gap between inner and outer sets. Those candidates between the inner and outer sets are thus fighting for the remaining positions: Positions(outer) = Positions(total) - Positions(inner). Rather than throw away the inner group after electing them, we want to keep them as placeholders for the Borda count - it's one thing to throw irrelevant alternatives away and quite another to throw relevant alternatives away. People who voted for a "gap" candidate above a winning candidate should have that vote count. We'll take the outer set and rank each candidate via Approval and Borda count. Approval is conducted pretty much as in the single-winner race - if there is a tie among all candidates in the gap the vote is not counted, otherwise the approval line is moved as little as possible so that at least one gap candidate is approved and at least one gap candidate is disapproved.
Now that we have lists for both Approval and Borda (and finally remove the placeholders), we count down the number of remaining positions from the top of each list. For example, if we have four spots left, we count down four spots on the Approval list and then four places down on the Borda list. Candidates that appear on both the Approval and Borda sublists are chosen - if all four (or however many) spots are the same, then those candidates are elected. If there are discrepancies between the lists, we could drop one place at a time until an appropriate number of candidates appear on both lists - and if there were a tie in the last candidate chosen, then run a Plurality test between them to see who gets the position. (One could of course run Plurality tests on all of the discrepancies in the first n places of both lists, where n is the number of positions remaining, but if there are more than two it becomes an ordeal.) Still another way would be to drop down the lists one place at a time until the number of different candidates on the two lists was at least equal to the number of positions remaining (you might have one candidate too many with two lists, but that's it). If you go over, you would first remove all the candidates that appeared on both lists - they would already have won. Of the remainder, we want to find the Condorcet loser and drop him; if there is no Condorcet loser, we can drop the Plurality loser, and fill the remaining positions.
Is the CAB method the best method of determining whom the voters want? Warren D. Smith would emphatically disagree, since it would make voting strategy intricate and mysterious, making it hard to test the effect of "strategic" voters (voters who rank candidates other than by honest preference in order to achieve a desired result) very difficult. He would argue that Range Voting is the natural choice. Dr. Donald G. Saari would disagree as well - his geometric arguments showing Borda to be the voting method of choice are quite interesting, although I feel it does not provide enough protection against strategic voting (Borda's exclamation that his method "is intended only for honest men!" rings true). Proponents of Instant Runoff Voting would ignore such a method with all of the intensity of person who, having finally decided he understands one voting method, does not want to put the effort into learning others that may be much better (I'm not being immodest here, many voting methods are better than IRVing). There are many proponents of Condorcet plus tiebreaker methods, which in most cases will choose the same winner mine does (for example, Black's method chooses the Condorcet winner, if any, and uses a Borda count if no Condorcet winner exists. Other than the explicit mention of the Smith set and the addition of an Approval cross-check for the tiebreaker, it is extremely similar, and should result in nearly-identical results except in rare cases).
So what properties should a good single-winner voting method have? In my opinion, it must select the Condorcet winner, if one exists. If a Condorcet winner does not exist, then it must select a winner from the Smith set. Since the Smith set generally contains more candidates than there are positions to fill, there has to be at least one tiebreaker method - and if there is more than one tiebreaker method, there has to be some method to choose between possible discrepancies. With two or more tiebreakers, it is important that each method differs in its "slant" or bias in selecting winners - that way, if the choice is unanimous, it almost certainly is right, and if it is split the winner will almost always be chosen by one method or another. It's always easier to select a winner if it is still part of a set rather than already discarded from it. To find the best tiebreakers, we could run Monte Carlo simulations using each method individually, then pairs of methods together with different discrepancy pickers, then possibly three or more tiebreaker methods and one or more discrepancy pickers. I would ignore the cases where methods chose the same winner (except perhaps as percentages) and look at where different winners are selected. If there is a set of methods that tends to choose better winners than the CAB method, then that method should be selected - maybe Range and Borda, Range and Nanson, or Nanson and Approval, cross-checking with Plurality.
Multi-winner voting methods should have similar constraints: Every member of the largest inner unbeaten Condorcet set should be elected, with the remainder of those elected chosen from the smallest outer unbeaten Condorcet set. In reality, then, we can restrict the action of "good" election rules on the candidate set C(outer)-C(inner), with no information at all from irrelevant candidates in the Condorcet loser set (we still might use information from the inner set of automatic winners, however). Depending on the difference in size between the inner and outer winning sets, the number of possible discrepancies may be quite small. It would be an interesting avenue of study to generate statistics on various voting choice methods acting on this group, and find the method or group of methods that tended to produce a more reasonable result. The modified CAB method above should produce a reasonable result, but perhaps (as in the single-winner method above) a Nanson-Approval, Range-Nanson, or Range-Borda may have more desirable characteristics.