by Michael Ossipoff
August 17, 2012
In this article, I would like to compare voting systems by strategy and criteria. I’d like to add to topics that I have already discussed on Democracy Chronicles. I might briefly refer, by name, to criteria, strategies or voting suggestions that I’ve mentioned in previous articles, while adding more of them here. Like before, I will start by introducing a table of contents for this article. It lists only the new items:
Table of Contents:
1. Some Criteria
a) Simpler FBC
b) Later-No-Harm and Later-No-Help
d) Smith Criterion
e) Condorcet Loser Criterion
2. Comparisons of a Few Methods by the Above Criteria:
b) Symmetrical ICT (introduced in this article)
c) Ordinary Unimproved Condorcet
d) Instant-Runoff (IRV)
3. The Unacceptable/Acceptable (u/a) Election and Its Strategy
4. Strategically Dealing With the “Chicken-Dilemma” in the Point Systems
(a) A Simpler Favorite-Betrayal-Criterion (FBC)
Let’s get started. To vote a candidate at top is to not vote anyone over that candidate. If no one wins other than the candidates you’ve voted at top, then raising an additional candidate to top on your ballot shouldn’t cause any candidate who is not voted at top on your ballot to win. That aforementioned wording is intended to cover the possibility of ties. But if we can assume that ties will be solved randomly to choose just one winner, and that we’re referring to that one winner, then the wording can be simplified further.
There is also a simpler FBC wording for randomly solved ties. If the winner is a candidate that you have voted at top, then raising an additional candidate to top on your ballot should not change the winner to a candidate who is not at top on your ballot. Henceforth, when I say “FBC”, I’m referring to the FBC that I’ve defined above. FBC is met by ICT, and by the point systems. The point systems are methods in which the voter assigns candidate a number of points from 0 to N. The value of N depends on the particular point system. The candidate receiving the most points wins.
Approval is the 0-1 point system, where N = 1. When N > 1, the point system is called “Cardinal Ratings”, or “Range Voting” (“Range” for short), or “Score Voting” (“Score”). Right now, “Range” is a popular name for point systems with N>1, but “Score” is the one now being used by advocates of those methods. Several popular “Score” versions are: 0-2 (sometimes referred to as the “Three-Slot Score Version”), 0-10 and 0-100. There will be more about those methods later.
I’ve already discussed the Condorcet Criterion (CC), in previous articles, therefore I will not go into greater detail other than to state that I’ve discussed why ICT (defined in previous articles) can fairly be said to meet CC.
b) Later-No-Harm (LNHa) and Later-No-Help (LNHe):
Adding more candidates to your ballot, voted below the ones (set S) that you’re already voting for–meaning that you’re voting S over them, and voting them over the other candidates–should never change the winner from a candidate who is in set S to a candidate who is not.
Adding more candidates to your ballot, voted below the ones (set S) you’re already voting for–meaning that you’re voting S over them, and voting them over the other candidates–should never change the winner from someone not in S to someone in S.
[end of LNHe definition]
ICT is a Condorcet method. Condorcet methods don’t meet LNHa. Approval is a point system. Point systems don’t meet LNHa. What meets LNHa? In a previous article I have discussed why Instant-Runoff (IRV) should not be recommended.
In previous articles, I’ve described what I have referred to as the “Co-operation-defection problem”. Actually, that problem can take a number of forms, and the form that it takes in voting situations is more specifically referred to as the Chicken-Dilemma, which I have defined and described in previous articles (as well as why ICT avoids that problem.)
That Chicken-Dilemma is the worst LNHa failure. I suggest that, by not having the Chicken-Dilemma, ICT doesn’t meaningfully fail LNHa, though it does technically fail it. As for compliance with LNHe, Condorcet methods, in general, including ICT, don’t meet LNHe. The point systems do meet LNHe.
(X>Y) is the number of voters ranking X over Y.
(Y>X) is the number of voters ranking Y over X.
(X=Y)T is the number of voters ranking X and Y both in 1st place.
“iff” means “if and only if”
“>”, by itself, means “greater than”
Unimproved Condorcet says:
X beats y iff (X>Y) > (Y>X)
X beats Y iff (X>Y) > (Y>X) + (X=Y)T
Remainder of Definition for ICT:
If only 1 candidate is unbeaten, then s/he wins. If all candidates, or no candidates, are undefeated, then the winner is the candidate who is ranked in 1st place on the most ballots. If some, but not all, candidates are undefeated, then the winner is the undefeated candidate who is ranked in 1st place on the most ballots.
As I said, Symmetrical ICT meets LNHe. And, like ordinary ICT, Symmetrical ICT meets FBC and is defection-resistant. So, in regard to FBC, LNHa and LNHe, Symmetrical ICT has the criterion compliances of the point systems, because it passes FBC and LNHe, but (technically) not LNHa.
No method meets all criteria, or has all of the desirable properties. However, I believe that meeting FBC and having defection-resistance are the important properties because those properties avoid the worst strategy problems–the strategy problems that can cause voters to seriously distort their preferences–resulting in disastrous effects on society.
I note here that Symmetrical ICT (at least for the criteria that I’ve mentioned) has everything that IRV has (because ICT and Symmetrical ICT–effectively, if not literally–meet LNHa), but additionally meets FBC and CC, as does ordinary ICT.
c) Majority Criterion (MC):
If a majority of the voters vote candidate X over all the other candidates, then X should win. MC is met by all of the proposed methods, including Plurality, the point systems, unimproved Condorcet, ICT, Symmetrical ICT, and IRV.
Smith Criterion: The Smith set is the smallest set of candidates, such that everyone in the set beats everyone outside of the set. The Smith criterion says that the winner should come from the Smith set. It probably isn’t possible for the Smith Criterion to be met by a method that meets FBC and is defection-resistant. In fact, the Smith Criterion is probably incompatible with FBC.
Likewise for “Condorcet Loser”, (defined below): The Condorcet Loser Criterion (CLC) states never to elect a candidate who has been defeated by all of the others. Now, upon first impression, failing that criterion might sound terrible. But how bad, really, can the Condorcet loser who is the favorite of the most people be? Or, who is acceptable to the most people? I suggest that, that would be a remarkably popular Condorcet loser.
For that reason, the fact that the point systems, ICT, and Symmetrical ICT fail Condorcet Loser doesn’t mean much, if anything at all.
2. Comparison of Some Methods by Those Criteria
FBC is met by the point systems, ICT, and Symmetrical ICT; FBC is failed by unimproved Condorcet, and by IRV. LNHa is met by IRV. But, due to their defection-resistance, ICT and Symmetrical ICT don’t meaningfully fail LNHa; they effectively meet LNHa, though they don’t technically meet it.
No Condorcet method technically meets LNHa. Unimproved Condorcet fails LNHa in a serious way, due to lack of defection-resistance. LNHe is met by the point systems and by Symmetrical ICT. IRV meets LNHa and LNHe. Because IRV meets LNHa, it is strongly defection-resistant. But IRV fails FBC and CC. IRV’s FBC failure is particularly flagrant and frequent, making IRV unacceptable.
3. The Unacceptable/Acceptable (u/a) Election and its Strategy:
To repeat, a u/a election is an election in which there are unacceptable candidates who could win. More specifically, in a u/a election, there are 2 sets of candidates, sets A and U, such that the merit difference within those sets is negligible in comparison to the merit difference between those sets. In other words, the only important thing is to try to ensure that the winner comes from A instead of from U.
I claim that our public political elections are u/a elections. Strategy is simpler in a u/a election. In Approval, approve all of the acceptables, but none of the unacceptables. In general, in the point systems, give maximum points to the acceptables, and minimum points to the unacceptables.
In Symmetrical ICT, rank all of the acceptables in 1st place. Don’t rank any of the unacceptables. (That amounts to bottom-ranking all of the unacceptables, ranking them in last place). In unimproved Condorcet, it’s very difficult, much worse: Though the need to top-rank all the acceptables remains, there is a risk: Moving a candidate to top can cause all of your top-ranked candidates to lose, and elect someone much worse.
No voting system is perfect. No voting system meets every desirable criterion. The point systems and Symmetrical ICT have the simplest u/a strategy, as described above. Now, there is one thing that can spoil the simple strategy of the point systems: The Chicken-Dilemma. Before I continue about that, let me re-emphasize that ICT and Symmetrical ICT don’t have the Chicken-Dilemma problem, because they are both defection-resistant (as described in previous articles).
Let me repeat my description of the Chicken-Dilemma in voting. Say (for simplicity) that there are 3 candidates, A, B, and C. The A voters and the B voters greatly prefer both A and B to C. In fact, they all despise C. The A voters and the B voters add up to a majority. It’s a u/a election, and C is the unacceptable. The A and B voters should therefore give maximum points to A and B, and minimum (zero) points to C.
But the problem is that the A voters and B voters don’t trust each other. What if the A voters give max to B, but the B voters give 0 to A? Then B wins, even if A’s faction is larger than B’s faction, merely because the B voters have taken advantage of the A voters’ willingness to co-operate. As I’ve described in previous articles, that isn’t a problem for ICT or Symmetrical ICT.
But it sounds like a serious problem for the point systems, including Approval. Actually, though, it turns out be well solvable, easily dealt with in all of the point systems. There are several solutions:
A: In a public political election, it’s impossible to keep secrets about how factions are going to vote. There will be conversations and discussions in all sorts of public places, in call-in shows, in panel discussions, etc. The A faction will know if the B faction is going to defect. In fact, there could even be an explicit between A and B, or the leaders of their factions or parties, to the effect that they’ll give max points to each other (or maybe some fraction of max, which I discuss further below).
B: The results of defection. If the B voters defect against the A voters, then the B voters can expect to lose the support of the A voters. Though they thereby win one election, the long term cost outweighs that gain, providing a deterrent against defection.
C: Confrontive Approach: Candidate merit arguments, principled vote-sharing refusal. It is not my first choice, but it’s among the choices available. The A voters could say, “Candidate A is the moral right choice. You know it too. We, as a matter of principle, will not compromise by supporting B. We will definitely not give points to B. The winner will be either A or C. Take your pick.”
That confrontive solution to the Chicken Dilemma is, of course, common in the animal kingdom (as well as among humans). Say that one cat is the home-territory-defender, and the other cat is the interloper.
The defender is a much stronger threat, compared to that of the interloper, because he is defending his territory; the defender is in the right and has much more to lose by not fighting. The interloper knows that.
So, undesirable though it sounds, the confrontive solution can’t be discounted or left out.
D: Tit-For-Tat. This is an improvement on b), above. Scientific American Magazine once described a Prisoner’s Dilemma tournament for computer programs. The Prisoner’s Dilemma is a co-operation/defection dilemma roughly similar to the Chicken-Dilemma, but worse. Each program in the tournament played repeated Prisoner’s Dilemma games with each one of the other programs–many games with each of, thousands of games.
|Cooperate||-1, -1||-10, 0|
|Defect||0, -10||-5, -5|
|The Prisoner’s Dilemma|
The winner was the computer program with the highest points total in all of its games. Though there were many complicated programs, the winner was a remarkably simple program, called “Tit-For-Tat”. Here is the Tit-For-Tat rule: Play the same way (co-operation or defection) that the other player played in the previous game.
That’s it; it’s that simple. So, if the B voters defect, then the A voters should defect in the next election. If the B voters co-operate, then the A voters should co-operate in the next election. That will quickly lead to co-operation in every election. With Tit-For-Tat, the Chicken-Dilemma is no longer a problem.
E: Strategic Fractional Ratings. But suppose we’re interested only in the current election. This could be the case in some polls, for instance. The following solution is due to Forest Simmons, a member of the election-methods forum. The idea is for the A voters to give, to B, just enough fraction of the maximum amount of points to help B to beat C, given some assumptions.
Below, I list three sets of assumptions that could be used. A, B, and C refer to the sizes of those candidate’s factions, the number of voters who consider them their favorite candidate.
First: B is at least as large as A, and:
1a) There are only 3 candidates. A and B comprise a certain fraction of the electorate called: N. The other option:
1b) A and B each = the average of what they were in some previous election or poll. C has the value that it had in that previous election or poll.
Second: A, B, and C all have the values that they had in some previous election or poll. As a guess, I’d rank the merit and usefulness of these assumptions in the order opposite to the order in which I listed them above. I’d rank them: 2>1b>1a. Or maybe 1b>2>1a. Below, I describe use of the assumptions in the order 1a, 2, 1b. Part of assumption 1a is that N > .5
f is the fraction of max points that the A voters should give to B.
By assumption, the A voters and B voters add up to a majority. Let’s be more specific, and say that N is the majority fraction of the electorate that they add up to. For example, if N = .6, then A and B add up to 60% of the electorate. Here’s the solution that Simmons suggested: The A voters could give to B, a fraction, f, of maximum points, which is just enough to enable B to beat C, if the B voters are more numerous than the A voters. That fraction, f that would barely achieve that (depending on N)
f = 2/N – 3.
Sometimes it might be useful to find N, in terms of f. Solving for N:
N = 2/(f+3).
For example, suppose that f is .1, meaning that the A voters give .1 of maximum support, which is 1/10 of maximum support, to B. Then B will have enough points to beat C if N > .645, or 65.4%
Maybe it isn’t even close to being a three-candidate election. And maybe, what we choose to go by are the results from the previous election, from which we estimate how A, B, and C (who might not be the only candidates) will do this time. So, A, B and C are the estimates for how many votes (or what percentage of the vote) those three candidates will get.
We want to know what fraction of max points the A voters should give to B, in order to barely make B win, if the candidates’ factions have the same sizes, or relative sizes, as before. This time, then, it should look like this:
B + Af = C (at least). Solving for f, then:
Af = C – B
f = (C-B)/A
In case someone is interested in calculating another of the variables, we can solve it for them:
A = (C-B)/f
B = C – Af
C = B + Af
Or maybe, for an assumption somewhat in between the previous two, we assume (as we initially did) that the A and B factions are equal, both having size equal to their average in some previous poll or election. And we assume that C will have the faction-size that it had in that poll or election. Let’s let “av” stand for the average of the votes for A and B in that previous poll or election. This leads to the following formula for f:
f = C/av – 1
Again, the solution for the other variables:
C = av(1+f)
av = C/(1+f)
Depending on which assumptions or estimates you prefer to use, any of those formulas for f, or for the other variables, could be used; I call it “Strategic Fractional Rating”, or SFR. SFR could be done unilaterally by the A voters, with no agreement. Or it could be done by agreement between the A voters and the B voters.
An advantage of SFR, for mutually distrustful factions, is that if the factions agreed to give each other max points, then they’d tend to tie. The winner among them would be the candidate with the most defecting voters. Dishonesty and un-cooperativeness, would determine the winner; however, that problem is neatly avoided if the agreement is for SFR. And, as I have said, SFR needn’t be by agreement. It could be effective and helpful as a unilateral strategy.
What about Approval, in which you don’t have a range of point, values to give? In Approval, either you give max or 0. In Approval, SFR can still be given, probabilistically. Suppose, for instance, that you want to give an f of 1/2. Each voter could flip a coin to decide whether to approve B. That would have the same overall effect as if each voter had given half of a point to B.
Want to give f of 1/3? Draw one of 3 pieces of paper from a bag and if you draw the one that says “Yes” then approve B. In general, you could do it as follows. For example, say you want to give an f of 15% to B:
Have, in the bag, pieces of paper numbered from 0 to 9. Draw one out. Write its number on a sheet of paper. Draw another piece of paper from the bag. Write its number on the sheet of paper, to the right of the first number. The 1st number is the 10s digit and the 2nd number is the 1s digit.
For instance, say the first number you drew was a 7, and the 2nd number you drew was a 2; then the number you’ve written on the sheet of paper is 72. Approve B if the number that you’ve written on the sheet of paper is less than one-hundred times the f that you’ve chosen, expressed as a percent. So, if you’ve chosen an f of 15% (.15), then you approve B if the number that you’ve written is less than 15. So if the number that you’ve written is 72 then you of course don’t approve B.
In summation, there are various good solutions to the Chicken-Dilemma in point systems and therefore the Chicken-Dilemma is not truly a problem in point systems, including Approval.
Thank you for your interest in voting systems.