Michel Balinski and Rida Laraki, in 2006, proposed a voting method quite similar to range voting, but using median scores rather than average scores, and with an interesting tie-breaking method. They wrote a paper (pdf) about it. An updated version of this paper has now appeared in the Proceedings of the National Academy of Sciences (USA).
B&L give an example.
The idea of median-based range voting ("highest median rating") had also been considered and discussed by several members of the electorama and range voting bulletin boards during the years 1996-2007. (Even something very similar, if not exactly equivalent, to B&L's tie-breaking scheme had been proposed, although I have not been able to track that particular post down, so it may have just been in emails.)
Here's a disturbing example created by Rob Lanphier in 1998. In this 99-voter election with 0-100 score range, B wins under the Balinski-Laraki voting method because B's median is 51 versus A's 50:
| #voters | their vote | |||
|---|---|---|---|---|
| 49 | A=100 | B=52 | C=0 | D=0 |
| 1 | A=50 | B=51 | C=100 | D=0 |
| 49 | A=49 | B=0 | C=0 | D=100 |
98 out of 99 voters strongly prefer A over B, but the single voter who slightly prefers B over A is the one that matters – B wins! Most people think that is not a good result; Lanphier considered it so disturbing that (he said at the time) median-ratings should be immediately dismissed from consideration as a voting system. In contrast, with ordinary average-based range voting A would win comfortably (A's average=74.25, B's average=26.25).
A related problem is Lomax's pizza-flavor choosing vote (0-9 score range), in which 3 friends want to buy a pizza, and one of them is a Jew who cannot eat pepperoni. All like mushroom, but the two non-Jews slightly prefer pepperoni:
| Pepperoni scores | 9 9 0 | median=9 |
|---|---|---|
| Mushroom scores | 8 8 9 | median=8 |
Another problem with median-based versus average-based range voting, is that the former is not "consistent with respect to partitioning into districts." For example (now with score-range 0-9):
| A scores | 9 8 7 4 0 | median=7 |
|---|---|---|
| B scores | 9 8 6 5 0 | median=6 |
| A scores | 9 4 3 2 0 | median=3 |
|---|---|---|
| B scores | 9 5 2 1 0 | median=2 |
| A scores | 9 9 8 7 4 4 3 2 0 0 | median=4 |
|---|---|---|
| B scores | 9 9 8 6 5 5 2 1 0 0 | median=5 |
In contrast, with average-based range voting, if A wins in both districts, then A always wins in the combined country.
Average-based range voting also has the advantage of being much simpler to describe than Balinski & Laraki's voting scheme, and it works on every voting machine in the USA, right now, with no reprogramming or modification required. (Median-based range voting can't say that.)
Dan Bishop pointed out another problem with the median-based system: election results can change if "all zero ballots" are added. As a simple example, consider this 3-voter election (0-9 score range):
| A's scores: | 1 6 9 | median=6 |
|---|---|---|
| B's scores: | 2 5 9 | median=5 |
If an "all zero ballot" A=0, B=0 is added, then, according to Balinski & Laraki's rules, B is now the winner. In contrast, with average-based range voting provided all voters score all candidates, adding an all-zero ballot never changes results. (This property would enable an interesting new kind of election fraud...)
Average-based range voting also outperforms median-based in graphical comparisons, in computer measurements of Bayesian regret, and in the eyes of the world's most experienced(?) election-practitioners, honeybees. Also, range voting has been used for decades by academic graders, whom I do not think would be willing to switch to medians.
Average-based range voting generalizes to a multiwinner proportional representation voting system called "reweighted range voting." (See papers 78 and 91 here.) But there currently is no known way to generalize median-based range voting to do that.
Average-based range voting is readily adapted to the case where voters have different "weights," e.g. type-A voters have 54+π votes whereas type-B voters have only 1 vote each. (Simply multiply the type-A votes by 54+π...) With median-based range voting it is a good deal harder to to handle that.
An extremely bothering problem with median-based range voting is its failure of "participation." Consider this situation:
| A's scores: | 9 6 5 3 0 | median=5 |
|---|---|---|
| B's scores: | 9 7 4 2 0 | median=4 |
| C's scores: | 9 4 3 2 0 | median=3 |
Now suppose you and your spouse (new voters) come. Your votes are both: A=9, B=7, C=0, preferring the current winner A over every opponent. These votes cause A to lose, because the new election is now
| A's scores: | 9 9 9 6 5 3 0 | median=6 |
|---|---|---|
| B's scores: | 9 7 7 7 4 2 0 | median=7 |
| C's scores: | 9 4 3 2 0 0 0 | median=2 |
In short, you and your spouse's decision to vote honestly, made the election result worse from your point of view. You would have been better off not voting at all ("no-show paradox"). In contrast, with average-based range voting, no-show paradoxes can never occur provided there are no blank (no opinion) votes. They can occur if blanks are allowed and are discarded before averaging. E.g. A scores: 9 0 and four blanks, avg = 4.5; B scores: 5 5 5 5 5 5, avg = 5; B wins; but you now add your vote A=8, B=9. That raises A's average to 17/3=5.667 and B's to 39/7=5.571 so now A wins; this example is valid if there is no quorum rule.
Balinski & Laraki pointed out the following enjoyable property of median-ratings (which had earlier been stated, albeit somewhat more clumsily worded, by electorama posters including Bart Ingles in 1999):
Theorem (Bart Ingles, then M.Balinski & R.Laraki): With median-rating, the winner W has the property that a majority of voters unanimously rates him ≥M, while no majority unanimously rates any opponent ≥M.
In contrast, average-based range voting does not enjoy that property, and indeed both average- and median-based range voting fail the traditional definition of the "Condorcet property." But both average-based and median-based range voting obey a nontraditional definition of that property and both, under reasonable assumptions about the behavior of strategic voters, will elect a Condorcet winner whenever one exists.
Also, we should remark that Ingles' theorem is not as nice as it sounds. Here's an example (0-9 score range, three candidates):
| A's scores: | 0 4 6 9 | median=4 |
|---|---|---|
| B's scores: | 0 3 7 9 | median=3 |
| C's scores: | 0 3 8 9 | median=3 |
As assured by the theorem, a majority of voters unanimously agree that A≥4, whereas no voter-majority unanimously agrees that B≥4 (or that C≥4). Sounds great at first, but... it is also true that a voter-majority unanimously agrees that A≤6, whereas no voter-majority unanimously agrees that B≤6 (or that C≤6). So by the same logic Balinski & Laraki used to conclude A was the best, we can conclude A is the worst! So it seems as though Balinski & Laraki's logic is self-contradictory. (About reversal failure; average-based range never suffers that.) Personally, in this situation I would prefer to elect C, not A, but that is just my opinion.
Finally, Balinski & Laraki's median-based scheme is more complicated than mean-based rnge voting.
Balinski & Laraki's proposal has a lot in common with ordinary mean-based range voting, but relatively speaking suffers numerous disadvantages. I cannot currently see how it can have enough relative advantages to compensate for that.
Hence I recommend ordinary range voting.