"STAR voting"

Warren D. Smith. Feb. 2018. PRELIMINARY (interesting comments still coming and remain to be processed).

People keep asking me what I think about "STAR voting." I'm biased against it and dubious, but (1) daresay it is superior to plain plurality and (2) suspect/hope in practice it may behave fairly similarly to the simpler plain range voting system.

What is "STAR voting"?

"STAR" stands for "score then automatic runoff." Perhaps a better name would have been STIR (score then instant runoff). It was invented by Rob Richie, Clay Shentrup, and Mark Frohnmayer. But later, Richie posted remarks which indicate he prefers Instant Runoff (IRV) over STAR (albeit his comments did not make much sense to me).

  1. Each voter, on her ballot, provides a score from {0,1,2,3,4,5} to each candidate.
  2. Call the top two candidates – meaning having the two highest summed scores – "A" and "B." If more voters prefer B over A than the reverse (i.e. more scored B above A than the reverse on their ballot) then B is elected.

Contrasting Definition of "Range + (genuine) Runoff" (R+R):

  1. do plain range voting,
  2. have a second round, where the top two finishers from round #1 are the only candidates (and we employ simple majority vote).

Initial comparison of STAR versus plain range and versus R+R

Both plain range voting (also called "score voting") and R+R are useful things to think about in contrast to STAR. Indeed, STAR is exactly an "instant" (single round) version of the 2-round R+R system, specialized to the {0,1,2,3,4,5} allowed-score set.

Based on that, it seems plausible STAR will be worse than plain range voting, and it definitely is more complicated, But there is a possible way out of that conclusion, pointed out to me by Jameson Quinn. Which is: R+R and STAR can hope to compensate for (what we call) "1-sided strategy." In strategic range voting, voters exaggerate scores to MAX and MIN hoping to increase their vote's impact. (This tends to effectively convert range – and also STAR – into approval voting.) In my previous computer simulations of this effect, I had, essentially, tossed a p-biased "coin," to decide whether each voter was going to be honest or strategic (if we wanted fraction p of the voter population to be strategists). Quinn instead wanted to consider voter pools in which the supporters of candidate A were more likely to be strategic than supporters of candidate B. (That could be modeled using two coins, with two different p-values.) In about 20 years of looking, I still have not seen any evidence that real human voter pools actually exhibit 1-sided strategy. But if they do, then range voting could be in trouble since it is comparatively vulnerable to 1-sided strategy. Which brings us to Quinn's point: the genuine runoff in R+R, and also its "instant" approximation in STAR, can protect us against those evil 1-sided exaggerators. Because if both A and B get into the runoff:

Neither R+R nor STAR will protect us if only one of {A,B} reaches the runoff, so they only confer partial protection. But this nevertheless might exert enough of a beneficial impact in practice so that STAR or R+R could be superior, in practice, to plain range.

Properties sacrificed by STAR voting. Risk of 2-party domination?

STAR is more complicated than plain range voting (aka score voting). It throws most of the nice theoretical properties of plain score in the garbage, e.g. monotonicity, participation property, not motivating Favorite Betrayal, and consistency upon partition into subdistricts. (We'll discuss those more later.)

2-party domination worry (or not). One property-sacrifice that worries me is that STAR fails the NESD* property. That suggests STAR voting might cause a country to fall irreversibly into 2-party domination. To understand this, consider an election where the voters exaggerate their scores for the two major-party contenders A and B (to make them sole-top or sole-bottom), in an effort to increase their impact, e.g.

51% vote "A5, B0, C4," while 49% vote "A0, B5, C4."

Here the best-for-society winner probably is C. So it would be bad, with STAR, if only A and B got into the runoff, whereupon A would win. Actually that does not happen, but the ultimate result is the same: The totals in the first round are A255, B245, C400, so the runoff is {A,C}. But then A wins that runoff 51-49.

The worry is: if there is 2-party domination, then voters may tend to exaggerate their scores for the two major party contenders to sole-MAX and MIN to gain impact in "the only battle that matters" – which in turn will guarantee a major-party victory – so the result would be self-reinforcing 2-party domination.

This nightmare is a plausible scenario because in Australia (which uses rank-order ballots), precisely this kind of sole-max exaggeration behavior is observed to occur on about 85% of all ballots, which is exactly why the Australian House has been massively 2-party dominated.

In contrast, with "range+runoff" where the second round is actually a separate 2-contender election conducted at a later date, all voting in the second round would be honest, not based on exaggerated scores, and furthermore the media (during the interval between the two rounds) would provide extensive analysis of C versus A giving both an equal shot to make their cases, free of other distractions. And then C would win.

That example illustrates both the superiority of R+R over STAR, and the former's ability to avoid 2-party domination.

However, STAR does satisfy the closely related (unstarred) "NESD property," as you can see by, in the above example, making enough voters raise C's score to 5. In that case C would win.

And further: while 3-candidate STAR violates "NESD* property," STAR in elections with ≥4 candidates actually does satisfy NESD*. To see that, consider the exaggerated-about-{A,B} votes
51% vote "A5, B0, C4, D4" while 49% vote "A0, B5, C4, D3."
The totals in the first round are A255, B245, C400, D351 so the instant runoff is {C,D} whereupon C wins, escaping 2-party domination.

In view of that there is hope that STAR, unlike the Australian House, might avoid the 2-party-domination trap. But the NESD* failure for 3-candidate STAR suggests there may be less hope than for either plain range voting, or for Range+Runoff.

Voter honesty. Range+runoff has the advantage that in the second round, all votes are going to be honest, since there is no strategic incentive for anybody to lie in their second round vote. This honesty can compensate for distortions caused by dishonest exaggerated scores in round #1. And indeed range+runoff was found in my computer simulations to be superior to plain range provided there were enough "strategic voters" in the population, more than about 75%.

But (a) based on poll data it looked dubious that this proviso would be satisfied by the real world, and (b) range+runoff does a lot worse than range for 100% honest voters, and only a little better than range with 100% strategic voters, and hence (and also considering the greater complexity) I preferred plain range.

STAR however does not enjoy range+runoff's theoretical property of inspiring voter honesty in the 2nd round, nor does it enjoy plain range voting's property of inspiring voter "semi-honesty" in the 3-candidate case.

That is: with 3-candidate range voting, you can always score A≥B whenever you honestly believe A>B, without ever sacrificing any strategic oomph. (I call that "semi-honesty.") In contrast, with STAR, you may find it beneficial to intentionally score your favorite candidate F zero, to prevent him getting into a runoff he would be doomed to lose; if only in that way you can hope for your second-choice S to be able to win over Hitler ("favorite betrayal"):
#voters their vote
7 H5, F4, S0
5 S5, F1, H0
2 S5, H4, F1
1 Fu, S1, H0 for some 2≤u≤5

The totals so far are H=7×5+2×4=43, S=7×5+1=36, F=7×4+7×1+u=35+u. If you (one additional voter who honestly prefers F>S≫H) are added to the picture, then you would be foolish to give score 5 to F, That would assure Hitler's victory (beats F by 9:6 in second round). Indeed, if you give F any positive number as score, then Hitler might win (depending on the value of u). And if you (as is your honest feeling) score F≥S in any manner whatever, that assures Hitlers victory. Only by scoring S5 F0, maximally dishonestly, can you assure Hitler's defeat.

So in a nutshell,

Voters are motivated to lie in their STAR votes.

What benefits does STAR provide to compensate for its simplicity and property sacrifices versus plain range? Well,

  1. The proponents of STAR claim that it has propaganda advantages. Basically, they claim "there are stupid arguments, made by idiots, against range voting. I will now make some stupid counterarguments against those, because I do not want to actually tell the truth; and in order to help me do so, I need STAR voting."
    Specifically: the stupid argument against range was this. It can happen with plain range voting that 51% of the ballots score Nixon sole-top, but somebody else wins. That phenomenon, when it happens, is generally a good thing. (E.g. if the other 49% like Cameron but hate Nixon, while the 51% who scored Nixon sole-top also score Cameron pretty high, then Cameron should win. That is the truth.) But some idiots worry that any voting method with this "flaw" is a horrible failure. Rather than telling that truth, the inventors of STAR voting ride to the "rescue," noting (correctly) that with STAR voting, anybody scored sole-top by >50% of the voters, is guaranteed to win.
    Incidentally, this "guarantee" is not even valid. It is correct for 3-candidate STAR elections, but as we saw above, not for ≥4 candidates.
  2. They claim they have done computer simulations, and STAR performs well on VSE, or equivalently Bayesian regret, despite its poor theoretical properties.

I find myself disgusted by (1), and while I like (2), I have some doubts about the assumptions inside their computer sims. Specifically I simply cannot believe STAR is better than range+runoff and if any sim says it is, that sim was just wrong. Indeed, with the kind of "strategic voting" in my sims, STAR should definitely perform worse than range+runoff with strategic, and perform exactly the same with honest, voters. And remember, my sims had found range+runoff is inferior to plain range if the electorate has a large enough percentage of honest voters as opposed to strategic ones. The crossover point was about 75% strategists.

Historically (based on Mark Frohnmayer was telling me at the time, anyhow), STAR was invented purely because of reason (1), which disgusted me, but later they came up with reason (2), which as I am saying sounds somewhat dubious. (Their sims, rightly or wrongly, employ assumptions different from mine, and most importantly some of their sims include "1-sided strategists.")

And what benefit does STAR provide in recompense for its honesty-sacrifice versus range+runoff? Just the fact it is a single-round system.

So hopefully STAR would in practice not do much worse than range+runoff, and enjoy the considerable advantage of being a single-round system.

Explicit election examples demonstrating some STAR problems

District-partitioning paradox: In the election below the country is divided into two halves: the West and East. (The West hates candidate D, but the East hates B.)

#VotersTheir STAR vote
5A=3, B=3, C=4, D=0, E=5
3A=5, B=5, C=3, D=0, E=0
1A=5, B=3, C=3, D=0, E=0
#VotersTheir STAR vote
5A=3, B=0, C=4, D=3, E=5
3A=5, B=0, C=3, D=5, E=0
1A=5, B=0, C=3, D=3, E=0

Consider the West alone. In the West, A has total score 35=5×3+3×5+1×5, B has total score 35=5×3+3×5+1×3, C has total score 32=5×4+3×3+1×3, D scores 0, and finally E scores 25=5×5. Therefore STAR's "instant runoff" is between A and B, which A wins by 1 vote, namely the vote in the bottom line of the western table.

Also A wins the East by the same calculation except with the roles of B↔D interchanged. Since A won in both halves of the country, he must win in the whole country, right? WRONG: C wins the countrywide STAR election. To verify that, we compute the countrywide total scores A=70, B=35, C=64, D=35, E=50; hence STAR's "instant runoff" is between A and C, which C wins 10:8.

Ordinary score voting cannot suffer this paradox (at least if every voter scores every candidate), and here would elect A in the whole country and both halves.

"No-show" paradox: In the following STAR election, A wins. (Totals: A=27=4×3+3×3+2×3, B=25=4×2+3×5+2×1, C=24=4×4+3×0+2×4; D=20=4×5+3×0+2×0; E=22=4×0+3×4+2×5; hence STAR's "instant runoff" is A vs B, which A wins 6:3.)

#VotersTheir STAR vote
4A=3, B=2, C=4, D=5, E=0
3A=3, B=5, C=0, D=0, E=4
2A=3, B=1, C=4, D=0, E=5

But now suppose one new voter comes, voting "A=5, B=0, C=2, D=0, E=0." This boosts C's score to 26, causing the runoff now to be A versus C; then C wins 6:4. Notice the new voter agreed that the old winner A was better than C, and said so in her vote. Her A>C vote made C win and made A lose. This is called a "no show" paradox because the new voter would have been better off "not showing up" rather than voting honestly.

Ordinary score voting cannot suffer this paradox (at least if every voter scores every candidate), and here would elect A both before and after the new voter shows up.

Remove-loser paradox: In the above 9-voter election won by A, suppose it was discovered on the day after the election that the loser B was a convicted multiple murderer and hence ineligible to run for office in that country. No problem: since B (fortunately) lost – and lost both in terms of score and in the A-versus-B STARian runoff – we can simply erase him from every ballot and re-count the STAR election, and A still will win, right? WRONG: now C wins by beating A by 6:3 in STAR's "instant runoff."

We also can regard B in this election as a "kingmaker" (or "spoiler"): even though B is doomed to lose, by the very act of running (or not) B has the power to control who becomes king.

With ordinary score voting spoilers and kingmakers cannot happen, in the sense that erasing a loser from all ballots cannot alter the winner.

Just one worked-through STAR election example demonstrating mucho goofiness

Here's a simple STAR election with 29 voters (of only 3 types) and 3 candidates named "A", "B", and "C".

#voters their vote
9 A5, B1, C0
12 B5, C1, A0
8 C5, A1, B0

With STAR, the score totals in the first round are B=12×5+9×1=69, A=9×5+8×1=53, C=8×5+12×1=52. So we eliminate C, and then A wins the "automatic instant runoff" by 17 to 12 over B. (Note 17=9+8.)

Notation: A vote "Jim5, Amy2, Bob0" means that voter prefers Jim more than Amy more than Bob, scoring Jim 5, Amy 2, and Bob 0.

FAILURE OF THE SNIFF TEST: First of all, without any analysis at all, who do you think ought to win this election? It sure looks to me like B is the "most correct" and "most democratic" winner. (B has the most top- and fewest bottom-rankings, and by far the greatest average score.) But STAR elects A.

Score voting considers this election an easy call. It would elect B if all voters gave score X to their first choice, Y to their second, and Z to their third, for any scores X≥Y≥Z, not all equal.

LESS IS MORE: What if 2 BCA voters lower the current loser B, e.g. changing their vote "B5, C1, A0" to "C5, A3, B0"? That makes B win with STAR! (A is eliminated.)

NO, MORE IS LESS: Suppose it suddenly dawns on five BCA voters that the current winner A is no longer their most-hated candidate; he actually is a god, as far superior over both B and C as a dog is versus an ameoba. I.e. two change their votes from "B5, C1, A0" to "A5, B1, C0" while the other three go even further, to "A5, B0, C0." That stops A winning (causes C, the new bottom-ranked guy in all these votes, to win).

YOU POOR SUCKERS VOTED AGAINST YOUR ENEMY? BIG MISTAKE: If 5 BCA voters instead of voting maximally-against the hated A, dishonestly vote maximally-for him (or just refuse to vote at all), that would have prevented A from winning and made their lesser-evil C win. (Same thing as the last one, just different viewpoint. Point is, STAR in this election incentivized voters to lie in their votes and made voting honestly be stupid.)

IT SURE WASN'T WISE TO VOTE A-BOTTOM, YOU CHUMPS SHOULD HAVE STAYED HOME: If we delete 5-10 BCA voters, then the old winner A loses (and C wins). So it was very foolish of them to vote A-bottom and honestly say they regarded A as the worst choice – that made A win – they should either have refused to vote, or lied in their vote.

IT SURE WAS CLEVER OF THOSE B-BOTTOM VOTERS TO STAY HOME: If 1-12 more CAB voters came, that'd make B win. Good job for them they stayed home and didn't let anybody know they hated B.

YOU PATSIES VOTED FOR YOUR FAVORITE? BIG MISTAKE: 1-12 BCA voters could have prevented their greatest evil A from winning by dishonestly lowering their favorite B (including in some cases, see "less is more" above, thus making their favorite B win!). E.g. if they changed their votes from "B5, C1, A0" to "C5, B1, A0."

WHOO-WHEE, I'M A KINGMAKER (also known as a SPOILER): After the voters had voted, it was belatedly discovered that B was a convicted criminal and hence ineligible to run. Fortunately, B had not (embarrassingly) won; and since all voters provided their full preference orders for all candidates, this didn't matter – we can now just erase B from all ballots and redo the count, which will of course still verify that A won and C finished in last place. Right? Wrong... now C wins.

SOME OTHER WAYS TO LOOK AT THE LAST TWO: B is a spoiler since by the act of running, he altered the winner from C to A; while if B drops out of the race at the last minute, he can alter the winner from A back to C. Also, this can be viewed as a tactical opportunity for B-voters: instead of honestly voting for their favorite B, they could dishonestly vote for their second choice C. The latter move works out better for them.

WHAT IF WE REVERSE all ballots? STAR, like plain range voting (but unlike IRV!) enjoys the property that reversing all ballots never (embarrassingly) causes the same winner to still win. (At least, if we ignore tied elections.) And indeed in our example now C wins (after B is eliminated).

To "reverse" the ballot "Jim5, Amy2, Bob0" we would change it to "Bob5, Amy3, Jim0," i.e. score X becomes 5-X.

PARTITION INTO DISTRICTS? DON'T MAKE ME LAUGH: If we split up the country into 3 districts in various ways, e.g.

#voters their vote
3 A5, B1, C0
4 B5, C1, A0
4 C5, A1, B0
#voters their vote
2 A5, B1, C0
3 B5, B1, A0
4 C5, A1, B0
#voters their vote
4 A5, B1, C0
5 B5, C1, A0


So "obviously" B wins the whole country, right? Wrong – A wins.

A "thwarted beats-all winner" example

#voters their vote
13 A5, C1, B0
12 C5, A2, B0
12 B5, C1, A0
6 B5, A1, C0

A wins this STAR election 25-18 after C is eliminated. But C defeats any rival pairwise: C beats A by 24-19 and B by 25-18. It seems odd for the proponents of STAR voting (who historically were so desperately concerned that majority-top candidates can lose under plain score voting) to suddenly stop caring when it comes to this problem. I happen to agree that in this election A probably, or at least arguably, should win (i.e. STAR, and also plain score, got the "right" winner here), but one might not have expected them to feel that way.

Does STAR enjoy any sort of monotonicity guarantee?


  1. If L loses a STAR election, and then some voters lower their scores for L while leaving everybody else's scores unaltered (in particular they do not "renormalize" their ballots to use the full range of allowed scores, as they did in the above "less is more" example), that cannot cause L to win.
    Proof: the top two, if they had not originally included L, still will not. If they did include L, then they either still will (in which case L still loses the 2nd round) or will not (in which case L still loses the STAR election).
  2. If W wins a STAR election, and then some voters raise their scores for W while leaving everybody else's scores unaltered (in particular they do not "renormalize" their ballots to use the full range of allowed scores, as they did in the above "more is less" example), that cannot cause W to lose.
    Proof: the top two still will include W, and W still will win.

Nevertheless, STAR elections can be nonmonotonic with a slightly wider sense of the word, as each of the next two election examples will show.

The relation between STAR and IRV or Nanson-Baldwin voting

In the special case when all scores are in {0,1,2} and no ballot gives two candidates equal scores, then 3-candidate STAR elections become Nanson-Baldwin elections. Hence any "paradox" suffered by Nanson-Baldwin in a 3-candidate election, also is suffered by STAR.

In the special case when all scores are in {0,1,N} and no ballot gives two candidates equal scores, then 3-candidate STAR elections become IRV elections in the limit N→∞. Hence most "paradoxes" suffered by IRV in 3-candidate elections, also are suffered by STAR. This need not always work because the STAR inventors demanded N≤5; but usually you can finagle your paradoxical IRV election example to make it still afflict STAR.

For example, here is a 60-voter 3-candidate Nanson-Baldwin or STAR election:

#voters their vote
23 B2, C1, A0
16 A2, C1, B0
11 C2, A1, B0
10 A2, B1, C0

Score-sums: C61, A63, B56. Eliminate B, whereupon C wins 34-26 over A.

However, suppose we change the 16 ACB ballots in the pink row to CAB, leaving all scores unaltered (i.e. the new ballots are "C2, A1, B0"). Then the score-sums are C77, B72, A47; we eliminate A; and B wins the runoff by 33-27 over C.

So in this election example, paradoxically, the ACB voters becoming CAB caused C to stop winning. This demonstrates non-monotonicity.

Here is a 17-voter 3-candidate IRV or STAR election:

#voters their vote
5 C5, B1, A0
4 A5, C1, B0
8 B5, A1, C0

Score-sums: B45, C29, A28. Eliminate A, whereupon C wins 9-8 over B.

However, suppose two of the BAC voters in the last row of the table change to "A5, B1, C0." The new score-sums are B37, A36, C29; we eliminate C; and B wins the runoff by 11-6 over A.

So in this election example, paradoxically, the BAC voters becoming ABC caused B to win. This is another kind of non-monotonicity.

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