## Examples in which best Range Voting strategy is "dishonest" approval-style

### Generalities about range voting strategy

Theorem 1. It is never strategically forced to give X less than the maximum possible score, if X is your true favorite (or one of your true favorite) candidates.

Proof. Increasing X's score will not change the winner (except if it changes to X).   Q.E.D.

Theorem 2. It is never strategically forced to misorder X>Y if you truly regard Y>X, in your range vote, if there are 3-or-fewer candidates. (Put another way: If you honestly feel Alice>Bob, there is never any motivation to vote as though Bob>Alice.)

Proof. Call the three candidates X, Y, and Z. By the preceding result we can assume wlog your favorite and most-hated candidates are respectively given the maximum- and minimum-allowed vote-scores. It then is not possible for you to vote X>Y if you honestly think UY>UX.
Q.E.D.

Theorem 3. If you know all the other voters' votes (or just know their totals) then there is always a strategically best vote for you to cast, which is "honest" in the sense that it never misorders X>Y if you truly regard Y>X.

Proof. Determine the best (in your view) candidate that your vote is capable of electing. Suppose his name is "X." Then vote maximum for X and every better candidate and minimum for all candidates worse than X.
Q.E.D.

But if there are 4 candidates, and you don't have complete information about the other votes, then we shall see via either of two examples, that dishonesty can be strategically forced.

### First example: correlation (highly ellipsoidal Gaussian)

There are two liberals L1,L2 and two conservatives C1,C2 running. You are pretty sure that L1,L2 will get a near-equal number of the other people's votes, ditto for C1,C2, but don't know whether the liberals or conservatives will be ahead. (This kind of situation can be modeled with a highly ellipsoidal Gaussian distribution being returned by a pre-election poll in which the correlation and covariance matrices are published, not just the estimated means. We admit that the sort of pre-election polls we have seen in newspapers and television never do publish covariance information, so this is somewhat unrealistic. However, this scenario as a whole nevertheless is tolerably realistically plausible. For example it could arise thus: L1 & L2 are regarded by all voters other than you as identical; C1 & C2 are regarded by all voters other than you as identical; the country's voters are 50% liberal and 50% conservative; rain will occur on election day, thus lowering turnout, but it is not known whether it will rain on the liberal or conservative part of the country.) Then your best strategy is to vote in the style

L1=1,  L2=0,  C1=1,  C2=0
even if you prefer both L's over both C's (or both C's over both L's). (We assume you slightly prefer the candidates with smaller subscripts, i.e. your true preferences are L1>L2>C1>C2.)

This example also can be generalized to 2N candidates falling into N ultra-correlated pairs. It apparently originally traces to S.J.Brams.

Simpler explanation [by Damien Sullivan]
The liberal set and the conservative set have roughly equal chances of winning, and within each set the two candidates are expected to have nearly equal votes. One votes for C1 to have influence in the case of the conservatives winning, and one votes L1 but not L2 to have preferred influence in the case of the liberals winning. The probability of vote-totals obeying L2 > C1 > L1 is expected to be much lower than for (L1, L2) > (C1, C2) or (C1, C2) > (L1, L2) because of the strong correlations. So if L2 > C1, it is highly likely that also L1 > C1, hence the 0 vote for L2.

Or another way: L1 is your favorite, so you naturally vote for him; C2 is your most-hated so you naturally vote against him. It is much more likely in this case that the front-runners will be L1 and L2 (so don't vote for L2, per your preference) or (with equal chance) C1 and C2 (so vote for C1, ditto) than any other pairs.

The optimum vote here will bring you about twice as much utility as more obvious/honest votes like L1=L2=C1=1, C2=0, since you have about twice as much chance to swing the election.

### Second example: no correlations (exactly spherical Gaussian)

Let the 4 candidates be A,B,C,D. Let the pre-election poll results say the vote totals (mean of the Gaussian) are A=100, B=80, C=70, D=10, (well more precisely, proportional to these; the totals are these times a number of order V) say, so A appears most likely to win, then B, then C, then D. We assume these 4 vote totals are samples from a spherically symmetric 4-dimensional Gaussian, i.e. with no correlations or anticorrelations among candidates. (It is easier to make one of our examples if correlations are assumed – as we just saw – but our point here is they are not needed.) There are V voters where we assume V=large (we work in this limit) and the Gaussian has mean of order V and stddev of order √V. Let your true utilities be:

UA=0,   UB=20,   UC=HUGE,   UD=25.
Here HUGE denotes some sufficiently large constant. I originally had in mind UC=70 but perhaps that was too generous (Larry Shepp once told me my problem was I confused the word "generous" with the word "stupid") so let us leave it unspecified.

Theorem 4. The unique strategic range vote, which maximizes your expected election-result-utility, is (A,B,C,D)=(0,1,1,0), which is dishonest for B and D.

Proof sketch. The top-two pair is AB – most likely to be tied for lead – other ties are exponentially less likely when V=large – so we give A=0 and B=1 in the vote to give us maximum chance to favorably break an AB tie. (Since UB>UA.)

Now AC is the next pair (second most likely to tie for the lead, but in the V=large limit, it is way exponentally less likely to be AC than it is to be AB; but AC is exponentially more likely than any other non-singleton set) so give C=1 to make C win over A in a situation where AC tied for lead and B and D's winning chances may be neglected. (Since UC>UA.)

Finally among sets involving D tied for the lead, the most likely to result in a lead-tie is A,B,C,D 4-way near-tie near 260/4=65 each. That is because any other location in the 4-space having D maximal has a larger sum-of-squares distance from (100,80,70,10) and hence is exponentially less likely in the V=large limit.

Now even if you don't believe the 4-way-near-tie business, fine. More precisely, all I want you to believe, and all you need to believe, is it is going to come down to either a DC, DB, or DA battle (conditioned on the assumption that D is involved in the lead-tie) with some fixed positive conditional probabilities for each, not necessarily 1/3, 1/3, 1/3, in the V=large limit. Call these probabilities PA, PB, PC. Basically what happens is as you move away from the magic (65,65,65,65) point, but preserving the assumed fact that D=maximal: as soon as you go a distance of order 1, i.e. order 1 vote worth, that causes a factor of order eorder1 falloff in probability. Hence when V=large there is negligible probability (conditioned on D leading or co-leading) that we are more than a distance of order (logV)/V away from the magic 65,65,65,65 point. You can see that using the formula defining the Gaussian (which we assume is still valid out here in the tail, perhaps somewhat unrealistically, but it is ok since we took the whole Gaussian model as an assumption, and anyhow I think it would be valid enough even out here in the tail, even without such an assumption) using stddev of order √V and distance-to-center of order V. That is the underlying reason for the positive PA, PB, PC existing. You may here worry that perhaps a 3-way or 4-way tie could be important, in which case the strategically best vote might not even be approval-style at all. However, if we agree in this example to make our spherical Gaussian have variance which is a sufficiently large constant or larger, then the conditional probability of a more-than-2-way tie can be made arbitrarily small, so this is not a worry. Arbitrarily large variances arise in practice if, e.g, the votes are postulated to come in blocs of order 1000 highly correlated votes. Also, a different reason why the optimum vote here must be approval-style, regardless of worries about 3-way near-ties, is given in the second proof in the preface-part of this.

So in this final case where we assume D has a chance (which is an assumption needed to make it worth voting about D at all) we should vote D=0 provided that 25 = UD < PAUA + PBUB + PCUC which by making UC=HUGE huge enough (since the values of PA,PB,PC exist and do not depend on UA,UB,UC), always happens. So we must vote D=0.

Q.E.D.

Sullivan-style simpler explanation:
You vote for C because it is your favorite and you always vote for the favorite, and vote for B because it is your preferred front-runner and the A-B contest is most likely to matter; so far this is just standard range/approval strategic voting.

If D has a chance to win, that means the pre-election polls were totally wrong (to allow D to get up about even with the top among A,B,C despite predictions). If they were that wrong, then we cannot trust them and need to assume D will be about even with A, and with B, and with C, i.e. all three scenarios occur with roughly equal chances. (We do not need to assume exactly equal chances.) If so then what matters is the fact that D is worse than the average utility for those 3, so you get more expected utility if you vote against D and hope for a random winner among {A,B,C}.

(This "we cannot trust them" argument was intuitive rather than rigorous, but it can be made rigorous under a certain normal-distribution model about how the pre-election polls err.)

I first devised this example and methodology in 1999 or 2000 albeit I did it rather badly and messily then and we probably want to avert our eyes. -WDS

The thinking in this example actually can be generalized to provide a way to determine your strategically optimum range-vote in any Gaussian-model scenario in the V=large limit.

But Mike Ossipoff comments: I want to re-emphasize that, although in extremely unlikely examples, reversing a preference can be optimal in Approval, one should never vote someone over one's favorite. One should always vote for one's favorite in Approval or Range voting. No exceptions.

One author familiar with the subject said that the kind of probability knowledge that would be needed to make preference-reversal optimal in Approval is so rare that such situations can be disregarded.

And (WDS further comments) even with such knowledge... these situations still are (empirically) quite rare. And even when they do occur they usually have a small probability of changing the winner, hence do not matter much. My computer simulations with 4- and 5-candidate elections indicated an expected social-utility change due to true-strategic voting versus "honest approval" (threshold strategy) voting of order 1 part in 1000.

### How frequent is this?

We explain how to get one crude estimate via computer simulation (and do it). The "naive zero info strategy" (NZIS) for range voting is, you approve the candidates with above-average utility (using unweighted average). Observe that this always produces a "semi-honest" vote.

The "moving average strategy" (MVAS), which is less-naive, is this. There is some pre-election perception that X>Y>Z>... where here X,Y,Z denote candidates and ">" denotes "has much greater chance to win than."

You (the voter) go through the candidates in that order X,Y,Z,... for each one approving or disapproving depending on whether its utility is greater or less than the average of the preceding candidates (except: for the first two X,Y, you approve the better and disapprove the worse). This can produce a vote different than NZIS, and which can be dishonestly misordered. The key word here is "preceding."

We now describe a simple computer simulation. Fix a number C≥2 of candidates. Generate a random voter by assigning random i.i.d. normal "personal utilities" Uk to each of the C candidates k. Find out for what percentage of such voters the MVAS produces a dishonestly-ordered vote – i.e. featuring two candidates P,Q for which the voter honestly thinks UP>UQ but approves Q while disapproving P.

Here are the results. Each datapoint is based on 222 to 233 Monte Carlo experiments:

C% who misorder
20
30
42.9248
57.2795
612.115
716.96
821.58
925.90
1029.87
2055.53
5081.1
10092.3
20097.7
50099.75
100099.97

I thank Rob LeGrand for stimulating correspondence.

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