Uncertainties in opinions

Humans are not perfect comparison machines, we are full of uncertainties and indifferences, so opinions are fuzzy.

To model this, assume that a voter's utilities $u_{i}$ for candidates $i$ are not exact numbers, but random variables drawn from distributions over some domain (e.g. issue space). When a voter casts a ballot, they're basing themselves on comparisons of these distributions, so rankings/scores themselves should be seen as representing underlying distributions of opinions/utilities.

Uncertainties under cardinal voting

In this way, scores are random variables drawn from certain (assumed unimodal) scoring distributions $S_{i}(s)$ for each candidate $i$ and score $s$ (e.g. $S_{A}(3)<math>istheprobabilitythatthisvoterscoresA=3).Ifyourepeatedtheelectionmanytimesandvoterskeptdrawingscoresfromtheirowndistributions,eachvoter'sscoredistributionforeachcandidatewouldapproximatelyreproducetheirtrueinnerscoringdistributioninthelimit,andthus,theunderlyingutilitydistributionasaccuratelyaspossibleunderthecircumstances(we'reboundbywhatevergoesintheballot).Nowimagineyourunascorevotingelectionusingthoseballot''distributions''themselves.Bylinearityofscorevoting,thiswouldbeaperfectrepresentationofthescoringdistributionofeveryoneinaggregate.Additionally,bylinearityofexpectedvalues,thewinnerofthiselectionbasedondistribution-ballotsisthesameexpectedwinnerfromaregularscorevotingelection.Bothdistributionsmatch,there'snodistortion.Soscorevotingelections,onaverage,arerobustunderopinionuncertaintyandstatisticalnoise.Inotherwords,thecriticismssomepeoplepointoutthatscoringisinconsistentoruncertainarenotreallyadequate.It'swhatmakesitwork.Cardinalvotingstatisticallyinternalizesthenaturaluncertaintyofopinionsinaconsistentway.==Uncertaintiesunderrankedvoting==Rankingsaretrickier.Saying<math>A>B$ is saying $u_{A}>u_{B}$ . But if these utilities are random variables, then $u_{A}>u_{B}$ is a probabilistic statement: there's a certain degree of certainty that a random utility drawn from $u_{A}(x)$ is larger than $u_{B}(x)$ . If the distributions overlap, this certainty decreases.

For example, say there are two Failed to parse (syntax error): {\displaystyle σ=1} normal distributions with means Failed to parse (syntax error): {\displaystyle μ_A = +1/2} and Failed to parse (syntax error): {\displaystyle μ_B = -1/2} . The probability $p(u_{A}>u_{B})\approx 75\%$ . So the ranking Failed to parse (syntax error): {\displaystyle A>B<math> is not just imprecise, as in the cardinal case, it is ''wrong'' 25% of the time due to the fuzziness of opinions. Sometimes the voter just feels like putting <math>B>A<math> on the ballot, even though A has a higher mean and would be better most of the time. Also note that two identical candidates (<math>μ_A = μ_B} , the distributions are identical) would have $p(u_{A}>u_{B})=50\%$ , in this symmetric distribution. Two very different candidates would have $p(u_{A}>u_{B})=100\%$ or $p(u_{A}>u_{B})=0\%$ , that is, the distributions are far apart but may be reversed in the 0% case.

But ranking systems assume this is all always 100% certain, as if the $u_{A}(x)andu_{B}(x)$ never have any overlap. (This is why I emphasize that it is rankings, not scores, that are assuming unreasonable infinite precision: they effectively assume the utility distributions $u_{i}(x)<math>areperfectlysharpDiracdeltas<math>u_{i}(x)=\delta (x-\mu _{i})<math>,entirelydefinedbytheirmeans.)Repeatingtheelectionmanytimesyou'llgetthesameballotdistributionforeachvoter(e.g.intheexample,B>A25\%ofthetime),justasinthescorecase.Theproblemarisesfromtheaggregation.Let'ssupposeasimple2candidateelectionbetweenAandB,majoritypreferencewins.(ThiscanbeapairwisematchinaCondorcetsystem,forexample).Let'sassumeeach<math>i$ of the $N$ voters has probability p( $u_{A}>u_{B}=p_{i}$ , due to whatever overlap (more overlap = p closer to 50%, if distributions are symmetric). We want to compute the probability that A wins the election, that is, the probability that a majority M > N/2 of voters has A>B against the minority's B>A.

For the sake of simplicity and understanding of the problem, and of a closed-form solution, we can assume all ps are the same. Since all voters are exchangeable, this is a Binomial distribution $B(N,n,p)$ , for a population of N, n the number of successes (number of A>B). We then want to compute: $p(Awins)=\sum _{n>N/2}^{N}B(N,n,p)$ . This is just the cumulative values of the upper half of the distribution on the $0\cdots N<math>interval,whichiswellknown(involvesahypergeometricfunction,butwe'llbequalitativehere).ForlargeN(manyvoters),thisfunctionis0\%for<math>p<0.5$ and rises sharply around $p=0.5<math>to100\%.ThismeanstheprobabilityofAwinningis''highly''sensitivetooverlappingdistributions.Recallthat<math>p=0.5$ is when the distributions are identical.

This means that under ranked voting for large N, any small variation from indifference is immensely amplified on average. So even if there's a 10-30% chance voters actually prefer B>A due to similarity, statistically the ranked election will elect B>A 0% of the time due to the way the individual probabilities are combined.

The ranked system is incapable of reproducing the underlying distributions of uncertainty on the average election. Therefore, small individual biases are amplified, and voters will on average betray their own interests significantly if there are similar candidates running and they are slightly uncertain about the candidates.

Note that this has nothing to do with aggregating the utility of the voters in a ranked system. Each voter is betraying their own interests by the aggregation of the rankings itself. This is a strictly ordinal model (beyond the internal cardinal utility, only used to create the $u_{A}>u_{B}$ comparisons) with no interpersonal comparisons of cardinal utility whatsoever.

Remarks

Now, the assumptions here may seem strong, like normal distributions, overlaps and identical voters, but one can make an argument from the Central Limit Theorem using the means of voters opinions. Basically, if you got utilities at random from voters of each faction, and took the mean of many such set of samples, their means would likely follow a normal distribution, and this would fit the above analysis well. But that would involve aggregating cardinal utilities. However, you could do the same analysis under the $p(A>B)+p(B>A)=1$ ballot distribution alone, instead of $u_{A}(x)<math>and<math>u_{B}(x)$ , which wouldn't have this problem.

These overlaps may seem insignificant to the typical ranked voter enthusiast, but they're really not. Indifference plays a large role in elections with multiple candidates, and always forcing perfect distinctions is extremely problematic and exacerbates the above problem. Consider also the role of clone candidates in elections and how we're trying to address that with better voting systems right now. Clones, by definition, have significant overlaps to other candidates. Also note that Condorcet cycles are likely to occur when the Smith set candidates have a lot of overlaps.

So this is a very relevant and realistic scenario.

Also, note that none of this relies on "strategical voting vs honest voting", and applies equally to both scenarios. Strategy can merely reshape the original opinion distributions.