Median-based range voting is "twice as resistant to 1-sided strategy" as average-based

Jameson Quinn suggested the following 1-parameter model about "one-sided strategy" in voting. Warren D. Smith then (April 2011) worked out the consequences of Quinn's model in closed form, finding a surprisingly elegantly simple result.

We had previously examined only the special ("unbiased," "honestly close election") case P=0 of this model. That special case turns out to behave unlike the P>0 cases. When P>0 we here argue median-based range voting is "twice as resistant to 1-sided strategy" as average-based. But when P=0 we previously had argued average-based range voting looked to have better resistance to 1-sided strategy than median-based! (Obviously, the P=0 special case is the most important point of this parameter space, but it seems dubious it should outweigh the whole rest of it.)

The model: Two candidates named "Gore" and "Bush." Large number of voters. Initially, each voter has honest score for each candidate that is an independent random-uniform real number in the real interval [0,1]. But now reverse a fraction P of the Bush>Gore votes, by applying the map x→1-x to its scores x. The result is a set of "honest" votes which, in net, favor Gore. These votes are described by a single parameter P with 0≤P≤1; the greater P is, the more pro-Gore the voters honestly are.

That situation will be our starting point. It has 100% honest voting. We will now consider the effect of pro-Bush 1-sided strategy.

This model implies there are a fraction (1-P)/2 Bush>Gore, and (1+P)/2 Gore>Bush, honest voters. The Gore>Bush voters have a triangular probability density of Gore scores: ProbDensity(x)=2x, with mean score 2/3 for Gore and mean=1/3 for Bush. The Bush>Gore voters have the same triangular probability density of Bush scores and mean score 2/3 for Bush and mean=1/3 for Gore.

Now suppose a fraction F, 0≤F≤1, of the Bush>Gore voters strategically exaggerate to Bush=1, Gore=0. We'll assume there is no counterstrategy by the Gore>Bush voters – they just stay honest.

THEOREM: In the above model with any P with 0<P<1, Gore wins with honest voting; but Bush wins with 1-sided exaggeration exactly when the fraction F of Bush>Gore voters who exaggerate obeys F>P/(1-P) for average-based range voting, and F>2P/(1-P) for median-based range voting. If P>1/2 and P>1/3 respectively, then Gore wins despite any amount of pro-Bush exaggeration. Approval voting (since it 'forces exaggeration') is entirely immune to 1-sided strategy, i.e. with it Gore wins, period.

PROOF: We first shall derive formulas for Bush & Gore's post-exaggeration median and average scores. (The pre-exaggeration results also will arise from our same formulas by taking F=0.) Note that there are three kinds A,B,C of voters, present in proportions

(1-P)F/2 : (1-F)(1-P)/2 : (1+P)/2       [which sum to 1],

who respectively rate Gore=0, Gore=honest, and Gore=honest. The type B∪C voters (combined set) have a trapezoidal probability density of scores with the two trapezoid leg-heights being (1-F)(1-P) and (1+P), up to a constant multiplicative scaling factor you need to choose to make the total probability-mass be 1, i.e. "normalized," at the Gore=0 and Gore=1 sides respectively. Gore's median score is then m where

(1-P)F/2 + (1-[1-P]F/2)∫0<x<m[(1-F)(1-P)+(2P+[1-P]F)x]dx / ∫0<x<1[(1-F)(1-P)+(2P+[1-P]F)x]dx = 1/2.

This simplifies to

(P+[1-P]F/2)m2 + (1-F)(1-P)m + (1-P)F/2 = 1/2

a quadratic equation whose solution m is

GoreMedian = ([F-1][1-P] + ([P+F]P+1-F)1/2) / ([1-P]F+2P).

Similarly Bush's median score is computed by noting the type-A,B,C voters rate Bush=1, Bush=honest, and Bush=honest. The type B∪C voters (combined set) have a trapezoidal probability density of scores with the two trapezoid leg-heights being (1+P) and (1-F)(1-P) (up to a constant multiplicative normalization factor) at the Bush=0 and Bush=1 sides respectively. We find analogously

0 + (1-[1-P]F/2)∫0<x<m[(1+P)-(2P+[1-P]F)x]dx / ∫0<x<1[(1+P)-(2P+[1-P]F)x]dx = 1/2.

This simplifies to

- ([1-P]F/2+P)m2 + (1+P)m = 1/2

a quadratic equation with solution

BushMedian = ([1+P] - ([P+F]P+1-F)1/2) / ([1-P]F+2P).

Meanwhile, Gore's average score is

0 + (1-[1-P]F/2)∫0<x<1[(1-F)(1-P)+(2P+[1-P]F)x]xdx / ∫0<x<1[(1-F)(1-P)+(2P+[1-P]F)x]dx

(note the integrals in the denominators are there merely to provide the right normalization-constant for the trapezoidal probability density). This simplifies to

GoreAverage = 1/2 - (1-P)F/6 + P/6 = (3+P+F-FP)/6.

Finally, Bush's average score is

(1-P)F/2 + (1-[1-P]F/2)∫0<x<1[(1+P)-(2P+[1-P]F)x]xdx / ∫0<x<1[(1+P)-(2P+[1-P]F)x]dx

which simplifies to

BushAverage = (3-P+F-FP)/6

Now we ask: what fraction F of Bush>Gore voters need to exaggerate in order to make Bush win? With averages, the answer is got by solving BushAverage=GoreAverage for F, with the very simple result

FExaggNeeded(Avg) = P / (1-P).

Note that if P>1/2 then average-based range voting becomes immune to strategy: no amount of exaggeration will suffice to make Bush win. I.e. this model yields such immunity when the honest ratio of Gore>Bush to Bush>Gore voters exceeds 3:1.

With medians, the answer is got by solving BushMedian=GoreMedian for F, with the amazingly simple result

FExaggNeeded(Med) = 2P / (1-P).

Note that median-based range is thus exactly twice as resistant to 1-sided strategy as average-based range, in this model, in the sense that exactly twice as many exaggerators are needed to make Bush win (regardless of P for 0<P<1).

Median-based range voting becomes immune to strategy in this model when P>1/3, i.e. when the honest ratio of Gore>Bush to Bush>Gore voters exceeds 2:1.

Finally, for approval voting, the votes are "already exaggerated" to the endpoints of the allowed score range, so further exaggeration has no effect. Hence if Gore wins, he wins.

Q.E.D.


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