Puzzle 46: Probability of unclear election winners
Puzzle:
In the random election model, with 3 candidates and
rankorder votes from V→∞ voters,

What is the probability that some two weighted positional voting methods
will disagree on the winner?

And what is the probability that every
weighted positional voting method's winner will agree but
disagree with the Condorcet winner?

Prove the latter problem cannot happen if
the model of voters instead is that voters and candidates are points on a line
and voters prefer candidates located closer to them.
Answer:
(a) D.G.Saari & M.M.Tataru:
The probability of dubious election winners,
Economic Theory 13 (1999) 345363,
claimed the answer was ≈69%.
But I don't understand their paper and my computer disagrees.
My computer found
47% disagreement between the plurality and antiplurality winners in 3candidate
random elections, in which case the answer should instead be 53%.
(Saari and Tataru claimed these two probabilities should be the same.)
Who is right?
William V. Gehrlein & Dominique Lepelley:
The probability that all weighted scoring rules elect the same winner,
Economic Lett. 66,2 (2000) 191197 did the same computer simulation I did and found the same
answer, so it appears I was right.
(b) Vincent Merlin, Maria Tataru, Fabrice Valognes:
On the likelihood of Condorcet's profiles,
Social Choice & Welfare 19,1 (2002) 193206,
said the probability of a situation such as the example below
in which every weightedpositional scoring voting system simultaneously
elects some unique winner different than the CW (and that CW exists) is about 1.808%
in 3candidate random elections with a large number of voters.
13voter Condorcet vs WP example by Warren D. Smith.
A is the unique CondorcetWinner.
But B wins in every weightedpositional voting system such as Plurality, Borda, and
AntiPlurality.
#Voters 
their vote 
5 
A>B>C 
4 
B>C>A 
2 
B>A>C 
2 
C>A>B 
(c) The possible votes are ABC, CBA, BCA, and BAC for three candidates A,B,C
located on a line in that order.
Suppose the numbers of these votes are respectively x,y,z and t.
The proof is then to consider the linear program
x≥0,
y≥0,
z≥0,
t≥0,
x+y+z+t=100
supplemented with either
2x+Qt≥2y+Qz,
2x+Qt≥Qx+Qy+2z+2t,
z+t+y≥x,
z+t+x≥y
(specifying that B is the Condorcet winner while A is the winner with a
weightedpositional system
with weights 2, Q, 0; there are an infinite set of Qcontaining inequalities,
one for each real Q with 0≤Q≤2)
or
2x+Qt≥2y+Qz,
2x+Qt≥Qx+Qy+2z+2t,
y+z≥x+t,
y≥x+z+t
(same but C is the Condorcet winner)
or
2t+2z+Qy+Qx≥2y+Qz,
2t+2z+Qy+Qx≥2x+Qz,
z+t+y≥x,
z+t+x≥y
(specifying that C is the Condorcet winner while B is the winner with a
weightedpositional system).
Either way, the linear program has only x=y=z=t=0 as a solution even if
we only employ Q in the set {0,1,2} rather than the full infinite set of Q.
In other words, we have proven that, in "1dimensional politics," it is impossible for Borda,
plurality, and antiPlurality voting all to agree on a winner, unless it is
a Condorcet winner.
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