Puzzle 46: Probability of unclear election winners

In the random election model, with 3 candidates and rank-order votes from V→∞ voters,

  1. What is the probability that some two weighted positional voting methods will disagree on the winner?
  2. And what is the probability that every weighted positional voting method's winner will agree but disagree with the Condorcet winner?
  3. 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.

(a) D.G.Saari & M.M.Tataru: The probability of dubious election winners, Economic Theory 13 (1999) 345-363, 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 3-candidate 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) 191-197 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) 193-206, said the probability of a situation such as the example below in which every weighted-positional scoring voting system simultaneously elects some unique winner different than the CW (and that CW exists) is about 1.808% in 3-candidate random elections with a large number of voters.

13-voter Condorcet vs WP example by Warren D. Smith. A is the unique Condorcet-Winner. But B wins in every weighted-positional 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 weighted-positional system with weights 2, Q, 0; there are an infinite set of Q-containing 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 weighted-positional 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 "1-dimensional 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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