Puzzle 46: Probability of unclear election winners
Puzzle:
In the random election model, with 3 candidates and
rank-order 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) 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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