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<font face="Helvetica, Arial, sans-serif">Forest – this isn’t an
answer, but a dissertation on a related topic.<br>
I recently read Condorcet's Essai, and then commentaries by
Todhunter, Black and Young. I concluded that Condorcet's
probability theory was all at sea, and that his commentators were
too kind to him. In my view you cannot make sense of voting using
pure probability theory (even turning a blind eye to the faults of
a jury model); you need to fall back on statistics. This makes it
possible to get away from Condorcet's untenable independence
assumption and also to correct another fatal error in his method.
He assumes that when a voter votes A>B>C, he is no likelier
to be right in placing A above C than in placing A above B. I
believe that this is impossible if erroneous rankings are treated
as random errors. But if you try to find the relation between the
likelihoods of one-place errors and two-place errors using pure
probability theory, you just can't get started. You have to use a
statistical model - eg. assume that true candidate merits are
Gaussianly distributed, estimated merits are contaminated by
Gaussian noise, and that a voter ranks candidates in decreasing
order of estimated merit. If you do this, you can find the
relative chances of one-place and two-place errors (subject to
suitable distributional assumptions).<br>
Having done this, you can say: "I will give candidate A one
point more than candidate B whenever he comes one place higher in
a ballot, and x points more whenever he comes two places higher";
x=1 gives the Condorcet criterion, x=2 gives the Borda count. By
my calculation, the optimal x is almost exactly 2 (and almost
independent of distributional assumptions), and the Borda count is
therefore almost optimal under a jury model with 3 candidates.
Condorcet thought his jury theorem showed his own criterion to be
optimal in the same case. <br>
So I would say that you have to recast your question: let x and
y be the candidate merits, distributed as N(0,1); let the noise be
distributed as N(0,sigmasquared); let sigmasquared be governed by
some fairly diffuse prior (maybe 1/sigmasquared). What is the
probability that x>y when each of 100 noisy estimates of x is
larger than the corresponding noisy estimate of y?<br>
If I have time, I might attempt the calculation.<br>
CJC<br>
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<div class="moz-cite-prefix">On 04/09/2022 04:43, Forest Simmons
wrote:<br>
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cite="mid:CANUDvfqheO5uSa8nOJ=ZDrPCQNXF8Q_PDQDng4mV1yh8A6VZBA@mail.gmail.com">
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<div>Actually, max likelihood analysis would say that the A>B
defeat was less likely to be reversed because max likelihood
estimation would estimate Prob(B>A)=0.
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<div dir="auto">But Baysian estimation would start with
Prob(B>A)=Prob(A>B)>0<Prob(A=B) prior
probabilities, and then adjust using Bayes' law to posterior
probabilities in the order
Prob(A=B)>Prob(A>B)>Prob(B>A)>0</div>
<div dir="auto"><br>
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<div dir="auto">and Prob(B>A) > Prob(G>F), so
according to Bayes, the F>G defeat is stronger (less
likely to be reversed in a parallel universe) than the
A>B defeat.</div>
<div dir="auto"><br>
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<div dir="auto">Somebody should do the precise Bayesian
calculations to verify (or refute) my statistical intuition.</div>
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