[EM] Defeat Strength

[Colin Champion]

Forest – this isn’t an answer, but a dissertation on a related topic.
    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).
    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.
    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?
    If I have time, I might attempt the calculation.
       CJC

On 04/09/2022 04:43, Forest Simmons wrote:
> 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.
>
> 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
>
> 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.
>
> Somebody should do the precise Bayesian calculations to verify (or 
> refute) my statistical intuition.
>

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