Consensus, Condorcet(0), and Condorcet(1/2)

Hugh R. Tobin htobin at
Fri Jul 26 07:34:41 PDT 1996

At 01:04 PM 7/25/96 -0700, Steve Eppley wrote::
>[A message I sent to EM was eaten when broke down last 
>week.  Here's the gist of it, hastily rewritten.]
>The discussion about using "1/2 votes against both" in pairwise 
>methods led me to think about what it is that the method ought to be 
>measuring, and ponder the following question:
>Which of the following is the greater violation of democratic 
>principle in a 2-candidate race:
>1. Electing candidate A even though A lost to B (46 to 54).
>2. Electing candidate A even though A lost to B (34 to 46, 
>     with 20 who have no preference between them).
>There's no clearcut answer, seems to me.  In case 1, A has a smaller
>margin of defeat and more supporters than in case 2.  But in case 2
>there are fewer voters who would be made unhappier by changing the 
>result from B to A than in case 1.
>Condorcet(0) implements the principle that case 1 would be a greater
>violation of democracy.  It tries to minimize the number of squawking 
>voters when breaking the circular tie.  I don't know what Condorcet(1/2) 
>>---Steve     (Steve Eppley    seppley at

But in the tiebreak we are comparing the results of at least three pairwise
The voter who ranked C higher than both A and B is clearly unhappy with
A's victory.  The question is whether it is fair or democratic that if C beat B
by 60-40 but lost to A 51-49,  A should win over C under your case 2 because
a large
block of the C voters considered A and B equally detestable, whereas if the same
block were split evenly between A and B (or engaged in vote-pairing to express
their indifference in the manner most favorable to C), their own candidate
would win.
Giving the victory to C in that case, by counting A as having 56 against,
seems more democratic
because it gives effect to the unhappiness of the C voters with A, without
giving them
any extra "votes-against" for tiebreak purposes due to their equal ranking
of A and B
(as in one odd variant mentioned recently as a straw man, where one could
cast a whole
vote against  each).  Where voters rank  candidates equally at the bottom of
their ballots,
counting one-half each way more truly captures the hypothetical unhappiness
of those voters.
The "zero" rule could be said to falsify their preferences by treating them
as if they had no
objection to either A or B.  

Where the equal rankings are at the top of a ballot, one might argue that
the voter would be totally
satisfied with either one and thus should not be forced to cast a half-vote
against each, with
the possible result of victory for her lowest-ranked candidate in case of a
circular tie.
Hence my modest proposal, posted earlier, which also reduces the pressure on
the C voter
to falsify a preference between A and B merely so as not to lose one of her
votes-against in the tiebreak.

-- Hugh Tobin
Registered ICC User
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