50-49 Tie

[Mike Ossipoff]

DEMOREP1 at aol.com writes:
> 
> Yet another Condorcet circular tie--

Presumably, when you say "Condorcet tie", you mean "circular tie".
A Condorcet tie is an instance where Condorcet's method returns
a tie. Circular ties aren't Condorcet ties, though a Condorcet
tie could occur in a small committee election, with or without
a cicular tie.

You word "Condorcet tie" dishonestly implies that it's a tie
returned by Condorcet's method, or that it somehow results from
Condorcet's method.

So, does your example result in a circular tie? No. A is
the beats-all winner. One of us mis-counted. At this time
I believe it was you who mis-counted.

> 48 AC
> 1   AB
> 
> 48 B
> 
> 1 CA
> 1 CB
> 
> 50 A > 49 B
> 50 B > 49 C
> 50 C > 49 A
> Does common sense say which candidate (and his/her voters) should lose to
> resolve the tie ?

If there were a circular tie, you'd have to do more than claim
common sense when saying who should win. Sometimes widely-held
principles answer the question. Other times they might not apply
to that particular election, but since all elections aren't the
same, it's desirable to have a method that will honor those
principles in general. In other words, some particular example
might not say anything about what the method should be. There
are times when there's no real case for saying who should win,
as in a completely symmetrical natural circular tie.
 

> 
> C should lose and his/her 2 first choice supporters.  The reason- the lowest
> number (i.e. smallest minority) of voters should have to drop their first
> choices to resolve the tie.

"Should" according to you?

> 
> Thus A should win.
> 
> The circular tie breaker applies in like manner with 4 or more candidates in
> the circular tie and then redoing the head to head math.
> 
> .-
> 


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