Circular tie solutions should be obvious

Sun Feb 20 19:40:22 PST 2000

>66 AB 33 (same as above)
>67 BC 32
>65 CA 35
>Is there still support for the single step tiebreaker using the fewest 
>against in the worst defeat (especially with 4 or more choices in a 
>tie) ?

Sure. The arrival of C doesn't change the fact that all of those
66 people still rank A over B, just as they did before. So A still
beats B with 66 people voting for that defeat. That's the beauty
of pairwise count methods. A still wins in the example, partly
because those 66 people are more numerous than the 65 voting C
over A. Condorcet's method is justified by the simple question
of what's the obvious thing to do when the pairwise defeats
conflict with eachother, so that we can't honor all of them,
and so we overrule the one with least support. It's also justified
by its criterion compliances, which I'll describe in a few days.

>My standard comment---  a choice should be getting a majority YES vote
>whether or not it is in a circular tie (in a public legislative body and/or
>in a public election for executive or judicial officers).

But people vote strategically. If you add a Y/N vote to a method,
voters will simply do their best to use that strategically too,
and so they won't use the Y/N vote to vote "Yes" on candidates
whom they absolutely want, and "No" on candidates whom they
absolutely don't want. They'll sometimes give a "Yes" to someone
whom they don't want, because he and someone they like less are
perceived to be the top rivals, for instance.

That's because, though some principled voters regard some candidates
as candidates whom they absolutely don't want & won't vote for
even strategically, a multicandidate election is fundamentally
not a question of "Yes" vs "No". It's a question of "Which?",
and most voters will strategize to get the best outcome they can,
just as they do now.

Additionally, adding Y/N to a method greatly complicates its

Mike Ossipoff

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