[EM] Problem solved (for pure ranked ballot

Abd ul-Rahman Lomax abd at lomaxdesign.com
Thu Jan 25 07:03:04 PST 2007

At 04:20 AM 1/25/2007, Michael Ossipoff wrote:
>But all kinds of methods are better than eachother in some sense or other.

Prove it.

(Hint: "better" is undefined for most of us, but Warren has a definition.)

I'm surprised that Ossipoff missed the problem. Range is subject to 
the same analysis as any other ranked method. Range is a ranked 
method plus *optional* information that Range uses if it is present 
(specifically, variations in preference strength). So if allegedly 
all ranked methods fail some criterion, with the proof being a 
specific election example, then Range will also fail the criterion, 
because Range ballots can be arranged to contain the same information 
as ranked ballots.

Only if the criterion considers preference strength will this not be 
true. Criteria that consider preference strength could be expected to 
show Range as superior, given a reasonable definition of "superior" 
and a reasonable criterion.

The flaw in Warren's proof, if I'm correct, is that you can show that 
Favorite Betrayal works with any method. I gave a brief proof on the 
Range list, it is repeated here, so that I can be doubly embarrassed 
if I'm suffering a brain fault.

>Now, if we can do what Warren did, then we can prove that no method, 
>including Range, satisfies ICC.
>Initial election consists of three voters voting a tie, each votes 
>A>B>C>A, but, of course, with their favorite omitted from it. We 
>will name the voters A, B, C, after their true favorite.
>If the initial condition is a tie, the voter can change the voter's 
>vote to vote as did the voters of their second preference. Indeed, 
>this creates a clone. The election now has two voters for one 
>candidate, and one for the remaining candidate. As long as the 
>method would elect a favorite of a 2/3 majority, it violates ICC.
>Range is a ranked method, allowing equal ranking, and possibly with 
>additional information about preference strength, which it uses. So 
>if voters choose to vote in such a way as to establish equally 
>separated ranks, which they generally would be able to do, they can 
>cause Range to violate any criterion that pure ranked methods would 
>violate, with the possible exception of criteria which consider 
>preference strength, thus allowing discrimination between Range and 
>the other ranked methods.
>If we apply this example to Range, The voters may rank their 
>favorite at max, their least favorite at min, and the remaining 
>candidate at midrange. Call this a range of -1, 0, +1. The sum of 
>votes for all candidates is initially zero. A tie.
>Now, *for whatever reason*, the one of the voters votes to reverse 
>his vote, to cause the second preference to win. Call it the C 
>voter. The initial votes were:
>A:  1,  0, -1
>B: -1,  0,  1
>C:  0, -1,  1
>it becomes
>A:  1,  0, -1
>B: -1,  0,  1
>C: -1,  0,  1
>B and C are clones, by the definition Warren is obviously using. A 
>wins, which was C's second preference.

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