[EM] Re: Condorcet package-wvx

[Daniel Bishop]

Ted Stern wrote:

>On 24 Feb 2005 at 18:45 PST, Daniel Bishop wrote:
>  
>
>>(quoting Ted):
>>    
>>
>>>Counting X1=X2=X3=...=X1000 as a fractional 0.001 vote for each candidate over
>>>every other is both impractical and nearly pointless.
>>>      
>>>
>>Don't you mean half a vote for each candidate over every other?  Of 
>>course, if you're using margins, it doesn't make any difference.
>>    
>>
>
>For 1000 equal ranked candidates, half a vote over every other is as bad as a
>full vote over every other.  You're still creating extra votes and breaking
>the "equal weight to each voter in each contest" rule.
>  
>
No, I'm not.

Casting a fully-ranked Condorcet ballot is equivalent to voting in each 
of the n*(n-1)/2 pairwise elections (with the restriction that 
non-transitive preferences are not allowed).  Consider the single 
pairwise election between X and Y:

* The ranking X>Y gives 1 vote for X.
* The ranking Y>X gives 1 vote for Y.
* The ranking X=Y, under the counting method I was referring to, gives 
1/2 a vote to X and 1/2 a vote to Y.  That's a total of 1 vote, just as 
if the voter did express a preference.

>>At least for single-winner Condorcet elections, I don't think it's 
>>necessary to explicitly count X=Y as (0.5 X>Y + 0.5 Y>X) as long as they 
>>are equivalent in the sense of
>>    
>>
>
>I'm not counting ranked ballots that way.  I'm arguing that what you say is
>wrong: don't count any votes for one candidate OR the other when they are of
>equal rank.  If it doesn't matter to you because you prefer margins, then
>don't hobble wv unnecessarily.
>  
>
You totally reversed the meaning of my statement.  I said that (0.5 X>Y 
+ 0.5 Y>X) had no advantage over simply ignoring X=Y.  This is an 
argument in favor of ignoring it: If the difference is unimportant, you 
should do things the easy way.

There are, however, other election methods in which it is not easy to 
deal with equal rankings by simply ignoring them.  For example, how 
would you count A=B>C in STV?  In this case, I think that counting the 
ballot as 0.5 A>B>C + 0.5 B>A>C is the most logical thing to do 
(although not perfect -- I'll write more on this tomorrow.)