[EM] MDDA: prefer deluxe, evaluate properties

[Kevin Venzke]

Warren,

--- Warren Smith <wds at math.temple.edu> a écrit :
> MDDA *if* all votes are full rank orderings, is just the "Smith set"
> and often yields a tied election.  In fact often the Smith set is the entire
> set of candidates.  (In most of Australia full rank orderings - i.e. none omitted -
> are demanded by law.)

If everyone approves all the candidates, then yes, there will likely be a
tie. But there is no connection with the Smith set.

> This seems a severe problem with MDDA and probably is a good reason to prefer
> "Deluxe MDDA" which will (in the limit of a large number of voters and with some randomness)
> generically not be tied, even with all voters expressing maximum information.  

I'm not a fan of an explicit approval cutoff for MDDA, since it makes burial
strategy risk-free unless your opponents also use burial strategy.

> Properties of Deluxe MDDA:
> It is fairly simply defined, though not as simple as range voting.
> 
> Deluxe MDDA seems plainly monotonic in both senses (ranking and approval thresh) simultaneously.
> 
> It elects a Condorcet Winner if one exists.
> 
> It refuses to elect a Condorcet loser.

(Deluxe) MDDA doesn't satisfy Condorcet or Condorcet Loser. It only considers
wins with more than half of the voters on the winning side.

> It is generically untied.
> 
> It is Clone-immune.

(Deluxe) MDDA doesn't satisfy Clone Independence. If candidate A wins, it is
possible that when A is replaced by a set of candidates in a cycle, the winner
won't be in this set.

>MDDA fails "add top".  That is, if you add some identical honest votes ranking A top,
>that can harm A (e.g. by creating a Condorcet winner [who is not A]
>who then wins, whereas previously there was a Condorcet cycle and A was the winner
>on approval counts).

MMDA fails Mono-add-top because adding A-top votes could cause a lower-ranked
candidate to lose a majority-strength loss, so that this candidate is the only
one not disqualified.

>Now this [mono-add-top failure] may not technically count as an FBC failure, because 
>I daresay there
>exists some way to rank A top and dishonestly order the remaining candidates,
>which still leaves A the winner.

True. Just rank the candidate (who would win with a sincere vote) last.

>   However, in practice, it may have a very similar
>effect to FBC failure.

I don't think so, because the problem is not caused by ranking A first.
In this situation, lowering A is not helpful.

Kevin Venzke



	

	
		
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