[EM] Re: majority rule, mutinous pirates, and voter strategy

[James Green-Armytage]

>
Hi Juho,
	My critique of your pro-minimax(margins) argument follows...

>I tend to see margins as "natural" and winning votes as something that 
>deviates from the more natural margins but that might be used somewhere 
>to eliminate strategic voting. (not a very scientific description but I 
>don't have any better short explanation available :-) )

	No, that's more or less how I think of it. However, when you say that wv
might be needed "somewhere" to reduce (not eliminate) strategic voting, I
suggest that most public elections will fall within the region of
"somewhere". (Please see my 3/14 post.)
>
copying your pirate example for reference:
101: a>b>x>c
101: b>c>x>a
101: c>a>x>b
100: x
...
>
>I meant that when X was the captain people wanted to change him to A, B 
>or C with a small margin of votes. But later when e.g. C became the 
>captain people wanted to change him to B with a large margin. Only a 
>minority wanted to change C to X. 

	I'm with you this far.

>But the point is that people 
>(majority of them) are now "less happy" 

	...you don't know how happy they are with any of these candidates...

>or "more mutinous" because of 
>the problematic B>C relationship. 

	Okay, let's get to the bottom of this. 
	No matter who wins, 202 pirates would rather have some other candidate in
particular. If X wins, this still holds, but 201 pirates strictly
disagree. In the other cases, e.g. A wins, 202 pirates would rather have
C, and only 101 pirates strictly disagree (the remaining 100 are
indifferent). 
	Your logic is as follows: If X wins, and a group of 202 pirates who
preferred another candidate rather than X wanted to mutiny, there would be
201 pirates ready to stand in their way, serving as an effective
deterrent. However, if A wins, and the 202 C>A pirates (101: B>C>X>A, 101:
C>A>X>B) mutiny in favor of C, there won't be sufficiently many pirates to
fight to defend A.
	Here's what I'd like you to consider: Let's say that A is the initial
winner, these 202 C>A pirates declare mutiny, and the 100 X pirates stay
neutral. There may or may not be a scuffle, but anyway the 101 A>B>X>C
pirates back down. Okay fine; C is the captain. But now the B>C pirates
will be emboldened to mutiny against C. The process repeats, and B is the
captain. Now it will be the A>B pirates' turn, and A will be captain once
more. This idiotic process could go on indefinitely, so that the captain
might shift several times in the duration of any given voyage, causing
general irritation. Or, it could result in serious violence, and there is
no guarantee that C will be on top when the dust settles.
	I suggest to you that this is a relatively intelligent bunch of pirates.
(This is evidenced by the fact they are using Condorcet's method to make
decisions.) If so, I suggest that the 202 C>A pirates will see the
risk/futility of their mutiny ahead of time. (I'm assuming that all the
pirates know each other's expressed ranked preferences, as would be the
case in any real public election.) Sure, they could oust A in favor of C
by force if the X voters sat on their hands. Maybe they could even kill
candidate A, so as to finalize his defeat. But if they did that, a pro-B
mutiny would be likely to follow, and perhaps this new coalition would
murder candidate C, for good measure. Half of the C>A voters (101:
B>C>X>A) would be all the more delighted with this second mutiny, but the
other half (101: C>A>X>B) would rather have A than B, and they would mourn
for C's death.
	So I ask you, would the B>C>X>A voters participate in the first mutiny
against A? I suggest that they would not, because they would realize that
a victory for C so reached would be unlikely to last.	In short, you
neglected to assign foresight to your imaginary pirates, and foresight
would prevent a mutiny against a Smith set member. Would foresight prevent
a mutiny against a non-Smith member, in favor of a Smith member? Not
necessarily! Example:

	Preferences:
35: R>S>T>Z
33: S>T>R>Z
32: T>R>S>Z
71: Z>R=S=T
	Pairwise comparisons:
R>S 67-33
S>T 68-32
T>R 65-35
R>Z 100-71
S>Z 100-71
T>Z 100-71

	Candidate Z is the minimax(margins) winner. However, he is in no wise the
most mutiny-proof candidate. If Z is the initial winner, then all 100 of
the R/S/T faction will have a common cause in ousting him. Perhaps if they
change the winner to R, there could conceivably be further mutiny, but no
matter what, such further mutiny will not lead to another result that the
R/S/T pirates like less than Z. (Hence they can happily mutiny against Z
without worrying that it will hurt them in the long run.) More likely,
however, there will be no further mutiny. The R/S/T faction would do well
to first choose whom they prefer among themselves (let's say that they
settle on R), and to then march over to the Z faction and announce the
change of leadership. The odds are running heavily in favor of the R/S/T
faction if a fight breaks out.
	Again, once Captain R (as in "ARRR!") takes over, any potential mutiny
coalition has to face the prospect of subsequent mutinies that cause a
result that they like less than Captain R. So I argue that Captain R would
suffer less risk of mutiny than Captain Z.
	I hope that I have disrupted your assumptions concerning the "risk of
mutiny" concept. 
>
>I think all the majorities are unambiguous (because that is what the 
>voters told us). A>X could be called "loopless", if we want to describe 
>how it is different from the others. Both electing X and electing A 
>violate a majority opinion. One can avoid violating A>X by not electing 
>X (= select one of the Smith candidates). But one can also avoid 
>violating e.g. A>B by not electing B. All of the individual preferences 
>are thus avoidable. And all the Smith loop violations can be avoided by 
>electing X. 

	If there is a majority rule cycle, then one cannot avoid ignoring at
least one majority preference. However, one can always avoid ignoring a
majority preference that is not contradicted by another majority
preference (via a cycle). 

>> In your pirate example, there are no compromise
>> candidates; the pirate electorate is very badly polarized.
>I agree. The basic setting is four parties of about equal size. I think 
>this situation is quite normal. 

	Four parties of equal size. Okay, that's not very common, but there's no
particular reason why it couldn't happen. What I'm calling your attention
to is not the relative size of the parties, but the intensity of the
polarization between them. We have intense political polarization in
countries that have voting systems that encourage polarization. In
Condorcet systems, we should not assume that this polarization will
remain; rather, it seems logical that compromise candidates will emerge,
which they haven't done in your example.
>
>I claim that 
>"mutiny" is one well defined criterion that is useful is some 
>situations and directly points out the correct voting method (MinMax 
>with margins).

	Please read and consider my recent post about strategic vulnerability in
"margins" methods before you state so unequivocally that it is "the
correct voting method". Actually, even then you might want to be careful
about calling anything "the correct voting method" without some sort of
qualification.
>
>Mutiny of everyone against one is one candidate for another real life 
>criterion. I think mutiny to replace one with one is however the most 
>useful and typical case (both in the ship and in politics). This 
>"mutiny for anyone else" would also give support to sticking to the 
>Smith set when electing the winner. 

	If your second criterion is to select the candidate who is not the first
choice of the fewest voters, this is equivalent to selecting the candidate
with the most first choice votes, a.k.a. plurality.

>That is not allowed :-). We had an election with four candidates. And 
>elections are not supposed to cause countries to break into separate 
>smaller countries. The best single winner election method must be 
>capable of electing one (the best) of these candidates. 

	Sure, but if all of the candidates are highly divisive (as they are in
your example), you can't blame the method for choosing a divisive
candidate. Based on the information available, A, B, and C are equally
good choices, which is to say that they are equally bad choices. X is a
slightly worse choice, because choosing X unnecessarily violates majority
rule.

all my best,
James
http://fc.antioch.edu/~james_green-armytage/voting.htm