[EM] an example of BTR-IRV (the name as now been changed to STV-ME)

raphfrk at netscape.net raphfrk at netscape.net
Thu Jun 8 17:43:01 PDT 2006


From: Anthony O'Neal <thasupasacfitinman at gmail.com>
>  Anthony O'Neal wrote:
> > (This is the election I pulled out of the Wikipedia article for
> CPO-STV.
> > I just didn't feel like making up an election where the results from
> > CPO-STV and STV differ right now. If you want to see how
> > the results for the CPO-STV and STV results were arrived upon, then
> > go to the article)
> >
> >(PS, does anyone know what the BTR part of BTR-IRV means?
> > Honestly, I can't figure it out, but that's what Warren on
> Rangevoting.com
> > calls it. I'm thinking of changing the name to Majority Elimination
> by IRV,
> > or ME-IRV, and ME-STV, but if BTR makes more sense...)It is 
probably "Bottom Two Runoff" or something similar.

>I guess I really need to change the name then, because it's not always 
a runoff between the bottom two, but a runoff >between the bottom N + 1 
(where N is the number of seats in the STV election that are currently 
needing to be filled). I'm >thinking Single Transferrable Vote with 
elimination by majorities, although this also doesn't really fit, 
because it's actually >elimination by the majorities of the Droop 
Quota.

I wonder what would be the best way to do your elimination.  Maybe
some kind of PR-STV that finds a loser ... highest ranked candidate
is eliminated and votes transferred until there is only one candidate
left.  Hmm, you could even use BTR-STV, though it would become
"Best" two runoff -- hows that for confusing the issue.

> > Now for this election, using the Hagenbach-Bischoff quota of
> votes/seats + 1,
> > the amount needed to get elected is 25 votes. So Andrea and Carter
> are
> > immediately declared elected, as their amount of votes exceeds the
> quota.
>
>> Why not use the Droop quota, that is much fairer?
>
>I believe you are confused.

I interpretted your formula as a c compiler would.

I thought:

>>Now for this election, using the Hagenbach-Bischoff quota of
>> votes/seats + 1,

meant (votes/seats) +1 rather than (votes)/(seats + 1)

According to wikipedia:

Droop quota is 1+ (Votes)/(Seats+1)
Hare is (Votes)/(Seats)

> You are talking about the Hare quota, which is a rather idealistic 
attempt to represent all voters.

Right, I was saying that if you use the Hare quota, there is a problem.

<snip you pointing out the same problem>

> However, in recommending quotas I'd probably go with Droop
> rather than Hagenbach-Bischoff, because in the extraordinarily
> unlikely event where the quotas would give different results in
> real elections, having a person elected because more votes
> were wasted than needed is quite a bit less catostrophic than
> having more people elected than there are seats.

Another option is to require Droop for a person to be elected, but then
work out the amount of transfer based on the Hagenbach-Bischoff
quota.


> This brings me to another point, though. Why even transfer surplus 
votes? Why not simply elect the person with the most > votes, then 
transfer all of that persons voters votes to the second person on their 
list at a value of V/M * 2 + 1 (where V is > the value of their ballot, 
and M is the number of candidates that person has had elected)?

Wouldn't V = 1 for all ballots ?

Is the formula V/(2*M + 1) ?

So, a ballot is worth a different amount based on how many candidates 
it has elected ?

0 -> 1
1 -> 1/3
2 -> 1/5
3 -> 1/7

etc.

How do you work out how many people a vote has elected ?
I assume it is all votes in the pile when the candidate is elected
(it could be you are considered to elect any elected candidate
placed higher than your current candidate).

A party with 30% and another with 70% would split a 4 seater as:

A1: 30%
A2: 0%
B1: 70%
B2: 0%
B3: 0%

B1 elected ( 70/3 = 23.3 passed to B2)

A1 elected ( 30/3 = 10.0 passed to A2)

B2 elected ( 70/5 = 14 passed to B3 )

B3 elected ( 70/7 = 10 passed to B? )

A: 1
B: 3

Now, if party A vote manage and split the vote evenly, it is:

A1: 15%
A2: 15%
B1: 70%
B2: 0%
B3: 0%

B1 elected ( 70/3 = 23.3 passed to B2 )

B2 elected ( 70/5 = 14 passed to B3 )

A1 elected ( 15/3 = 5 passed to A2 )

A2 elected ( 5/5 + 15/3 = 6 passed to A? )

Fairness requires that party B gets 3 seats as they have
3 full Droop quotas.  They definately shouldn't have
equal representation to a party which only obtained
43% as many votes as they did.  The optimal strategy
is to evenly split the vote over the number of candidates
who are to be elected as transferred votes don't count
at full strength.

If both parties perfectly split their vote, the result would be:

A1: 15%
A2: 15%
B1: 17.5%
B2: 17.5%
B3: 17.5%
B4: 17.5%

B gets 4 elected.

So, hmm, it would seem that the optimal strategy is not quite
so simple.  However, it does seem like vote management matters.

Maybe, including the option to for candidates to withdraw would
help/solve the problem.
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