[EM] Sainte-Lague vs d'Hondt for party list PR
email9648742 at gmail.com
Fri Jul 6 21:10:30 PDT 2012
Looking again at your Sainte-Lague splitting-strategy example, I don't
think that the situation is quite as bad as you said.
The smaller group, with 39% of the voters ends up with only 53% of the
seats, unless I've made an error, which is quite possible.
Here's the scenario as I find it:
I'll express votes in terms of percent. So we have 26 "parties" with 1.5
each, and one with 61.
I'll use .5, 1.5, 2.5 for the SL denominators
Dividing votes by denominators, we get the scores:
Small parties: 1.5/.5 = 3.
Large party: Until it gets 20 seats, its score is at least 3.1282. So it
gets 20 seats before the small parties get any.
Then its score is down to 2.97561
Now the small parties have a higher score. They all get another seat,
resulting in new scores of 1.
Now, 46 seats have been given, and the large party has the highest score.
For it to have a score as low as the small parties, it would have to have
61 seats. So the large party gets the rest of the seats.
So, the small parties, together, have 26 seats, and the large party has 23
The small party, with 26 seats, has 53% of the seats.
The small parties, by their devious strategy, have gotten an undeserved
majority. Even though it isn't 80%, it's still a majority.
Your suggestion of modified Sainte-Lague, with the 1st seat a d'Hondt seat,
by using 2 instead of 1 as the first denominator, would be good, especially
if combined with the transfer of votes from unseated parties, or the
bottom-end STV you suggested.
Of course the problem could still happen, to a lesser degree, even with
That might explain the popularity of d'Hondt. But d'Hondt has two serious
1. A PR parliament is supposed to be a scale model of the voting
population. d'Hondt's great large-bias distorts that scale model.
2. d'Hondt can strategically force people to vote for a compromise party
instead of their favorite, in order to maximize their weight in parliament.
What other solutions are there? Largest-Remainder. In your example, In LR,
the large party immediately gets 29 quota seats, and then the first
remainder seat. The small parties get the rest of the remainder seats.
Large party: 30
Small "parties": 19
Very different. LR is a valid and good contingency plan if SL should ever
turn out to still have a splitting-strategy problem even with the Modified
Sainte-Lague. I'd definitely prefer LR to d'Hondt.
I've criticized LR's random deviations from proportionality. And you might
say that it's ridiculous that a party that has gained votes could lose a
seat to a party that has lost votes. But those things aren't important in
the long run, compared to LR's neat avoidance of the problem that you
described, and its unbias, under the assumption of uniform probability
I couldn't get the rest of the text to delete, but all of my comments are
above this point.
On Mon, Jul 2, 2012 at 6:58 AM, Raph Frank <raphfrk at gmail.com> wrote:
> Another possibility is "alternative-vote" based PR.
> You rank up to 2 parties. Something like,
> Use divisors 2, 3, 5, 7, 9, ...
> This is Websters but is d'Hondt-like for the first seat.
> The seats would be allocated using that rule, and any party which got no
> seats would be "eliminated" and the votes given to the alternative party.
> I don't think it matters if you use normal or the modified version of
> Webster for the 2nd stage. This eliminates "micro" parties getting seats
> due to the round up from 0.5 to 1. If 0.5 to 1.0 sized parties get a
> boost, then parties would try to aim for those sizes (splitting if needed),
> so it breaks the uniform distribution assumption.
> For example, 26 parties at 1.5% and one party at 61% for a 49 seat
> parliament would split the seats, 20 for the large party and 29 for split
> between the micro parties. The micro parties get 59% of the seats for 39%
> of the vote.
> This pretty much has the same effect as a threshold, but at least votes
> can be reassigned. You could also have a random threshold. For example,
> if there was 100 seats, then the threshold could be a random number between
> 1% and 2% of the vote. This would make it hard to game the threshold
> You could also go down the IRV path, and keep eliminating the weakest
> party one at a time, until all remaining parties get at least 1 seat. The
> ranked ballot could also have more than 2 slots, but 2 slots is probably
> enough. The voters would be recommended to pick a party with > 2-3 seats
> for their alternative choice.
> Also, has anyone look into picking the number of seats in a range, so as
> to minimise bias. Use Websters, but pick a house size +/- 10 from nominal
> and have some measure. Does that add or remove bias? That might require a
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