Effective number of parties
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Author Topic: Effective number of parties  (Read 4346 times)
Хahar 🤔
Xahar
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« on: May 25, 2010, 01:01:20 AM »

This might not belong here, O Boardbashi, so I humbly beg you move this thread if its presence distrurbs the tranquility of this board.

This looks interesting:



Perhaps someone who speaks math could interpret? It's lifted from Page 331 of this article (JSTOR access required).

Graphs from the same article:

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Lief 🗽
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« Reply #1 on: May 25, 2010, 02:47:02 AM »

Effective number of parties is actually a pretty standard comparative politics calculation. I can't read math, so maybe that article has some fancy new formulas, but as far as I remember it, the calculation is 1 divided by the sum of the square of the percentages (as a decimal) of each party. You can do it for both the popular vote and the seat percentage, which you can compare ((Nv - Ns) / Nv * 100) to get the percentage reduction in proportionality created by the electoral system.

Anything between 1 and 3 is a two-party system, 3 to 5 or 6 is moderately multi-party, and more than six is extreme multi-party. Most of the time the results are pretty obvious, but it's a handy way to quantify what can be somewhat subjective.
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Harry Hayfield
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« Reply #2 on: May 25, 2010, 03:46:43 AM »

Here in Ceredigion, we have the following parties who contest and have councillors elected
(All tallies as of last locals in 2008)

Conservatives 805 votes (from 11 candidates) winning 0 councillors
Labour 885 votes (from 2 candidates) winning 1 councillor
Liberal Democrats 6,655 votes (from 26 candidates) winning 10 seats
Plaid Cymru 12,141 votes (from 34 candidates) winning 19 seats
There are also 12 Independents (who may or may not be members of other parties as well who polled 7,371 votes, two Greens who polled 139 votes between them and a member of the Popular Alliance who polled 117 votes as an Independent)
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platypeanArchcow
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« Reply #3 on: May 25, 2010, 11:19:53 AM »
« Edited: May 25, 2010, 09:49:49 PM by platypeanArchcow »

So yeah, the formula calculates essentially the amount of choice voters have.  If you have $n$ parties which each get $1/n$ of the vote, then you get $1/(n(1/n^2))=n$ for the effective number of parties.  If the parties are significantly imbalanced, that decreases the effective number.  E.g. if there are two parties but one wins 2/3 of the vote, the effective number is $1/((2/3)^2+(1/3)^2)=9/5=1.8$.  If anything, this overestimates this effective number, since if one party wins two-thirds of the vote, that's essentially a one-party system.  Or if there are three parties which get 1/2, 1/4, and 1/4, we get $1/(1/4+1/16+1/16)=8/3$, even though that's essentially a two-party system.  But maybe there's some sort of theory that makes this formula meaningful.

Btw, I suspect, Xahar, that you're hobbling yourself by claiming to be unable to do math.

(Edit: oh how fail I at counting!)
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