Jobst Heitzig <[EMAIL PROTECTED]> wrote: >Hello Ralph! > >You in reply to me wrote: >>>Can you prove this? Seems not so obvious to me... >> >> It is based on the assumption that the "cost" in votes to get elected >> is roughly the same in every election. If it always costs 1000 votes >> to get elected and a candiate gets 300 votes, the candidate is 30% of >> the way to get a seat. > >Sorry to insist, but can you prove the proportionality? I tried some >minutes but failed, so at least it is not obvious... What is obvious is >that more votes lead to being elected more often (=monotonicity), and >having non-zero votes leads to being elected eventually.
If the number of votes needed to be elected is constant, then it follows automatically. A candidate who gets 10% of this value every election will win the seat every 10th election. Someone who gets 25% of this value will win the election every 4th election. The question comes down to how constant that threshold is. It isn't perfectly constant so the system isn't perfectly proportional. This is due to a combination of the fact that voter turnout isn't constant, and also some inherent randomness in the process. Simulations indicate that candiates between 10 and 40% support received within 2% of the correct number of seats over 100 elections and mostly were within 1%. There is an upper limit on how high the threshold can be. It is equal to the total excess held by 2 candidates from the previous election plus the total vote divided by 2. This occurs if all the votes are voted are for just 2 candidates and one wins by only one vote. The maximum excess that a candidate can hold is the threshold from the previous election (as otherwise the candidate would be elected) T(n) = threshold for the nth election V(n) = votes cast in the nth election E(n) = total excess carried forward from the nth election T(n) <= T(n-1) + V(n)/2 A second effect is that the sum of all the thresholds must be less than the sum of all the votes cast. This is due to the fact that that is the only way votes can leave the system. Sum(V(n)) = Sum(T(n)) + E(n) giving V(n) = T(n) + [E(n) - E(n-1)] This gives a long term average with the threshold equal to the turnout in each election with some additional high frequency noise. The more candidates who take part in the election, the less noise as each candidate would get a small number of votes and would all be spread evenly from 0 to the threshold. The totals for the top two would be barely change from election to election. If the threshold is to be set to be constant to remove this noise, it would need to be a multi seat district or people would need to be willing to accept that sometimes they get no representative and other times they get 2. In a large multi seat district, the threshold would likely not move much at all. One possible issue is if lots of new candidates run in an election, this pushes down the threshold as they all start with 0 excess. It might be worth putting in a rule for a minimum threshold or something. __________________________________________________________________ Switch to Netscape Internet Service. As low as $9.95 a month -- Sign up today at http://isp.netscape.com/register Netscape. Just the Net You Need. New! Netscape Toolbar for Internet Explorer Search from anywhere on the Web and block those annoying pop-ups. Download now at http://channels.netscape.com/ns/search/install.jsp ---- election-methods mailing list - see http://electorama.com/em for list info
