[EM] average over time proportionality election method

Jobst Heitzig heitzig-j at web.de
Mon Mar 6 11:28:01 PST 2006

Dear Raphael!

> It would be the number of votes + excess (effective total) that the
> 2nd place candidate has.  The winning candidate has that number of
> votes subtracted from his total as he needs at least that number to
> win.

OK, understood. What I don't get is why you subtract just this number
from the winner's votes.

I would instead find the following variant natural:

Start with "excess" 0 for all voters. In each election, add to each
voter's excess the fraction of the vote she received in that election
(that is, a number between 0 and 1) to get the new excess. Elect the
candidate whose so-computed excess is largest and subtract 1 from this
voter's excess.

In this way, a candidate's "excess" is exactly the number of times that
candidate should have won by the proportionality requirement, minus the
number of times the candidate did actually win. Since it is easy to see
that these excesses sum up to 0 and are always between -1 and the number
of candidates, it follows directly that in the long run the quotient
between wins deserved and wins received converges exactly to 1 for each
candidate, with the order of convergence being 1/#elections. And: at no
time is there a candidate who won more than one time more often than she

So, what was the motivation for your version?

Yours, Jobst

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