On 10/04/2012 07:05 PM, Peter Zbornik wrote:
Dear all,
A simple extention of IRV to two rounds IRV would be the following:
1. In the first round have no quota (i.e. no transfer of surpluses).
2. The two candidates who are eliminated last go to the second round
3. In the second round the two candidates meet in a normal IRV election.
Question: Will this method always generate the same winner as one-round
IRV, in the case that the preference orderings of the voters are the
same in both rounds?
I believe yes and it seems trivial, but cannot prove it just like that.
Yes, at least if you disregard ties. Say that after every round but the
last has been run, A and B remain. Then A will win in IRV iff a majority
of the ballots that express a difference between A and B, ranks A above
B. Now consider a genuine runoff between A and B. The voters can only
give their vote for either A or B (or abstain), and if the majority
interested in that contest votes for A, then A wins, and same for B.
Since the preference orderings are the same, this A-vs-B contest is also
the same as in the final round of IRV.
I am considering proposing this method for use among the Czech Greens as
an improved IRV and a natural two-round alternative to the Run-off
elections we use today.
It might be better than Plurality runoff - I haven't heard of anybody
using IRV+runoffs before, so I don't know its performance for sure.
If you're looking at variants of IRV, you could ensure the CW stays in
the contest (when there is one) by not eliminating the Plurality loser
at each stage of IRV, but the one who loses a one-on-one between the two
candidates with least Plurality score for the round in question. That
may be of use if you want to have a runoff and at the same time make
sure the Condorcet winner stays in it.
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