Equal defeats in RP(m)

[Blake Cretney]

If you want a result that doesn't depend on a random tie-breaker,
then, I would do the following.  Find all the possible winners (based
on Zavist and Tideman's method, and with arbitrary TBRC).  That is,
don't assume that the TBRC has to be drawn from the ballots, a
possible winner is a winner for any TBRC.  I can help if you have any
trouble calculating this.

For the purposes of Voter's Choice, score this as approval for all the
possible winners, and, of course, for anyone the voter likes more.

Since I assume you will find this easier to carry out, this is the
tiebreaker I now give for the purposes of this Voter's Choice
election.

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Blake Cretney   http://www.fortunecity.com/meltingpot/harrow/124/path

Ranked Pairs gives the ranking of the options that always reflects 
the majority preference between any two options, except in order to
reflect majority preferences with greater margins. 
(B. T. Zavist & T. Tideman, "Complete independence  of clones in the 
ranked pairs rule", Social choice and welfare, vol 6, 167-173, 1989)