# [EM] Discounting ties, how can MinMax differ from Ranked Pairs? ooops!

Tue Jun 11 22:37:33 PDT 2019

```They can't when there are three candidates.  But plain MinMax doesn't
meet Clone-Winner

You can have a trio of clones in a cycle who all narrowly pairwise beat
another single candidate, but
whose strongest defeats (which are only to each other) are stronger than
any of that single (Condorcet Loser) candidate's
defeats (to all of them).

But Smith//MinMax doesn't have that problem and will nearly always give
the same winner as Ranked Pairs
(and Schulze and River).

A very small consolation prize for MinMax(Margins) is that unlike those
methods it meets Mono-add-Top (like IRV).

Chris Benham

On 12/06/2019 2:25 pm, robert bristow-johnson wrote:
>
> I said some things wrong.  First of all, if there are N candidates,
> the number of Pairwise Defeats is N⋅(N-1)/2, not double that value.
>
> Secondly, I did the scenario with the cycle wrong. I wanted several
> N-3 values in the decreasing sequence, not just one. But I hope you
> have the idea.
>
> r b-j
>
>
>
> ---------------------------- Original Message ----------------------------
> Subject: [EM] Discounting ties, how can MinMax differ from Ranked Pairs?
> From: "robert bristow-johnson" <rbj at audioimagination.com>
> Date: Tue, June 11, 2019 9:46 pm
> To: election-methods at electorama.com
> --------------------------------------------------------------------------
>
> >
> >
> >
> > Let's assume Margins, but I think this works for Winning Votes also.
> > Begin with all candidates marked as "Plausible Winner" (which means
> they are not yet marked as "beaten") and order all N⋅(N-1) candidate
> Pairwise Defeats from strongest defeat strength to weakest defeat
> strength.  Call that ordered list the "Original List".
> > Now create another ordered list of Pairwise Defeats from the above
> Original List.
> >
> > MinMax starts with
> > the entire Original List with N⋅(N-1) entries and
> successively eliminates entries from the bottom up.
> > Referring to a quote in this Wikipedia
> article: https://en.wikipedia.org/wiki/Minimax_Condorcet_method#Variants_of_the_pairwise_score ,
> MinMax starts at the bottom and:
> >      "Disregards the weakest Pairwise Defeat until one candidate is
> unbeaten."
> >
> >
> >
> > Similarly Ranked Pairs starts with an empty list and successively
> includes entries from the Original List from the top down.
> > Cannot Ranked Pairs be concisely stated as:
> >      "Include the strongest Pairwise Defeat until only one candidate
> is unbeaten."   ?
> >
> > Is that not an accurate description of both
> > methods?
> > Now, a decreasing sequence of numbers (*not* strictly decreasing) of
> length N⋅(N-1)+1 that represents the number of "unbeaten" candidates
> (of all Pairwise Defeats above it) having
> > value of N at the top, that can be thought of as having each of the
> N⋅(N-1) Pairwise Defeats inserted between these numbers of decreasing
> integer value.
> > Assuming no ties, the value of this sequence at the bottom
> > must be either 0 or 1.  Also, assuming no ties, this decreasing
> sequence can only be decremented by 1 or 0.
> > If the value of this sequence is 1 at the bottom, there is a
> Condorcet Winner and all Pairwise Defeats are considered.  But
> regardless of whether MinMax or RP is used, this
> > ordered list is the same decreasing sequence of integer values.  If
> the value of this sequence of numbers is 0 at the bottom, there is no
> Condorcet Winner.  But, for all Pairwise Defeats having "1" below
> them, would not the undefeated candidate be the same
> > candidate?
> > This might look like:
> >
> > Plausible Winners ----- Pairwise Defeat
> > N
> >                              A>D
> > N-1
> >                              A>C
> > N-2
> >                              B>D
> > N-2
> >                              B>A
> > N-3
> >                              B>C
> > N-3
> >                              C>D
> > N-3
> >
> > Now if N=4 then the bottom N-3 is 1 and we have a Condorcet Winner.
> > Now suppose we have a cycle:
> > Plausible Winners ----- Pairwise Defeat
> > N
> >                              A>D
> > N-1
> >                              A>B
> > N-2
> >                              B>D
> > N-2
> >                              C>D
> > N-2
> >                              B>C
> > N-3
> >                              C>A
> > N-4
> >
> > Now if N=4 then the bottom N-4 is 0 and we don't have a Condorcet
> Winner.  But the unbeaten candidate is A whether we include the B>C
> Pairwise Defeats or not.  Assuming no ties, how can the unbeaten
> candidate be different?
> > You guys may have discussed
> > this before when I wasn't paying attention, but it seems to me that
> if there are no ties and all of the Defeat strengths are unequal
> values, the ordering of this list must be the same and the number of
> Plausible Winners must decrease from N at the top to 1 at the bottom
> if there is a CW (or N to 0
> > if no CW).  How can MinMax and Ranked Pairs elect a different candidate?
> > Thank you for any attention and thought put to this.
> > --
> >
> >
> > r b-j                         rbj at audioimagination.com
> >
> >
> >
> > "Imagination is more important than knowledge."
> >
> >
> >
> >
> >
> > ----
> > Election-Methods mailing list - see https://electorama.com/em for
> list info
> >
>
>
> --
>
> r b-j                         rbj at audioimagination.com
>
> "Imagination is more important than knowledge."
>
>
> ----
> Election-Methods mailing list - see https://electorama.com/em for list info

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