Richard Lung voting at ukscientists.com
Fri Jul 13 11:56:51 PDT 2018

```  I took ranked pairs to mean Condorcet pairing. No?

Richard Lung.

On 13/07/2018 17:37, Kristofer Munsterhjelm wrote:
> On 2018-07-13 08:53, Richard Lung wrote:
>>
>> It is ironic that the world seems to have this electoral reform
>> battle of the big-enders versus the little-enders (from Gullivers
>> Travels). That is to say the collectivist Europeans and their
>> outliers, back a proportional count with party lists - such as
>> sabotages individual choice, without a ranked choice for voters.
>> While the good ole US of A does not forget individual representation
>> with a ranked choice, but forgets equality of representation with a
>> proportional count. The only exception is Cambridge Mass. using STV,
>> and one or two minor cases of STV in Minnesota, I believe.
>> One EM member says STV is BAD. As the ignored inventor of FAB STV, I
>> know the limitations of traditional STV but essentially it is on the
>> right lines, laid down by the original inventors, Carl Andrae and
>> Thomas Hare, namely the quota-preferential method, as the Aussies
>> pithily describe its essence.
>>
>> As for ranked pairing, my understanding is that it is not an
>> independent method at all, but a means of cross-referencing a ranked
>> choice electoral system. As previously mentioned to this email group,
>> I wrote a supplemetary chapter on this, in my book FAB STV: Four
>> Averages Binomial Single Transferable Vote.
>
> If you mean Ranked Pairs, that is a method of its own. It goes like this:
>
> 1. Construct a pairwise matrix (call it d) from the set of ranked
> ballots. Usually this is defined as c[x, y] = number of voters who
> preferred x to y, and d[x, y] = c[x, y] if c[x, y] > c[y, x],
> otherwise d[x, y] = 0. (wv) Margins is also possible; see below.
>
> 2. Make a list of pairwise contests and sort it from greatest value of
> d, to least value of d. A pairwise contest entry is of the form "X
> over Y" and has strength equal to d[x, y].
> E.g. if d[a, b] = 100 and d[b, c] = 80, then "A over B" comes before
> "B over C".
>
> 3. Go down the list and affirm a pairwise contest unless it would
> contradict what has already been affirmed. "Affirming X over Y" means
> that the final outcome must rank X over Y.
>
> 4. Return the candidate who has nobody affirmed over him; he is the
> winner.
>
> Strictly speaking, what I've described above is MAM (without
> tiebreakers). Ranked Pairs "proper" uses margins (d[x, y] = c[x, y] if
> c[x, y] > c[y, x], otherwise d[x, y] = c[x, y] - c[y, x]).
>
> An example instance of step 3 could go like:
>
> If the list is:
>     40: A over B
>     34: B over C
>     30: C over A
>      0: B over A
>      0: C over B
>      0: A over C
>
> Then step three proceeds this way:
>     Affirm A over B: A must be higher ranked than B in the outcome
>     Affirm B over C: B must be higher ranked than C in the outcome
>     C over A: Impossible because we've affirmed A over B and B over C,
> which means we indirectly have A over C, and C over A would contradict
> it.
>     B over A: Contradicts A over B
>     C over B: contradicts B over C
>     Affirm A over C: A must be higher ranked than C in the outcome
> (though we already knew that).
>
> And then the outcome is: A first, B second, C third. A is the winner.
>

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