# Cycle Combinations tie breaker

Mike Ossipoff dfb at bbs.cruzio.com
Wed Aug 7 02:38:38 PDT 1996

```DEMOREP1 at aol.com writes:
>
> Assume that there is a Condorcet cycle with 4 candidates such as A>B>C>D>A.
>
> There are 24 combinations of the candidates (1 of the 4 for first, 1 of the
> remaining 3 for second, 1 of the remaining 2 for third and the last remaining
> candidate-- or 4 x 3 x 2 x 1 or 4 factorial (4 ! ) ).

What you mean there is that there are 24 _orders_ in which 4
candidates can be ordered. Combinations isn't the word, because
combinations refers to sets taken from the 4 candidates, without
regard to order within those sets. A way that mathematicians might
put it is that you're referring to the number of permutations of
4, in 4 objects (candidates). That's written 4P4 (where the 4s are
subscripts, something that e-mail doesn't easily allow).

But the most convenient way for us to put that is to say that
there are 24 orders in which 4 objects can be ordered.

>
> Should the tie breaker be the one combination (if any) that the most voters
> agree with ?

Depends on what people want from the voting system. What are the
advantages? Properties? Criteria met?

>
> The votes in each X versus Y pairing in each of the 24 combinations would
> have to be added together.
>
> Example-
> Assume the combination D>B>A>C has the highest sum (that is, the votes for D
> in the D-B, D-A and D-C pairings plus the votes for B in the B-A and B-C
> pairings plus the votes for A in the A-C pairing).

But, if the object is to find a single winner, then surely the
only pairings relating to D's qualification to win are the pairings
that include D.

What if everyone were indifferent about every pairing not
including D? Why should that reduce D's count score?

And if everyone ranks all the candidates, then, for the purpose
of picking a single winner, doesn't your proposal give the
same results as if we only counted, in D's score, the pairings
involving D?

And if we scored D according to how many votes he received over
all of his pairings, that's the Borda system. I've talked about
the problems of the Borda system. It retains the lesser-of-2-evils
problem; you can fully vote Favorite over Middle, or you can fully
vote Middle over Worst, but you can't do both. And, as I said,
Borda is the only rank-balloting method that I've heard of
that can fail to elect an alternative ranked 1st by a full majority
of all the voters. It fails the Majority Winner Criterion, not
to mention the other majority & LO2E criteria.

Borda, it seems to me, is usually defined by saying that, if there
are N candidates, your 1st choice gets N-1 points from you, your
2nd choice gets N-2 points from you, etc. If everyone ranks
all the candidates, without equal rankings, then this is the
same as how I defined Borda earlier in this letter.

> Note 1- no voter may have actually voted for such final one combination.
> Note 2- all of the 12 head to head pairings from the head to head (Condorcet)

Therea are only 6 head-to-head pairings, with 4 candidates. But
you may be referring to what could be called "half-pairing counts",
such as the count of how many people rank A over B, in the A vs B
pairing.

> matrix are used.
> Note 3- If a majority coalition exists (such as D and B), then such coalition
> should defeat the minority candidates (A or C).
> -------
> With the standard 3 candidate cycle (A>B>C>A), there are 3 x 2 x 1 or 6
> combinations.

6 possible orderings for the 3 candidates.

> Do any folks who have done 3 candidate examples want to do the 6 combinations
> math in their examples ???
>

As I said, when you introduce a new method, it's up to _you_ to
determine its consequences, to tell what it offers, and why you
recommend it. Sure, others could check it out, to find faults,
or to find advantages, and they may or may not do that, but the
burden is on you to demonstrate how your method does, and what
it offers.

Mike

>

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