# [EM] Condorcet cyclic drop rule

Tom Ruen tomruen at itascacg.com
Tue Mar 27 14:23:21 PST 2001

```I've been curious about the Condorcet drop rule with cycles. Previously I've
just computed a pair win table but not implemented rules to handle cycles.

At: http://russp.org/ElectionMethods.org/CondorcetEx.htm
****
* Plain Condorcet (PC)
* The Plain Condorcet method of ambiguity resolution is the
* simplest method: drop the weakest (smallest magnitude)
* defeat, repeating if necessary until one of the candidates
* is unbeaten.
****

It seems like there are two ways to define weakest defeat:
1. The defeat with the smallest difference.
2. The defeat with the smallest ratio (closest to one).

Example ballots: AC=5, BA=4, CB=3

Pair elections:
A:B=5:7 - diff=2 ratio=1.4
B:C=4:8 - diff=4, ratio=2
A:C=9:3 - diff=6, ratio=3

A<B is the weakest defeat by both rules, so the result is A>C>B, and A wins.

However if some voters bullet vote:
Example ballots: AC=3, A=2, BA=4, CB=3

Pair elections:
A:B=5:7 - diff=2, ratio=1.4
B:C=4:6 - diff=2, ratio=1.5
A:C=9:3 - diff=6, ratio=3

Which is the weakest defeat? (With some work I could come up with a case
that the difference and ratio pick different defeats, but a tied difference
is still interesting.)

I would judge the weakest defeat as the one in the election with more voters
participating. Therefore the 5:7 defeat is "weaker" than the 4:6 defeat.

Do these examples look correct? Might weakest ratio be a better criterion
than weakest difference?

Tom Ruen

```