# [EM] Copeland DMTBR (DMTCBR) failure

Kristofer Munsterhjelm km_elmet at t-online.de
Sun Jun 26 06:28:48 PDT 2022

```Here's a failure example for Copeland, showing that Copeland//X or
Copeland,X can't pass DMTBR, no matter what method X is. This should
work with ties being counted at half a point, as well as ties counting zero:

4: A>B>C>D
4: B>A>C>D
1: C>D>A>B
1: D>A>B>C

A is the CW and DMT candidate and thus wins. Then the B faction buries:

4: A>B>C>D
4: B>C>D>A <-- burial
1: C>D>A>B
1: D>A>B>C

and Copeland elects B, hence violating DMTCBR (and DMTBR).

According to Rob LeGrand's voting calculator, the example also works for
Black, Borda, Bucklin, and Small's method: first A is elected, then B is
elected after the burial.

The Copeland variant where ties count as much as a victory (which
apparently Llull proposed?) doesn't seem to have any "simple" DMTBR
failures for four candidates, unless my simulator again is bad. (By
"simple" I mean that there's a single winner, then the burial happens,
and then there's another single winner, i.e. there are no ties either
before or after.)

If it really has no such failures, that's interesting, but it yields to
five candidates:

1: B>A>C>D>E
2: C>E>D>B>A
2: D>E>C>A>B

C is the CW and DMTC and wins. Then bury:

1: B>A>C>D>E
2: C>E>D>B>A
2: D>E>A>B>C <-- burial

and D wins.

Point out where I made a mistake if I did, of course :-)

-km
```