# [EM] CIBR examples, and its CC failure

Bishop, Daniel J daniel-j-bishop at neo.tamu.edu
Fri May 27 09:54:08 PDT 2005

```Simmons, Forest wrote:

>This looks promising.  I like this kind of creativity.
>
>Three Questions:
>
>1.  Exactly how do you define correlation?

My suggestion is this:

The "absolute Borda difference" (ABD) between two candidates on one
ballot is the absolute value of the difference of their Borda scores on
that Ballot.

The "total absolute Borda difference" (TABD) between two candidates is
the sum of their ABDs on all ballots.  "Correlation" is the inverse of
the TABD.

(I had intended to send this on Wednesday night, only to find out that
neo.tamu.edu no longer lets me log in as "dbishop".  In the meantime, Ken
posted his suggestion.  It looks like mine is identical, but simpler.)

EXAMPLE

Consider the Tennessee capital election from Wikipedia.  (C=Chattanooga,
K=Knoxville, M=Memphis, N=Nashville)

42: M>N>C>K
26: N>C>K>M
15: C>K>N>M
17: K>C>N>M

Nashville has 194 points, Chattanooga has 173, Memphis 126, and
Knoxville 107.

On the M>N>C>K ballots:
ABD(C, K) = 1
ABD(C, M) = 2
ABD(C, N) = 1
ABD(K, M) = 3
ABD(K, N) = 2
ABD(M, N) = 1

On the N>C>K>M ballots:
ABD(C, K) = 1
ABD(C, M) = 2
ABD(C, N) = 1
ABD(K, M) = 1
ABD(K, N) = 2
ABD(M, N) = 3

On the C>K>N>M ballots:
ABD(C, K) = 1
ABD(C, M) = 3
ABD(C, N) = 2
ABD(K, M) = 2
ABD(K, N) = 1
ABD(M, N) = 1

On the K>C>N>M ballots:
ABD(C, K) = 1
ABD(C, M) = 2
ABD(C, N) = 1
ABD(K, M) = 3
ABD(K, N) = 2
ABD(M, N) = 1

Therefore,
TABD(C, K) = 1×42 + 1×26 + 1×15 + 1×17 = 100
TABD(C, M) = 2×42 + 2×26 + 3×15 + 2×17 = 215
TABD(C, N) = 1×42 + 1×26 + 2×15 + 1×17 = 115
TABD(K, M) = 3×42 + 1×26 + 2×15 + 3×17 = 233
TABD(K, N) = 2×42 + 2×26 + 1×15 + 2×17 = 185
TABD(M, N) = 1×42 + 3×26 + 1×15 + 1×17 = 152

Chattanooga and Knoxville are the most correlated pair, which shouldn't
be too surprising -- they're clones.  Knoxville has fewer Borda points
and is eliminated.  Then the ballots become:

42: M>N>C
26: N>C>M
32: C>N>M

Nashville has 126 Borda points, Chattanooga has 90, and Memphis has 84.

If correlations are not re-calculated after eliminations (which has the
advantage of making the method second-order summable), then Chattanooga
and Nashville are the most correlated.  If correlations are
recalculated, then the new values are:

TABD(C, M) = 174
TABD(C, N) = 100
TABD(M, N) = 126

Either way, Chattanooga is eliminated, and then Nashville beats Memphis,
58-42.

In this particular example, the Condorcet Winner is elected.  However, this is
not always the case.

EXAMPLE

1: A>C>B
2: B>A>C
2: B>C>A
4: C>B>A

The Condorcet Winner is C, who beats A 5-4 and beats B 6-3.  However, the
Borda ranking is B (12) > C (11) > A (4).  The most-correlated pair is B and
C, so C is eliminated, and then B wins.

```