[EM] strange 2-dim. fun example

[Jobst Heitzig]

Hi folks!

Here's some fun example showing that, even with a 2-dimensional model
and only 4 candidates, classical voting methods can give completely
different winners and orders...

In the example, 100 voters from all over the globe elect the earth's
capital from the following candidates (which build an almost regular
tetrahedron):


CANDIDATES:

   candidate     phi  theta
   |             |    |  place
   |             |    |  |

   a           145  -38  Melbourne
   b          -123   38  San Francisco
   c           -58  -35  Buenos Aires
   d            53   30  Persepolis


In this, phi and theta are angular coordinates with respect to Greenwich
and to the equator, in degrees.

The voters' preferences are simply by distance, and all approve the
nearest two cities:


VOTERS:

   voters
   | preferences (according to distance)
   | and approval cutoffs:
   | |           phi  theta
   | |           |    |  place
   | |           |    |  |

   4 a>b>>c>d  178  -17  Fidji Islans
   3 a>b>>d>c  147   -9  Port Moresby
   8 a>c>>b>d  173  -44  Christchurch
   5 a>d>>b>c  134  -23  Alice Springs
   6 a>d>>c>b   78  -39  St. Paul Island

   7 b>a>>c>d -159   22  Honolulu
   8 b>c>>d>a  -83   23  Havana
   8 b>d>>c>a -113   53  Edmonton

   5 c>a>>b>d  -68  -50  Piedrabuena
   3 c>a>>d>b  -27  -76  Halley Bay
   8 c>b>>a>d  -91    0  Galapagos Islands
   3 c>b>>d>a  -60   -3  Manaus
   4 c>d>>a>b   18  -34  Cape Town
   1 c>d>>b>a  -28  -20  some small island east of Rio de Janeiro

   6 d>a>>b>c  114   23  Hong Kong
   5 d>a>>c>b   78    4  Male
   5 d>b>>a>c   69   42  Tashkent
   3 d>b>>c>a   38   56  Moscow
   8 d>c>>b>a    4    7  Lagos

   ---
   100


Now let us see what different methods do with this information:


ANALYSIS:

   ------------------------------
                   c   b   d   a
   ------------------------------
   1st ranks      24  23  27  26  --> plurality order: d>a>c>b
   2nd ranks      24  26  24  26
   ------------------------------
   approval       48  49  51  52  --> approval order:  a>d>b>c
   ------------------------------
   3rd ranks      33  33  17  17
   4th ranks      19  18  32  31  --> "skating" order: a>d>b>c
   ------------------------------
   Borda score   153 154 146 147  --> Borda order:     b>c>a>d
   ------------------------------
   above c         -  49  49  49
   above b        51   -  46  49
   above d        51  54   -  49  --> transitive defeats,
   above a        51  51  51   -      minmax order:    c>b>a>d
   ------------------------------
   Copeland score  3   2   1   0  --> Copeland order:  c>b>d>a
   ------------------------------
   IRV            24  23  27  26
                  31      35  34
                          51  49  --> IRV order:       d>a>c>b
   ------------------------------
   random ballot and DFC
   winning probabilities:
                 .24 .23 .27 .26
   ------------------------------


If I didn't mess the numbers up, the classical methods
Approval/Borda/Condorcet/Plurality elect A/B/C/D, respectively.

As it turns out, each of the cities can get each rank from 1 to 4
depending on the method, but there is not a single defeat circle which
would indicate any problems...


Aren't such examples a good reason to always consider more than just one
piece of the information (like approval or ranks or defeats or direct
support) and instead combine them to get the whole picture?

Yours, Jobst