Jobst's example could be something like a vote to determine where the olympic games or some world wide conference was to take place.
Notice that the question, "Which is the best city for this purpose?" has different answers for different voters, as opposed to some voters being right and some being wrong. If (in order to share costs) all the participants were to contribute (to a transportation pool) the average cost of getting to the winning city, then it would be to their economic advantage to choose the city minimizing the average voter distance (assuming that flight costs were proportional to distance). But that's a pretty big "if." Since in most cases it's every man for himself, it would probably be more practical to achieve a measure of fairness by rotating the location among the cities. Democratic Fair Choice would accomplish this fairly, over time. Forest From: Jobst Heitzig <[EMAIL PROTECTED]> Subject: [EM] strange 2-dim. fun example To: [email protected] Message-ID: <[EMAIL PROTECTED]> Content-Type: text/plain; charset=us-ascii 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
<<winmail.dat>>
---- Election-methods mailing list - see http://electorama.com/em for list info
