Selected topics in finite mathematics/Condorcet criterion
[Give a very very brief overview of the criteria?]
Objectives
editTo satisfy the Condorcet criterion, a candidate that is favored to beat ever other candidate in a pairwise race wins. Therefore, if a candidate can defeat every other candidate, then they should win.
Details
edit[State the definition of the criteria?]
[Rephrase the criteria into if-then form?]
[Give a prose-explanation of the criteria?]
Examples
editThe Condorcet method satisfies the Condorcet criteria. Indeed, this is its namesake: if a candidate wins all pairwise races, that candidate wins by the Condorcet method.
Sequential pariwise elections also satisfies the Condorcet criterion. In particular, suppose A wins in all pairwise races. Then as soon as A comes up in the agenda, A will win every [pairwise] race until declared the winner.
Nonexamples
editPlurality fails to satisfy the Condorcet criterion. Consider the following election. A wins in all pairwise elections, but instead B wins by plurality.
1st choice | A | B | C |
---|---|---|---|
2nd choice | B | A | A |
3rd choice | C | C | B |
Number of votes | 2 | 3 | 2 |
Borda count fails to satisfy the Condorcet criterion. Consider the following election. One on one A wins all pairwise races. However, in this election as it is, B wins.
1st choice | B | E | D |
---|---|---|---|
2nd choice | C | A | A |
3rd choice | F | B | B |
4th choice | A | C | E |
5th choice | E | D | C |
6th choice | D | F | F |
Number of votes | 1 | 1 | 1 |
Sequential runoffs fail to satisfy the Condorcet Criterion. Consider the following election. Indeed one on one A could defeat either candidate. But in the election as a whole, A is eliminated first.
1st choice | A | B | C |
---|---|---|---|
2nd choice | B | A | A |
3rd choice | C | C | B |
Number of votes | 2 | 3 | 3 |
FAQ
editHomework
edit[Create two elections: one which satisfies the criteria and one which does not?]
[You can add the solution to the problem here!]
[You can add the solution to the problem here!]