Bluffton University > Mathematics > Darryl Nester > Java[Script]: Voting Method Demos
This page is intended to demonstrate the voting methods described in Chapter 9 of For All Practical Purposes.
As in that book, an election is described by each voter's preference list. This simply lists the candidates in order from most to least preferred. For example, if there are 4 candidates (A,B,C,D), and a voter's preference list is CBAD, then that voter would most like C to be chosen, then B, then A, then D. More specifically, if any two candidates were running (because the others had dropped out of the race), that voter would make his or her choice based on which candidate appears first on his/her preference list.
In any election, we would like the voting method used to have certain properties. Unfortunately, Arrow's impossibility theorem says that (when there are three candidates), there is no voting method that can have all of those desirable properties.
A voting method satisfies the Condorcet Winner Criterion if that method will choose the Condorcet winner (described below) when one exists.
Suppose that every voter ranks candidate A higher than B (that is, in a one-on-one election between the two, A would get all the votes). A voting method satisfies the Pareto condition if a candidate B would not be among the winners.
Suppose that we hold an election in which candidate A is one of the winners, and candidate B is one of the losers. Then one voter (say "X") alters his/her preference list, and we hold the election again. If A is now higher on X's preference list, the voting method satisfies monotonicity (or "is monotone") if it is impossible for A to become one of the losers. If B is moved up ... In other words:
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Number of candidates (3-7): Agenda (for sequential pairwise voting): Click the bars below for details of election results. |
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Under plurality, every voter casts a vote for his/her most preferred candidate. Here are the tallies:
The winner is Candidate C.
A Condorcet candidate is one who would win every two-person race. Not every set of preference lists has a Condorcet candidate. If one exists, we can find him/her by considering every possible two-way race, and adding up the total number of wins for each candidate.
The table below shows all 6 two-way election results. In practice, it is not always necessary to find the result of every such election. For example, if B and C have each lost (or tied) a race, then neither can be the Condorcet candidate, so we would not care about the outcome of the B vs. C election.
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Overall (with ties counting as half a win):
The Condorcet winner is B.
For the Borda count, each candidate earns 3 points for being at the top of a preference list, 2 points for being second, ..., and 0 points for being at the bottom. The tallies are:
The result is a 2-way tie between Candidates B and C.
Using the Hare system, we remove the candidate(s) who is/are at the top of the fewest preference lists. Then we promote the remaining candidates in all the preference lists, and repeat until we have one or more winners. (That is, we stop if we have one candidate, or all remaining candidates are tied.)
In the first round of voting, we can remove all candidates with 0 votes, as well as those with the smallest positive tally (unless they are also tied for first!). After the first round, every remaining candidate will have a positive tally.
The winner is Candidate A.
To choose a winner under sequential pairwise voting, we hold a series of two-way elections based on the order specified in the "agenda" (entered above). The winner of each of these elections advances to the "next round."
For the agenda ABCD:
The winner is Candidate B.