Ranked pairs

Procedure

The RP procedure is as follows:

1. Tally the vote count comparing each pair of candidates, and determine the winner of each pair (provided there is not a tie)
2. Sort (rank) each pair, by the largest strength of victory first to smallest last.
3. "Lock in" each pair, starting with the one with the largest number of winning votes, and add one in turn to a graph as long as they do not create a cycle (which would create an ambiguity). The completed graph shows the winner.

RP can also be used to create a sorted list of preferred candidates. To create a sorted list, repeatedly use RP to select a winner, remove that winner from the list of candidates, and repeat (to find the next runner up, and so forth).

Tally

To tally the votes, consider each voter's preferences. For example, if a voter states "A > B > C" (A is better than B, and B is better than C), the tally should add one for A in A vs. B, one for A in A vs. C, and one for B in B vs. C. Voters may also express indifference (e.g., A = B), and unstated candidates are assumed to be equally worse than the stated candidates.

Once tallied the majorities can be determined. If "Vxy" is the number of Votes that rank x over y, then "x" wins if Vxy > Vyx, and "y" wins if Vyx > Vxy.

Sort

The pairs of winners, called the "majorities", are then sorted from the largest majority to the smallest majority. A majority for x over y precedes a majority for z over w if and only if one of the following conditions holds:

1. Vxy > Vzw. In other words, the majority having more support for its alternative is ranked first.
2. Vxy = Vzw and Vwz > Vyx. Where the majorities are equal, the majority with the smaller minority opposition is ranked first.

Lock

The next step is to examine each pair in turn to determine the pairs to "lock in". This can be visualized by drawing an arrow from the pair's winner to the pair's loser in a directed graph. Using the sorted list above, lock in each pair in turn unless the pair will create a circularity in the graph (for example, where A is more than B, B is more than C, but C is more than A).

This step can be somewhat more complicated if there are equal-weighted pairs which create a cycle: A naïve approach will result in the outcome depending on the (unspecified) method of resolving ties in the sort order. One way to resolve this issue is to allow cycles if they are needed to resolve ties (i.e., if a single new edge would not create a cycle, but multiple tied edges would), and then define the winners as the resulting Schwartz set.

Winner

In the resulting graph, the source corresponds to the winner. A source is bound to exist because the graph is a directed acyclic graph by construction, and such graphs always have sources. In the absence of pairwise ties, the source is also unique (because whenever two nodes appear as sources, there would be no valid reason not to connect them, leaving only one of them as a source).

Imagine that Tennessee is having an election on the location of its capital. The population of Tennessee is concentrated around its four major cities, which are spread throughout the state. For this example, suppose that the entire electorate lives in these four cities and that everyone wants to live as near to the capital as possible.

The candidates for the capital are:

The preferences of the voters would be divided like this:

The results would be tabulated as follows:

Tally

First, list every pair, and determine the winner:

Note that absolute counts of votes can be used, or percentages of the total number of votes; it makes no difference since it is the ratio of votes between two candidates that matters.

Sort

The votes are then sorted. The largest majority is "Chattanooga over Knoxville"; 83% of the voters prefer Chattanooga. Nashville (68%) beats both Chattanooga and Knoxville by a score of 68% over 32% (a tie, unlikely in real life for this many voters). Since Chattanooga > Knoxville, and they are the losers, Nashville vs. Knoxville will be added first, followed by Nashville vs. Chattanooga.

Thus, the pairs from above would be sorted this way:

Lock

The pairs are then locked in order, skipping any pairs that would create a cycle:

In this case, no cycles are created by any of the pairs, so every single one is locked in.

Every "lock in" would add another arrow to the graph showing the relationship between the candidates. Here is the final graph (where arrows point away from the winner).

In this example, Nashville is the winner using RP, followed by Chattanooga, Knoxville, and Memphis in second, third, and fourth places respectively.

Ambiguity resolution example

For a simple situation involving candidates A, B, and C.

In this situation we "lock in" the majorities starting with the greatest one first.

Therefore, A is the winner.

Summary

In the example election, the winner is Nashville. This would be true for any Condorcet method.

Using the First-past-the-post voting and some other systems, Memphis would have won the election by having the most people, even though Nashville won every simulated pairwise election outright. Using Instant-runoff voting in this example would result in Knoxville winning even though more people preferred Nashville over Knoxville.

Criteria

Of the formal voting system criteria, the ranked pairs method passes the majority criterion, the monotonicity criterion, the Condorcet criterion, the Condorcet loser criterion, and the independence of clones criterion. Ranked pairs fails the consistency criterion and the participation criterion. While ranked pairs is not fully independent of irrelevant alternatives, it still satisfies local independence of irrelevant alternatives.

Independence of irrelevant alternatives

Ranked pairs fails independence of irrelevant alternatives. However, the method adheres to a less strict property, sometimes called independence of Smith-dominated alternatives (ISDA). It says that if one candidate (X) wins an election, and a new alternative (Y) is added, X will win the election if Y is not in the Smith set. ISDA implies the Condorcet criterion.