Maximize Affirmed Majorities

From Academic Kids

Maximize Affirmed Majorities (MAM) is a voting method developed by Stephen Eppley that selects a single winner using votes that express preferences. MAM can also be used to create a sorted list of winners.

If there is a candidate who is preferred over the other candidates, when compared in turn with each of the others, MAM guarantees that that candidate will win. Because of this property, MAM is (by definition) a Condorcet method. Note that this is different from some other preference voting systems such as Borda and Instant-runoff voting, which do not make this guarantee.

MAM is a variation of the Ranked Pairs (RP) voting method, with additional refinements to break ties using as much information as available from a set of votes. Another voting method that is a variation of ranked pairs is Maximum Majority Voting (MMV). One of the primary "competing" Condorcet methods is Cloneproof Schwartz Sequential Dropping (CSSD).



The MAM procedure is as follows:

  1. Tally the vote count comparing each pair of candidates, and determine the winner of each pair (if there is one)
  2. Compute the Tiebreaker (this is unnecessary if there are many voters)
  3. Sort (rank) each pair, using the MAM ranking method (primarily by the number of winning votes)
  4. Affirm 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.

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


To tally the votes, consider each voters' 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.

Compute Tiebreak

Compute a tiebreak ordering, in case there are exactly the same number of votes for a situation. Do this by repeatedly retrieving an unused random vote, using its order to define an ordering of the candidates, until all candidates are ordered or all votes are used up. Once an order is determined, don't change it later. If all votes are used up, assign the remaining ranks randomly (this can only occur if no voter expressed a preference; if no voter expressed a preference, voting methods don't have anything to work with). This process is called the Random Voter Hierarchy procedure.

If there is a privileged alternative (e.g., the status quo), or a privileged voter (e.g., a chairperson), that alternative and/or voter could be used first before the random choices. The original developer of MAM does not consider that normally a part of MAM.


The pairs of defeats, 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 at least 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.
  3. Vxy = Vzw and Vwz = Vyx and Tiebreak ranks w over y.
  4. Vxy = Vzw and Vwz = Vyx and w = y and Tiebreak ranks x over z.


The next step is to examine each pair in turn to determine which pairs to "affirm". Using the sorted list above, affirm each pair in turn unless the pair will create a circularity in a graph (e.g., where A is more than B, B is more than C, but C is more than A). Then use the tiebreaker information, if necessary, to disambiguate any cases.

An Example

The Situation

Imagine an election for the capital of Tennessee, a state in the United States that is over 500 miles east-to-west, and only 110 miles north-to-south. Let's say the candidates for the capital are Memphis (on the far west end), Nashville (in the center), Chattanooga (129 miles southeast of Nashville), and Knoxville (on the far east side, 114 northeast of Chattanooga). Here's the population breakdown by metro area (surrounding county):

  • Memphis (Shelby County): 826,330
  • Nashville (Davidson County): 510,784
  • Chattanooga (Hamilton County): 285,536
  • Knoxville (Knox County): 335,749

Let's say that in the vote, the voters vote based on geographic proximity. Assuming that the population distribution of the rest of Tennessee follows from those population centers, one could easily envision an election where the percentages of votes would be as follows:

42% of voters (close to Memphis)
1. Memphis
2. Nashville
3. Chattanooga
4. Knoxville

26% of voters (close to Nashville)
1. Nashville
2. Chattanooga
3. Knoxville
4. Memphis

15% of voters (close to Chattanooga)
1. Chattanooga
2. Knoxville
3. Nashville
4. Memphis

17% of voters (close to Knoxville)
1. Knoxville
2. Chattanooga
3. Nashville
4. Memphis

The results would be tabulated as follows:

Pairwise Election Results
Memphis Nashville Chattanooga Knoxville
BMemphis[A] 58%
[B] 42%
[A] 58%
[B] 42%
[A] 58%
[B] 42%
Nashville[A] 42%
[B] 58%
[A] 32%
[B] 68%
[A] 32%
[B] 68%
Chattanooga[A] 42%
[B] 58%
[A] 68%
[B] 32%
[A] 17%
[B] 83%
Knoxville[A] 42%
[B] 58%
[A] 68%
[B] 32%
[A] 83%
[B] 17%
Pairwise election results (won-lost-tied): 0-3-0 3-0-0 2-1-0 1-2-0
Votes against in worst pairwise defeat: 58%N/A68%83%
  • [A] indicates voters who preferred the candidate listed in the column caption to the candidate listed in the row caption
  • [B] indicates voters who preferred the candidate listed in the row caption to the candidate listed in the column caption
  • [NP] indicates voters who expressed no preference between either candidate


First, list every pair, and determine the winner:

Memphis (42%) vs. Nashville (58%) Nashville 58%
Memphis (42%) vs. Chattanooga (58%) Chattanooga 58%
Memphis (42%) vs. Knoxville (58%) Knoxville 58%
Nashville (68%) vs. Chattanooga (32%) Nashville 68%
Nashville (68%) vs. Knoxville (32%)Nashville 68%
Chattanooga (83%) vs. Knoxville (17%) Chattanooga: 83%

Note that absolute counts of votes can be used, or percentages of the total number of votes; it makes no difference.

Compute Tiebreak

Tiebreak computation is only useful if there are only a few voters, and is unlikely to be needed to determine a winner of a statewide election. Indeed, in this example the winner will be the same regardless of the tiebreak value. However, for example purposes, presume a tiebreak was found by pulling a specific vote; the most likely of the options is a voter close to Memphis, and if such a vote is selected the tiebreak would be "Memphis > Nashville > Chattanooga > Knoxville".


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% (an exact tie, which is unlikely in real life for this many voters). Since Chattanooga > Knoxville, and they're the losers, Nashville vs. Knoxville will be added first, followed by Nashville vs. Chattanooga.

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

Chattanooga (83%) vs. Knoxville (17%) Chattanooga 83%
Nashville (68%) vs. Knoxville (32%)Nashville 68%
Nashville (68%) vs. Chattanooga (32%) Nashville 68%
Memphis (42%) vs. Nashville (58%) Nashville 58%
Memphis (42%) vs. Chattanooga (58%) Chattanooga 58%
Memphis (42%) vs. Knoxville (58%) Knoxville 58%


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

  • Affirm Chattanooga over Knoxville.
  • Affirm Nashville over Knoxville.
  • Affirm Nashville over Chattanooga.
  • Affirm Nashville over Memphis.
  • Affirm Chattanooga over Memphis.
  • Affirm Knoxville over Memphis.

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

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

Missing image

In this example, Nashville is the winner using MAM.


In this election, the winner is Nashville. This would be true for any a Condorcet method. Using the first-past-the-post system 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.

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