In social choice theory, the independence of clones criterion says that adding a clone, i.e. a new candidate very similar to an alreadyexisting candidate, should not spoil the results.^{[1]} It can be considered a very weak form of the independence of irrelevant alternatives (IIA) criterion.
More precisely, the process of cloning a candidate involves taking an existing candidate C, then replacing them with several candidates C1, C2... These candidates are then "slotted" into the original ballots in the same spot as C was, with the candidates being arranged in any order. In other words, a group of candidates are called clones if they are always ranked "together" or "sidebyside" by every voter, i.e. no voter ranks any of the nonclone candidates between or equal to the clones. If a set of clones contains at least two candidates, the criterion requires that deleting one of the clones must not increase or decrease the winning chance of any candidate not in the set of clones.
A voting method may satisfy the independence of clones criterion while still being vulnerable to spoiler effects and strategic nomination.^{[2]}
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Avoiding Arrow's Impossibility (Alternative Voting Criteria)

Should we always elect the Condorcet winner?

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Clone negative and clone positive
Election methods that fail independence of clones can be clone negative (the addition of a similar candidate decreases another candidate's chance of winning) or clone positive (the addition of a similar candidate increases another candidate's chance of winning).
The Borda count is an example of a clonepositive method; in fact, the method is so clonepositive that an arbitrary candidate can simply "clone their way to victory", with the winner being the coalition that runs the most clones. Plurality voting is an example of a strongly clonenegative method because of votesplitting.
A method can also fail the independence of clones method without being clonepositive or clonenegative. This happens if the method changes its decision about the winner when a nonwinning candidate is cloned, but the new winner is not the candidate that was cloned. This effect is called crowding. Copeland's method is an example of a method that exhibits crowding.
Compliant methods
Ranked pairs, the Schulze method, and any system that satisfies independence of irrelevant alternatives such as range voting or majority judgment satisfies the criterion. Whether or not instantrunoff voting passes depends on how equalranked ballots are defined, with the more common definition (splitIRV) failing the criterion.^{[3]}
The Borda count, minimax, Kemeny–Young, Copeland's method, Bucklin voting without skipped ranks, plurality, the tworound system, and approval voting with zeroinformation rational voters. Voting methods that limit the number of allowed ranks also fail the criterion, because the addition of clones can leave voters with insufficient space to express their preferences about other candidates. For similar reasons, ballot formats that impose such a limit may cause an otherwise cloneindependent method to fail.
Examples
Borda count
Consider an election in which there are two candidates, A and B. Suppose the voters have the following preferences:
66%: A>B  34%: B>A 
Candidate A would receive 66% Borda points (66%×1 + 34%×0) and B would receive 34% (66%×0 + 34%×1). Thus candidate A would win by a 66% landslide.
Now suppose supporters of B nominate an additional candidate, B_{2}, that is very similar to B but considered inferior by all voters. For the 66% who prefer A, B continues to be their second choice. For the 34% who prefer B, A continues to be their least preferred candidate. Now the voters' preferences are as follows:
66%: A>B>B_{2}  34%: B>B_{2}>A 
Candidate A now has 132% Borda points (66%×2 + 34%×0). B has 134% (66%×1 + 34%×2). B_{2} has 34% (66%×0 + 34%×1). The nomination of B_{2} changes the winner from A to B, overturning the landslide, even though the additional information about voters' preferences is redundant due to the similarity of B_{2} to B.
Similar examples can be constructed to show that given the Borda count, any arbitrarily large landslide can be overturned by adding enough candidates (assuming at least one voter prefers the landslide loser). For example, to overturn a 90% landslide preference for A over B, add 9 alternatives similar/inferior to B. Then A's score would be 900% (90%×10 + 10%×0) and B's score would be 910% (90%×9 + 10%×10).
No knowledge of the voters' preferences is needed to exploit this strategy. Factions could simply nominate as many alternatives as possible that are similar to their preferred alternative.
In typical elections, game theory suggests this manipulability of Borda can be expected to be a serious problem, particularly when a significant number of voters can be expected to vote their sincere order of preference (as in public elections, where many voters are not strategically sophisticated; cite Michael R. Alvarez of Caltech). Small minorities typically have the power to nominate additional candidates, and typically it is easy to find additional candidates that are similar.
In the context of people running for office, people can take similar positions on the issues, and in the context of voting on proposals, it is easy to construct similar proposals. Game theory suggests that all factions would seek to nominate as many similar candidates as possible since the winner would depend on the number of similar candidates, regardless of the voters' preferences.
Copeland
These examples show that Copeland's method violates the Independence of clones criterion.
Crowding
Copeland's method is vulnerable against crowding, that is the outcome of the election is changed by adding (nonwinning) clones of a nonwinning candidate. Assume five candidates A, B, B_{2}, B_{3} and C and 4 voters with the following preferences:
# of voters  Preferences 

1  A > B_{3} > B > B_{2} > C 
1  B_{3} > B > B_{2} > C > A 
2  C > A > B_{2} > B > B_{3} 
Note, that B, B_{2} and B_{3} form a clone set.
Clones not nominated
If only one of the clones would compete, preferences would be as follows:
# of voters  Preferences 

1  A > B > C 
1  B > C > A 
2  C > A > B 
The results would be tabulated as follows:
X  
A  B  C  
Y  A  [X] 1 [Y] 3 
[X] 3 [Y] 1  
B  [X] 3 [Y] 1 
[X] 2 [Y] 2  
C  [X] 1 [Y] 3 
[X] 2 [Y] 2 

Pairwise election results (wontiedlost):  101  011  110 
 [X] indicates voters who preferred the candidate listed in the column caption to the candidate listed in the row caption
 [Y] indicates voters who preferred the candidate listed in the row caption to the candidate listed in the column caption
Result: C has one win and no defeats, A has one win and one defeat. Thus, C is elected Copeland winner.
Clones nominated
Assume, all three clones would compete. The preferences would be the following:
# of voters  Preferences 

1  A > B_{3} > B > B_{2} > C 
1  B_{3} > B > B_{2} > C > A 
2  C > A > B_{2} > B > B_{3} 
The results would be tabulated as follows:
X  
A  B  B_{2}  B_{3}  C  
Y  A  [X] 1 [Y] 3 
[X] 1 [Y] 3 
[X] 1 [Y] 3 
[X] 3 [Y] 1  
B  [X] 3 [Y] 1 
[X] 2 [Y] 2 
[X] 2 [Y] 2 
[X] 2 [Y] 2  
B_{2}  [X] 3 [Y] 1 
[X] 2 [Y] 2 
[X] 2 [Y] 2 
[X] 2 [Y] 2  
B_{3}  [X] 3 [Y] 1 
[X] 2 [Y] 2 
[X] 2 [Y] 2 
[X] 2 [Y] 2  
C  [X] 1 [Y] 3 
[X] 2 [Y] 2 
[X] 2 [Y] 2 
[X] 2 [Y] 2 

Pairwise election results (wontiedlost):  301  031  031  031  130 
Result: Still, C has one win and no defeat, but now A has three wins and one defeat. Thus, A is elected Copeland winner.
Conclusion
A benefits from the clones of the candidate he defeats, while C cannot benefit from the clones because C ties with all of them. Thus, by adding two clones of the nonwinning candidate B, the winner has changed. Thus, Copeland's method is vulnerable against crowding and fails the independence of clones criterion.
Teaming
Copeland's method is also vulnerable against teaming, that is adding clones raises the winning chances of the set of clones. Again, assume five candidates A, B, B_{2}, B_{3} and C and 2 voters with the following preferences:
# of voters  Preferences 

1  A > C > B > B_{3} > B_{2} 
1  B > B_{2} > B_{3} > A > C 
Note, that B, B_{2} and B_{3} form a clone set.
Clones not nominated
Assume that only one of the clones would compete. The preferences would be as follows:
# of voters  Preferences 

1  A > C > B 
1  B > A > C 
The results would be tabulated as follows:
X  
A  B  C  
Y  A  [X] 1 [Y] 1 
[X] 0 [Y] 2  
B  [X] 1 [Y] 1 
[X] 1 [Y] 1  
C  [X] 2 [Y] 0 
[X] 1 [Y] 1 

Pairwise election results (wontiedlost):  110  020  011 
Result: A has one win and no defeats, B has no wins or defeats so A is elected Copeland winner.
Clones nominated
If all three clones competed, the preferences would be as follows:
# of voters  Preferences 

1  A > C > B > B_{3} > B_{2} 
1  B > B_{2} > B_{3} > A > C 
The results would be tabulated as follows:
X  
A  B  B_{2}  B_{3}  C  
Y  A  [X] 1 [Y] 1 
[X] 1 [Y] 1 
[X] 1 [Y] 1 
[X] 0 [Y] 2  
B  [X] 1 [Y] 1 
[X] 0 [Y] 2 
[X] 0 [Y] 2 
[X] 1 [Y] 1  
B_{2}  [X] 1 [Y] 1 
[X] 2 [Y] 0 
[X] 1 [Y] 1 
[X] 1 [Y] 1  
B_{3}  [X] 1 [Y] 1 
[X] 2 [Y] 0 
[X] 1 [Y] 1 
[X] 1 [Y] 1  
C  [X] 2 [Y] 0 
[X] 1 [Y] 1 
[X] 1 [Y] 1 
[X] 1 [Y] 1 

Pairwise election results (wontiedlost):  130  220  031  031  031 
Result: A has one win and no defeat, but now B has two wins and no defeat. Thus, B is elected Copeland winner.
Conclusion
B benefits from adding inferior clones, while A cannot benefit from the clones because he ties with all of them. So, by adding two clones of B, B changed from loser to winner. Thus, Copeland's method is vulnerable against Teaming and fails the Independence of clones criterion.
Plurality voting
Suppose there are two candidates, A and B, and 55% of the voters prefer A over B. A would win the election, 55% to 45%. But suppose the supporters of B also nominate an alternative similar to A, named A_{2}. Assume a significant number of the voters who prefer A over B also prefer A_{2} over A. When they vote for A_{2}, this reduces A's total below 45%, causing B to win.
A 55%  A 30% 
A_{2} not present  A_{2} 25% 
B 45%  B 45% 
Range voting
Range voting satisfies the independence of clones criterion.
Voters changing their opinion
However, like in every voting system, if voters change their opinions about candidates if similar candidates are added, adding clone candidates can change the outcome of an election. This can be seen by some premises and a simple example:
In range voting, to raise the influence of the ballot, the voter can give the maximum possible score to their most preferred alternative and the minimum possible score to their least preferred alternative. In fact, giving the maximum possible score to all candidates that are over some threshold and giving the minimum possible score to the other candidates, will maximize the influence of a ballot on the outcome. However, for this example it is necessary that the voter uses the first simple rule, but not the second.
Begin by supposing there are 3 alternatives: A, B and B_{2}, where B_{2} is similar to B but considered inferior by the supporters of A and B. The voters supporting A would have the order of preference "A>B>B_{2}" so that they give A the maximum possible score, they give B_{2} the minimum possible score, and they give B a score that's somewhere in between (greater than the minimum). The supporters of B would have the order of preference "B>B_{2}>A", so they give B the maximum possible score, A the minimum score and B_{2} a score somewhere in between. Assume B narrowly wins the election.
Now suppose B_{2} isn't nominated. The voters supporting A who would have given B a score somewhere in between would now give B the minimum score while the supporters of B will still give B the maximum score, changing the winner to A. This violates the criterion. Note, that if the voters that support B would prefer B_{2} to B, this result would not hold, since removing B_{2} would raise the score B receives from his supporters in an analogous way as the score he receives from the supporters of A would decrease.
The conclusion that can be drawn is that considering all voters voting in a certain special way, range voting creates an incentive to nominate additional alternatives that are similar to one you prefer, but considered clearly inferior by his voters and by the voters of his opponent, since this can be expected to cause the voters supporting the opponent to raise their score of the one you prefer (because it looks better by comparison to the inferior ones), but not his own voters to lower their score.
Kemeny–Young method
This example shows that the Kemeny–Young method violates the Independence of clones criterion. Assume five candidates A, B_{1}, B_{2}, B_{3} and C and 13 voters with the following preferences:
# of voters  Preferences 

4  A > B_{1} > B_{2} > B_{3} > C 
5  B_{1} > B_{2} > B_{3} > C > A 
4  C > A > B_{1} > B_{2} > B_{3} 
Note, that B_{1}, B_{2} and B_{3} form a clone set.
Clones not nominated
Assume only one of the clones competes. The preferences would be:
# of voters  Preferences 

4  A > B_{1} > C 
5  B_{1} > C > A 
4  C > A > B_{1} 
The Kemeny–Young method arranges the pairwise comparison counts in the following tally table:
All possible pairs of choice names 
Number of votes with indicated preference  

Prefer X over Y  Equal preference  Prefer Y over X  
X = A  Y = B_{1}  8  0  5 
X = A  Y = C  4  0  9 
X = B_{1}  Y = C  9  0  4 
The ranking scores of all possible rankings are:
Preferences  1. vs 2.  1. vs 3.  2. vs 3.  Total 

A > B_{1} > C  8  4  9  21 
A > C > B_{1}  4  8  4  16 
B_{1} > A > C  5  9  4  18 
B_{1} > C > A  9  5  9  23 
C > A > B_{1}  9  4  8  21 
C > B_{1} > A  4  9  5  18 
Result: The ranking B_{1} > C > A has the highest ranking score. Thus, B_{1} wins ahead of C and A.
Clones nominated
Assume all three clones compete. The preferences would be:
# of voters  Preferences 

4  A > B_{1} > B_{2} > B_{3} > C 
5  B_{1} > B_{2} > B_{3} > C > A 
4  C > A > B_{1} > B_{2} > B_{3} 
The Kemeny–Young method arranges the pairwise comparison counts in the following tally table (with ) :
All possible pairs of choice names 
Number of votes with indicated preference  

Prefer X over Y  Equal preference  Prefer Y over X  
X = A  Y = B_{i}  8  0  5 
X = A  Y = C  4  0  9 
X = B_{i}  Y = C  9  0  4 
X = B_{1}  Y = B_{2}  13  0  0 
X = B_{1}  Y = B_{3}  13  0  0 
X = B_{2}  Y = B_{3}  13  0  0 
Since the clones have identical results against all other candidates, they have to be ranked one after another in the optimal ranking. More over, the optimal ranking within the clones is unambiguous: B_{1} > B_{2} > B_{3}. In fact, for computing the results, the three clones can be seen as one united candidate B, whose wins and defeats are three times as strong as of every single clone. The ranking scores of all possible rankings with respect to that are:
Preferences  1. vs 2.  1. vs 3.  2. vs 3.  Total 

A > B > C  24  4  27  55 
A > C > B  4  24  12  40 
B > A > C  15  27  4  46 
B > C > A  27  15  9  51 
C > A > B  9  12  24  45 
C > B > A  12  9  15  36 
Result: The ranking A > B_{1} > B_{2} > B_{3} > C has the highest ranking score. Thus, A wins ahead of the clones B_{i} and C.
Conclusion
A benefits from the two clones of B_{1} because A's win is multiplied by three. So, by adding two clones of B, B changed from winner to loser. Thus, the Kemeny–Young method is vulnerable against spoilers and fails the independence of clones criterion.
Minimax
This example shows that the minimax method violates the Independence of clones criterion. Assume four candidates A, B_{1}, B_{2} and B_{3} and 9 voters with the following preferences:
# of voters  Preferences 

3  A > B_{1} > B_{2} > B_{3} 
3  B_{2} > B_{3} > B_{1} > A 
2  B_{3} > B_{1} > B_{2} > A 
1  A > B_{3} > B_{1} > B_{2} 
Note, that B_{1}, B_{2} and B_{3} form a clone set.
Since all preferences are strict rankings (no equals are present), all three minimax methods (winning votes, margins and pairwise opposite) elect the same winners.
Clones not nominated
Assume only one of the clones would compete. The preferences would be:
# of voters  Preferences 

4  A > B_{1} 
5  B_{1} > A 
The results would be tabulated as follows:
X  
A  B_{1}  
Y  A  [X] 5 [Y] 4  
B_{1}  [X] 4 [Y] 5 

Pairwise election results (wontiedlost):  01  10  
worst pairwise defeat (winning votes):  5  0  
worst pairwise defeat (margins):  1  0  
worst pairwise opposition:  5  4 
 [X] indicates voters who preferred the candidate listed in the column caption to the candidate listed in the row caption
 [Y] indicates voters who preferred the candidate listed in the row caption to the candidate listed in the column caption
Result: B is the Condorcet winner. Thus, B is elected minimax winner.
Clones nominated
Now assume all three clones would compete. The preferences would be as follows:
# of voters  Preferences 

3  A > B_{1} > B_{2} > B_{3} 
3  B_{2} > B_{3} > B_{1} > A 
2  B_{3} > B_{1} > B_{2} > A 
1  A > B_{3} > B_{1} > B_{2} 
The results would be tabulated as follows:
X  
A  B_{1}  B_{2}  B_{3}  
Y  A  [X] 5 [Y] 4 
[X] 5 [Y] 4 
[X] 5 [Y] 4  
B_{1}  [X] 4 [Y] 5 
[X] 3 [Y] 6 
[X] 6 [Y] 3  
B_{2}  [X] 4 [Y] 5 
[X] 6 [Y] 3 
[X] 3 [Y] 6  
B_{3}  [X] 4 [Y] 5 
[X] 3 [Y] 6 
[X] 6 [Y] 3 

Pairwise election results (wontiedlost):  003  201  201  201  
worst pairwise defeat (winning votes):  5  6  6  6  
worst pairwise defeat (margins):  1  3  3  3  
worst pairwise opposition:  5  6  6  6 
Result: A has the closest biggest defeat. Thus, A is elected minimax winner.
Conclusion
By adding clones, the Condorcet winner B_{1} becomes defeated. All three clones beat each other in clear defeats. A benefits from that. So, by adding two clones of B, B changed from winner to loser. Thus, the minimax method is vulnerable against spoilers and fails the independence of clones criterion.
STAR voting
STAR voting consists of an automatic runoff between the two candidates with the highest rated scores. This example involves clones with nearly identical scores, and shows teaming.
Clones not nominated
Scores  

# of voters  Amy  Brian  Clancy 
2  5  2  1 
4  4  2  1 
11  0  1  1 
The finalists are Amy and Brian, and Brian beats Amy pairwise and thus wins.^{[4]}
Clones nominated
Scores  

# of voters  Amy  Amy's clone  Brian  Clancy 
2  5  5  2  1 
2  4  3  2  1 
2  4  5  2  1 
11  0  0  1  1 
The finalists are Amy and her clone, and Amy's clone wins.^{[5]}
See also
References
 ^ T. Nicolaus Tideman, "Independence of clones as a criterion for voting rules", Social Choice and Welfare Vol. 4, No. 3 (1987), pp. 185–206.
 ^ J. GreenArmytage (2014). "Strategic voting and nomination". Social Choice and Welfare. 42 (1). Springer: 111–138. doi:10.1007/s0035501307253. ISSN 01761714. JSTOR 43663746. S2CID 253847024. Retrieved 20240223. Figure 4 on page 137 shows Hare voting having exit incentive despite being clone independent.
 ^ Delemazure, Théo; Peters, Dominik (20240417). "Generalizing Instant Runoff Voting to Allow Indifferences". arXiv:2404.11407 [cs.GT].
 ^ Larry Hastings (20230602). "Test election: demonstrate cloneproofness 1". GitHub. Retrieved 20240224.
 ^ Larry Hastings (20230602). "Test election: demonstrate cloneproofness 2". GitHub. Retrieved 20240224.