Later-no-help criterion

Anti-plurality elects the candidate the fewest voters rank last when submitting a complete ranking of the candidates.

Later-No-Help can be considered not applicable to Anti-Plurality if the method is assumed to not accept truncated preference listings from the voter. On the other hand, Later-No-Help can be applied to Anti-Plurality if the method is assumed to apportion the last place vote among unlisted candidates equally, as shown in the example below.

Assume four voters (marked bold) submit a truncated preference listing A > B = C by apportioning the possible orderings for B and C equally. Each vote is counted ${\displaystyle {\tfrac {1}{2}}}$ A > B > C, and ${\displaystyle {\tfrac {1}{2}}}$ A > C > B:

Result: A is listed last on 3 ballots; B is listed last on 2 ballots; C is listed last on 6 ballots. B is listed last on the least ballots. B wins. A loses.

Now assume that the four voters supporting A (marked bold) add later preference C, as follows:

Result: A is listed last on 3 ballots; B is listed last on 4 ballots; C is listed last on 4 ballots. A is listed last on the least ballots. A wins.

The four voters supporting A increase the probability of A winning by adding later preference C to their ballot, changing A from a loser to the winner. Thus, Anti-plurality fails the Later-no-help criterion when truncated ballots are considered to apportion the last place vote amongst unlisted candidates equally.

Coombs' method repeatedly eliminates the candidate listed last on most ballots, until a winner is reached. If at any time a candidate wins an absolute majority of first place votes among candidates not eliminated, that candidate is elected.

Later-No-Help can be considered not applicable to Coombs if the method is assumed to not accept truncated preference listings from the voter. On the other hand, Later-No-Help can be applied to Coombs if the method is assumed to apportion the last place vote among unlisted candidates equally, as shown in the example below.

Assume four voters (marked bold) submit a truncated preference listing A > B = C by apportioning the possible orderings for B and C equally. Each vote is counted ${\displaystyle {\tfrac {1}{2}}}$ A > B > C, and ${\displaystyle {\tfrac {1}{2}}}$ A > C > B:

Result: A is listed last on 4 ballots; B is listed last on 4 ballots; C is listed last on 6 ballots. C is listed last on the most ballots. C is eliminated, and B defeats A pairwise 8 to 6. B wins. A loses.

Now assume that the four voters supporting A (marked bold) add later preference C, as follows:

Result: A is listed last on 4 ballots; B is listed last on 6 ballots; C is listed last on 4 ballots. B is listed last on the most ballots. B is eliminated, and A defeats C pairwise 8 to 6. A wins.

The four voters supporting A increase the probability of A winning by adding later preference C to their ballot, changing A from a loser to the winner. Thus, Coombs' method fails the Later-no-help criterion when truncated ballots are considered to apportion the last place vote amongst unlisted candidates equally.

This example shows that Copeland's method violates the Later-no-help criterion. Assume four candidates A, B, C and D with 7 voters:

Assume that the two voters supporting A (marked bold) do not express later preferences on the ballots:

The results would be tabulated as follows:

Result: Both A and B have two pairwise wins and one pairwise tie, so A and B are tied for the Copeland winner. Depending on the tie resolution method used, A can lose.

Now assume the two voters supporting A (marked bold) express later preferences on their ballot.

The results would be tabulated as follows:

Result: B now has two pairwise defeats. A still has two pairwise wins, one tie, and no defeats. Thus, A is elected Copeland winner.

By expressing later preferences, the two voters supporting A promote their first preference A from a tie to becoming the outright winner (increasing the probability that A wins). Thus, Copeland's method fails the Later-no-help criterion.

Dodgson's' method elects a Condorcet winner if there is one, and otherwise elects the candidate who can become the Condorcet winner after the fewest ordinal preference swaps on voters' ballots.

Later-No-Help can be considered not applicable to Dodgson if the method is assumed to not accept truncated preference listings from the voter. On the other hand, Later-No-Help can be applied to Dodgson if the method is assumed to apportion possible rankings among unlisted candidates equally, as shown in the example below.

Assume ten voters (marked bold) submit a truncated preference listing A > B = C by apportioning the possible orderings for B and C equally. Each vote is counted ${\displaystyle {\tfrac {1}{2}}}$ A > B > C, and ${\displaystyle {\tfrac {1}{2}}}$ A > C > B:

Result: B is the Condorcet winner and the Dodgson winner. A loses.

Now assume that the ten voters supporting A (marked bold) add later preference C, as follows:

Result: There is no Condorcet winner. A is the Dodgson winner, because A becomes the Condorcet Winner with only two ordinal preference swaps (changing B > A to A > B). A wins.

The ten voters supporting A increase the probability of A winning by adding later preference C to their ballot, changing A from a loser to the winner. Thus, Dodgson's method fails the Later-no-help criterion when truncated ballots are considered to apportion the possible rankings amongst unlisted candidates equally.

For example, in an election conducted using the Condorcet compliant method Ranked pairs the following votes are cast:

A is preferred to C by 70 votes to 30 votes. (Locked)
B is preferred to A by 42 votes to 28 votes. (Locked)
B is preferred to C by 42 votes to 30 votes. (Locked)

B is the Condorcet winner and therefore the Ranked pairs winner.

Suppose the 28 A voters specify second choice C (they are burying B).

The votes are now:

A is preferred to C by 70 votes to 30 votes. (Locked)
C is preferred to B by 58 votes to 42 votes. (Locked)
B is preferred to A by 42 votes to 28 votes. (Cycle)

There is no Condorcet winner and A is the Ranked pairs winner.

By giving a second preference to candidate C the 28 A voters have caused their first choice to win. Note that, should the C voters decide to bury A in response, B will beat A by 72, restoring B to victory.

Similar examples can be constructed for any Condorcet-compliant method, as the Condorcet and later-no-help criteria are incompatible.