Truncated Ballot Profile

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.

Adding Later Preferences

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.

Conclusion

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.

Truncated Ballot Profile

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.

Adding Later Preferences

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.

Conclusion

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.

Truncated preferences

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.

Express later preferences

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.

Conclusion

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.

Truncated Ballot Profile

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.

Adding Later Preferences

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.

Conclusion

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.