Smith//IRV

(Redirected from Smith,IRV)

The method

1. Eliminate all candidates not in the Smith set.

2. Perform an IRV tally among remaining candidates.[1]

Example:

```49: A>B
3: B
48 C>B```

B pairwise beats A (51 to 49) and C (52 to 48), so B is the only candidate in the Smith set i.e. is the Condorcet winner. Therefore, A and C are eliminated, and B, being the only remaining candidate, wins.

Notes

When there are only 3 candidates in the Smith set, Smith//IRV elects the pairwise winner between the two candidates with the most 1st choices after eliminating everyone outside of the Smith set and redistributing support.

Smith//IRV passes ISDA (Independence of Smith-dominated Alternatives) but fails mono-add-plump (adding in ballots that bullet vote the winner shouldn't make the winner lose), which is the opposite of several other Condorcet-IRV hybrids.[2]

Smith//IRV's precinct-summability depends on how large the Smith Set is: while a non-summable method like IRV requires O(c!) space in the worst case, Smith//IRV only requires O(k!), where k is the number of candidates in the Smith set. However, it's not possible to decide beforehand how large the Smith set will be; any such restriction will cause the method to violate universal domain. Thus, Smith//IRV is non-summable in full generality. However, once the precincts report the pairwise table and the Smith Set is known, at most the precincts only have to do a second pass of the ballots (if the Smith Set is 3 candidates or larger). In general, if the Smith Set has n candidates in it, then each precinct only has to report the top n-2 preferences each voter has over the Smith Set candidates. When the Smith Set has 3 candidates (which is expected to be the case most of the time that it is larger than 2 candidates), only each voter's 1st choice in the Smith Set need be known, since IRV will eliminate the candidate with the fewest 1st choices, and then there will be only 2 candidates remaining; the winner in IRV when only 2 candidates remain is guaranteed to be the winner of the pairwise matchup between the two candidates, so only the pairwise table need be consulted then to determine the winner. If there are 4 candidates in the Smith Set, each voter's 1st and 2nd choice in the Smith Set must be known, since once the candidate with the fewest 1st choices is eliminated, 3 candidates will remain, and then the 2nd choices of the voters who ranked the just-eliminated candidate 1st must be known to determine who to eliminate next, then there will be only 2 candidates remaining, and their pairwise matchup in the table determines the winner, etc. If the Smith Set was rather large (say, 10 candidates or greater), only then would it likely be necessary to conduct centralized counting as is done with IRV.

One hybrid of Smith//IRV and Benham's method would be "eliminate everyone outside the Smith Set, then do IRV, but before each elimination, elect the Condorcet winner (based solely on pairwise matchups between uneliminated candidates) if there is one."

Some discussion on Smith//IRV and other Condorcet-IRV hybrids (names differ in the linked article):[3]

It is also possible to do Condorcet//IRV: "elect the Condorcet winner if there is one, otherwise elect the IRV winner."

Neither Smith//IRV nor Condorcet//IRV passes dominant mutual third burial resistance,[4] though both pass when the DMT set consists of a single candidate. Both methods elect from the resistant set, granting them greater resistance to burial strategy.