Droop quota

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In the study of electoral systems, the Droop quota (sometimes called the Hagenbach-Bischoff quota^[a]) is the minimum number of votes needed for a party or candidate to guarantee themselves one extra seat in a legislature in modern STV and other voting systems. It is the preferred quota, being known to be less likely than the Hare quota, to give majority of seats to a minority party.^[1] It is the smallest portion of votes that elects the correct number of members to fill the seats, but no more than that number.

It generalizes the concept of a majority to multiple-winner elections: just as a majority (more than half of votes) guarantees a candidate can be declared the winner of a one-on-one election, having more than one Droop quota's worth of votes measures the number of votes a candidate needs to be guaranteed victory in a multiwinner election.

Besides establishing winners, the Droop quota is used to define the number of excess votes (votes not needed for a candidate that is declared elected). In proportional systems such as STV, CPO-STV, and proportional approval (or score) voting, these excess votes are transferred to other candidates, if possible, preventing them from being wasted.

The Droop quota was first devised by the English lawyer and mathematician Henry Richmond Droop (1831–1884), as an improvement to the earliest proposals for the single transferable vote (using the Hare quota). It was later independently used by Swiss physicist Eduard Hagenbach-Bischoff for efficient calculation by the D'Hondt method.

Today the Droop quota is used in almost all STV elections, including those in the Republic of Ireland, Northern Ireland, Malta, and Australia. It is also used in South Africa to allocate seats by the largest remainder method.

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- WELCOME TO A LESSON ON HAMILTON'S METHOD OF APPORTIONMENT. ALEXANDER HAMILTON PROPOSED THE METHOD THAT NOW BEARS HIS NAME. HIS METHOD WAS APPROVED BY CONGRESS IN 1791, BUT WAS VETOED BY PRESIDENT WASHINGTON. IT WAS LATER ADOPTED IN 1852, AND USED THROUGH 1911. SINCE HE WAS INTERESTED IN THE QUESTION OF CONGRESSIONAL REPRESENTATION, WE'LL USE THE LANGUAGE OF STATES AND REPRESENTATIVES. HAMILTON'S METHOD PROVIDES A PROCEDURE TO DETERMINE HOW MANY REPRESENTATIVES EACH STATE SHOULD RECEIVE. SO THE FIRST STEP IN HAMILTON'S METHOD IS TO DETERMINE HOW MANY PEOPLE EACH REPRESENTATIVE SHOULD REPRESENT. WE DO THIS BY DIVIDING THE TOTAL POPULATION OF ALL THE STATES BY THE TOTAL NUMBER OF REPRESENTATIVES. THIS ANSWER IS CALLED THE STANDARD DIVISOR, OR DIVISOR. STEP TWO, WE DIVIDE EACH STATE'S POPULATION BY THE DIVISOR TO DETERMINE HOW MANY REPRESENTATIVES IT SHOULD HAVE. WE WOULD RECORD THIS ANSWER TO SEVERAL DECIMAL PLACES, AND THIS ANSWER IS CALLED THE QUOTA. SINCE WE CAN ONLY ALLOCATE WHOLE REPRESENTATIVES, HAMILTON RESOLVES THE WHOLE NUMBER PROBLEM AS FOLLOWS: STEP THREE, WE CUT OFF THE DECIMAL PARTS OF ALL THE QUOTAS, AND THESE ARE CALLED THE LOWER QUOTAS. THEN WE ADD THE LOWER QUOTAS. THIS SUM WILL ALWAYS BE LESS THAN OR EQUAL TO THE TOTAL NUMBER OF REPRESENTATIVES. STEP FOUR, ASSUMING THAT THE TOTAL FROM STEP THREE WAS LESS THAN THE TOTAL NUMBER OF REPRESENTATIVES, ASSIGN THE REMAINING REPRESENTATIVES ONE EACH TO THE STATES WHOSE DECIMAL PART OF THE QUOTA WERE LARGEST UNTIL THE DESIRED TOTAL IS REACHED. WE DO WANT TO MAKE SURE THAT EACH STATE ENDS UP WITH AT LEAST ONE REPRESENTATIVE. LET'S TAKE A LOOK AT OUR FIRST EXAMPLE. AGAIN, THE FIRST STEP, WE WANT TO FIND THE DIVISOR, OR STANDARD DIVISOR. SO WE TAKE THE SUM OF THE POPULATION FROM ALL THE STATES, WHICH IS 189,000, AND DIVIDE BY THE TOTAL OF SEATS IN CONGRESS, WHICH IS 30. NOTICE HOW THIS GIVES US A DIVISOR OF 6,300. AND NOW, TO FIND THE QUOTAS WE TAKE EACH STATE POPULATION AND DIVIDE BY 6,300. SO FOR STATE "A" THE QUOTA WOULD BE 27,500 DIVIDED BY 6,300. IF WE ROUND TO FOUR DECIMAL PLACES, THE QUOTA FOR STATE "A" WOULD BE APPROXIMATELY 4.3651. FOR STATE B WE WOULD HAVE 38,300 DIVIDED BY 6,300, SO THE QUOTA FOR STATE B WOULD BE APPROXIMATELY 6.0794, AND SO ON. TO SAVE SOME TIME WE WON'T SHOW ALL THIS DIVISION. SO HERE ARE OUR QUOTAS TO FOUR DECIMAL PLACES, AND NOW FOR THE INITIAL APPORTIONMENT OR INITIAL ASSIGNMENT OF THE SEATS IN CONGRESS, WE REMOVE THE DECIMAL PARTS OF THE QUOTA. WHICH MEANS STATE "A" WOULD RECEIVE 4, STATE B WOULD RECEIVE 6, STATE C WOULD RECEIVE 7, AND STATE D WOULD RECEIVE 12. WE NOTICE HOW THE SUM HERE IS 29, AND WE HAVE A TOTAL OF 30 SEATS. WE NOW ASSIGN THE REMAINING SEAT TO THE STATE WHOSE QUOTA HAS THE LARGEST DECIMAL PART. NOTICE STATE "A" HAS THE LARGEST DECIMAL PART AT .3651, AND THEREFORE STATE "A" RECEIVES ONE MORE SEAT. WHICH MEANS FOR THE FINAL APPORTIONMENT STATE "A" RECEIVES 5 SEATS, STATE B RECEIVES 6, STATE C RECEIVES 7, AND STATE D RECEIVES 12. NOTICE HERE THE TOTAL IS 30. WE'VE USED ALL THE SEATS IN CONGRESS. AND HERE'S THE RESULT IN A NICE TABLE. LET'S TAKE A LOOK AT A SECOND EXAMPLE. HERE A TEACHER WISHES TO DISTRIBUTE 10 IDENTICAL PIECES OF CANDY AMONG 4 STUDENTS, BASED UPON HOW MANY PAGES OF A BOOK THEY READ LAST MONTH, USING HAMILTON'S METHOD. THE TABLE BELOW LISTS THE TOTAL NUMBER OF PAGES READ BY EACH STUDENT. SO IN THIS PARTICULAR QUESTION WE'RE ASKED TO FIND THE DIVISOR, THE QUOTA FOR ANTONIO, AND THE INITIAL APPORTIONMENT FOR ANTONIO. BUT WE'LL ACTUALLY GO AHEAD AND GO THROUGH THIS ENTIRE PROCESS FOR THIS PROBLEM. SO THE FIRST STEP IS TO FIND THE DIVISOR, SO WE'LL SET THIS UP AS A TABLE AS WE SEE HERE. NOTICE HOW WE ALREADY FOUND THE SUM OF THE TOTAL NUMBER OF PAGES, WHICH IS 1,120, AND THERE ARE 10 PIECES OF CANDY TO APPORTION. SO OUR DIVISOR IS 1,120 DIVIDED BY 10, OR 112. SO TO FIND THE QUOTA WE'LL TAKE THE NUMBER OF PAGES EACH STUDENT READ AND DIVIDE BY 112. SO THE QUOTE FOR ALAN WOULD BE 580 DIVIDED BY 112, WHICH WOULD BE APPROXIMATELY 5.1786. FOR ANTONIO THE QUOTA WOULD BE 230 DIVIDED BY 112, GIVING A QUOTA OF APPROXIMATELY 2.0536, AND SO ON. SO AGAIN, HERE ARE OUR QUOTAS. WE FOUND FROM THE PREVIOUS SLIDE THE DIVISOR IS 112, AND THE QUOTA FOR ANTONIO WE NOW KNOW IS 2.0536. AND NOW FOR THE INITIAL APPORTIONMENT WE REMOVE THE DECIMALS FROM THE QUOTA. SO ALAN WOULD RECEIVE 5 PIECES, ANTONIO WOULD RECEIVE 2, ALEX WOULD RECEIVE 1, AND LUCAS WOULD ALSO RECEIVE 1. SO THE INITIAL APPORTIONMENT FOR ANTONIO, WHICH IS WHAT OUR QUESTION ASKED FOR, IS 2. BUT NOTICE HOW THERE ARE 10 PIECES OF CANDY, AND THIS ONLY ADDS TO 9. SO GOING BACK TO THE ORIGINAL QUOTAS, SINCE ALEX HAS THE LARGEST DECIMAL PART IN HIS QUOTA, HE WOULD RECEIVE THE EXTRA PIECE OF CANDY. AND THEREFORE THE FINAL APPORTIONMENT WOULD BE 5, 2, 2, AND 1. NOTICE HOW HERE THE TOTAL IS 10 PIECES, SO WE'VE USED ALL THE CANDY. AND AGAIN, HERE IS THE RESULT IN A NICE TABLE. HAMILTON'S METHOD DOES SATISFY WHAT IS CALLED THE QUOTA RULE. THE QUOTA RULE SAYS THAT THE FINAL NUMBER OF REPRESENTATIVES A STATE GETS SHOULD BE WITHIN ONE OF THAT STATE'S QUOTA. SINCE WE'RE DEALING WITH WHOLE NUMBERS FOR OUR FINAL ANSWERS, THAT MEANS THAT EACH STATE SHOULD EITHER GO UP TO THE NEXT WHOLE NUMBER ABOVE ITS QUOTA OR GO DOWN TO THE NEXT WHOLE NUMBER BELOW ITS QUOTA. AND THEN FINALLY, THERE IS SOME CONTROVERSY WHEN USING HAMILTON'S METHOD. AFTER SEEING HAMILTON'S METHOD, MANY PEOPLE FIND THAT IT MAKES SENSE, AND IT'S NOT THAT DIFFICULT TO USE. SO WHY WOULD ANYONE WANT TO USE ANOTHER METHOD? WELL, THE PROBLEM IS THAT HAMILTON'S METHOD IS SUBJECT TO SEVERAL WHAT WE CALL PARADOXES. THREE OF THEM HAPPENED ON SEPARATE OCCASIONS WHEN HAMILTON'S METHOD WAS USED TO APPORTION THE UNITED STATES HOUSE OF REPRESENTATIVES. AND THOSE THREE PARADOXES ARE NUMBER ONE, THE ALABAMA PARADOX, TWO, THE NEW STATES PARADOX, AND THREE, THE POPULATION PARADOX. WE'LL TALK ABOUT EACH OF THESE IN FUTURE LESSONS. I HOPE YOU FOUND THIS HELPFUL.

Standard Formula

The exact form of the Droop quota for a $k$ -winner election is any amount larger than the formula:^[b]

{\frac {\text{total votes}}{k+1}}

For simplicity and regarding voting sytems that use whole votes (not fractional votes), Droop is one more than the total number of valid votes divided by one more than the number of seats. H.R. Droop put it this way: "mV n+1 + i votes will be sufficient to elect one representative."^[2]

In the case of a single-winner election, this reduces to the familiar simple majority rule. Under such a rule, a candidate can be declared elected as soon as they have strictly more than 50% of the vote, i.e one more than. ${\textstyle {\frac {\text{total votes}}{2}}}$ .

Sometimes, the Droop quota is written as a share (i.e. percentage) of the total votes.

Sometimes Droop is written as one more than $.mw-parser-output .frac{white-space:nowrap}.mw-parser-output .frac .num,.mw-parser-output .frac .den{font-size:80%;line-height:0;vertical-align:super}.mw-parser-output .frac .den{vertical-align:sub}.mw-parser-output .sr-only{border:0;clip:rect(0,0,0,0);clip-path:polygon(0px 0px,0px 0px,0px 0px);height:1px;margin:-1px;overflow:hidden;padding:0;position:absolute;width:1px}1⁄k+1$ , in which case the winner is a candidate who has a number of votes equal to or greater than the quota.

Derivation

The Droop quota can be derived by considering what would happen if $k$ candidates (called "Droop winners") have exceeded the Droop quota; the goal is to identify whether an outside candidate could defeat any of these candidates.

In this situation, each quota winner's share of the vote exceeds $1 ⁄ k +1$ , while all unelected candidates' share of the vote, taken together, is $1 ⁄ k +1$ votes or fewer. Thus, even if there were only one unelected candidate who held all the remaining votes, they would not be able to defeat any of the Droop winners.

Example in STV

The following election has 3 seats to be filled by single transferable vote. There are 4 candidates: George Washington, Alexander Hamilton, Thomas Jefferson, and Aaron Burr. There are 102 voters, but two of the votes are spoiled.

The total number of valid votes is 100, and there are 3 seats. The Droop quota is therefore ${\textstyle {\frac {100}{3+1}}+1=26}$ . These votes are as follows:

	45 voters	20 voters	25 voters	10 voters
1	Washington	Burr	Jefferson	Hamilton
2	Hamilton	Jefferson	Burr	Washington
3	Jefferson	Washington	Washington	Jefferson

First preferences for each candidate are tallied:

Washington: 45 Y
Hamilton: 10
Burr: 20
Jefferson: 25

Only Washington has strictly more than 25 votes. As a result, he is immediately elected. Washington has 19 excess votes that can be transferred to their second choice, Hamilton. The tallies therefore become:

Washington: 26 Y
Hamilton: 29 Y
Burr: 20
Jefferson: 25

Hamilton is elected, so his excess votes are redistributed. Thanks to Hamilton's support, Jefferson wins 28 votes to Burr's 20 and is elected.

It can happen that two candidate might be tied although with thousands of votes used in real STV elections this is extremely rare.. The tiebreaking rules discussed below indicate that the candidate with more first-preference votes is declared the winner.

Incorrect versions

Off-by-one errors

There is a great deal of confusion among legislators and political observers about the exact definition of the Droop quota. At least six different mistaken versions appear in various legal codes or definitions of the quota, all varying from the above definition by at most one or two votes.

(Part of the difference is produced by some applications saying a winner must take more than the quota, while others say the winner must take at least the quota. Droop himself presented the Droop quota as votes/(number of seats plus 1), plus 1.)^[3]

Some of the nonstandard formulations shown above have been justified by claiming the exact Droop quota can elect more candidates than there are seats, or that it can result in ties. However, this is incorrect, so long as candidates are only considered to be elected when their vote total is strictly greater than the Droop quota. In addition, tied votes can occur with any quota.

The first two variants, $L1$ and $L2$ , approximate the Droop quota by rounding up (to avoid decimals), and are sometimes called the rounded Droop quota.^[b] These versions are sometimes used by legislators who believe a quota of votes must be a whole number. The $L3$ quota is caused by mistakenly ignoring the floor function in $L1$ .

The origins of the third variant, $C1$ , are not clear, as this variant is not original to Droop.^[4] Variant $S2$ is sometimes smaller than the actual Droop quota, and Variant $S1$ is always no larger than the correct formula. In cases where they are smaller, it would be possible for them to result in too many candidates being elected.

{\begin{array}{rlrlrl}{\text{L1:}}&&{\Bigl \lfloor }{\frac {\text{total votes}}{{\text{seats}}+1}}+1{\Bigr \rfloor }&&{\text{L2:}}&&\left\lceil {\frac {\text{total votes}}{{\text{seats}}+1}}\right\rceil &&{\text{L3:}}&&{\frac {\text{total votes}}{{\text{seats}}+1}}+1\\{\text{C1:}}&&{\phantom {\Bigl \lfloor }}{\frac {{\text{total votes}}+1}{{\text{seats}}+1}}{\phantom {\Bigr \rfloor }}&&{\text{S1:}}&&\left\lfloor {\frac {\text{total votes}}{{\text{seats}}+1}}\right\rfloor &&{\text{S2:}}&&\left\lfloor {\frac {\text{total votes}}{{\text{seats}}+1}}+{\frac {1}{2}}\right\rfloor \end{array}}

Spoiled ballots should not be included when calculating the Droop quota; however, some jurisdictions fail to specify this in their election administration laws.

Handling ties

Whenever two candidates are tied in an STV election, some systems say that ties should be broken by ignoring ballots transferred from previous winners. In other words, candidates should be ordered first by their total number of votes, and then by the number of votes they have that have never used to elect a winner. (This should not be confused with ordering candidates by their number of first-preference votes, as votes transferred after a candidate has been eliminated should still be included in the vote total.)

Other systems say in case of a tie, the candidate with the most first preference votes should be declared the winner.

This rule has the advantage of minimizing the number of voters with no representation (i.e. whose ballots are not used to elect any candidate). It can also be justified by taking the right-hand limit of seat apportionments as the quota approaches the exact Droop quota from above, an approach that allows for calculating additional tiebreakers when needed (in favor of the least well-represented voters).^[c]

Confusion with the Hare quota

The Droop quota is sometimes confused with the more intuitive Hare quota.

Some of the confusion between the two quotas may result from a fencepost error, caused by forgetting that sometimes votes are left with unelected candidates at the end of the counting process. As well, under STV, it is possible for candidates to be elected with less than quota at the end of the count when the field of candidates is thinned to the number of remaining open seats. It is sometimes said that those sub-quota winners are more common under systems that use the Hare quota but Droop system can also produce them.^[5] In vote counts that fill seats with sub-quota winners, there are no candidates who are neither elected nor eliminated. Waste of votes by un-used votes is decreased but at expense of equality of winners' vote tallies. But systems that elect multiple winners based on quota, such as Droop, are more fair and more proportional than systems where single winners are elected in isolated, separate contests.

The Droop quota is today the most popular quota for STV elections.^{[citation needed]}

Notes

^ Some texts distinguish between a "Droop quota" and a "Hagenbach-Bischoff quota." (Here, the Droop quota is defined as the Hagenbach-Bischoff quota rounded to the next-largest integer).
^ ^a ^b Some authors use the terms "Hagenbach-Bischoff quota" or "exact Droop quota" to refer to the quantity described in this article, and reserve the term "Droop quota" for the rounded Droop quota (the original form in the works of Henry Droop).
^ This procedure gives a "leximax" ordering, ranking candidates by the number of ballots previously used to elect only (0, 1, 2...) candidates.

References

^ Humphreys (1911). Proportional Representation. p. 138.
^ Droop, H. R. (June 1881). "On Methods of Electing Representatives". Journal of the Statistical Society of London. 44 (2): 141. doi:10.2307/2339223.
^ Humphreys (1911). Proportional Representation. p. 138.
^ Dančišin, Vladimír (2013). "Misinterpretation of the Hagenbach-Bischoff quota". Annales Scientia Politica. 2 (1).
^ A Report on Alberta Elections 1905-1982.

v t e Types of majority and other electoral quotas
Single winner (majority)	Plurality (relative majority) Majority (absolute majority) Supermajority Double majority
Multi-winner (electoral quotas)	Hare quota Droop quota Imperiali quota Hagenbach-Bischoff quota Huntington–Hill method
Other	Electoral threshold

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