# Multiple At-Large Candidates

If there are multiple candidates to be elected (e.g., a school board) without voting districts, their are two natural voting procedures: Everyone casts one vote, or everyone casts as many votes as there are candidates. There are also compromises between those two alternatives: In my home town, everyone cast six votes for the nine positions on the school board (in accordance with the two party system, each party ran six candidates, so the winning party got six persons elected, and the losing party three). We shall investigate the implications of each voter voting for one candidate, each voter voting for as many candidates as will be elected, and each voter voting for a specified intermediate number of candidates for minority representation. We shall consider two factions rather than parties, since minorities may not be identified with either party. We shall use n to denote the number of voters and k to denote the number of candidates to be elected.

Recall the notation from the previous webpage:
n = the number of voters
k = the number of candidates which will be elected (which is the number of voting districts if voting districts are used)
v = the number of votes which each voter casts (for at-large voting, v=1 for voting districts).
x = the number of voters in your party (hence n-x is the number of candidates in the oppositon party, we are assuming two parties or factions).
w = the number of candidates which your party elects (hence k-w is the number of candidates which the oppositon elects, we are assuming two parties or factions).

## As many votes as candidates

The case where each voter casts as many votes as candidates (v=k) will be elected is the most trivial. If one faction has a majority, they can all vote for a set of k candidates and those candidates will be elected. Hence a minority faction cannot be assured of any representation. To be more explicit, if you need a majority of the voters (x > n/2) in order to win one representative, but can indeed win all (k) the representativs in you have a majority of the voters.

## One vote

A faction could get 2 candidates elected if it had 2002 voters, 3 if it had 3003 voters, etc. If there are only two factions, dividing the number of voters in each by 1001 [n/(k+1) rounded up] and then then rounding the resultant quotients down will usually account for all 9 [k] candidates to be elected. (The rounding down reduces k+1 to k candidates that are elected.) The other cases may entail ties which we shall not consider.

## An intermediate specified number of votes

Let v < k be the number of votes each voter casts. If v > k/2, the strategy for a minority faction is simple, everyone votes for the same v candidates (we shall not consider the problem of deciding ties, since the ties will be among minority candidates and it does not matter who wins). The problem lies with the majority faction which must determine how many candidates it can give more votes than the minority faction has. If v < k/2, the minority faction may also need to split its votes among more than v candidates.

There are two ways to approach this problem: 1) how many candiates can a party get elected, if their number of voters is known, and 2) how many voters do they need to get a specified number of candidates elected. We shall first consider the first case where the number of voters in each party is known. This can addressed by constructing a table.

• x vs. min(n-x, (v × (n-x))/k
• min(x, (v × x)/2) vs. min(n-x, (v × (n-x))/(k-1)
• min(x, (v × x)/3) vs. min(n-x, (v × (n-x))/(k-2)
• . . .
• min(x, (v × x)/(k-2)) vs. min(n-x, (v × (n-x))/3
• min(x, (v × x)/(k-1)) vs. min(n-x, (v × (n-x))/2
• min(x, (v × x)/k) vs. n-x
The minimum function (min) indicates that a faction may split their votes among more than z > v candidates in a systematic fashion so that each candidate gets v × x / z votes, but can never give more than x votes to one candidate. (v × x is the total number of votes a faction casts.

An example of spreading votes among more than 6 candidates which is used below follows. It is for spreading 6 × 5500 votes among 7 candidates. In this case there cannot be just two blocs of majority voters, but overlapping candidate sets. For example, if Alfred, Bob, Celeste, Daphne, Eunice, Frank and George are to be elected; 642 voters vote for ABCDEF, 643 voters vote for BCDEFG, 643 voters vote for ACDEFG, 643 voters vote for ABDEFG, 643 voters vote for ABCEFG, 643 voters vote for ABCDFG, and 643 voters vote for ABCDEG.

As an example, let n=10000, x = 2500, v = 6, k = 9
looking at the first juxtaposition (vs.) above, you can only give 2500 votes to a candidate, but the opposition can spread its 6 × 7500 = 45000 votes among 9 candidates giving 5000 votes to each. Hence the opposition cangive 9 candidates 5000 votes each, but you can only give 2500 votes to a candidate, and the opposition can elect 9 candidates.

Next consider n=10000, x = 4500, v = 6, k = 9
The second juxtaposition above gives 4500 > 4125 = 6 × 5500 / 8, which means you can get 2 candidates elected if they spread their votes among 8 candidates.
however, the third juxtaposition above gives 4500 < 4714.28 = 6 × 5500 / 7, which means that the opposition could give more votes to 7 candidates than you could give to three, so they can get 7 candidates elected.

Note that these comparisons are base on a total of k+1 candidates. The faction that can give more votes to their block of candidates can elect at least that many candidates. You will always reach a point where the inequality sign reverses as you go down the table (unless one party can elect all k candidates), and the numbers of candidates each party can elect will sum to k (for one party the number will be before the inequality reversed, and for the other it will be after).

The majority party can always elect at least v candidates, since no voter can cast more than one vote for a candidate. Further, if the majority party has M voters, it can always elect at least min(k, v × M/ (n-M)) because it can spread its votes among that many candidates and give them more votes than (n-M). However, if v < k/2 both the minority and majority parties may need to spread their votes among more than v candidates, and further analysis is required.

In the above examples v > k/2 (6 > 9/2), and the above bound provideds the results. With M = 7500, 6 × 7500/2500 = 18, min(9, 18) = 9. With M = 5500, 6 × 5500/4500 = 7.33, and the majority can elect 7 candidates.

The thrust of these calculations is that in order for a minority faction to gain representation, they must be a large minority (near 50%) and/or have few votes per voter.

The second question is how many voters a party needs to get a specified number of candidates elected. If v=1, the answer is 1+n/(k+1), rounding up is often necessary. As indicated above, twice as many voters are are needed to elect two candidates, thrice as many to elect three candidates ... .

if v=k, then the majority takes all, one needs a majority to get one candidate elected, in which case on can get all k candidates elected.

The case 1 < v < k is less trivial. [I believe the next three paragraphs are a nice solution to this question and provide a complementary perspective on the table above, but I shall only expect you to know the cases w=1, w=k, and w=majority. The cases w=1 and w=k have been treated above. The case w=majority can be solved by invoking symmetry: one faction must win a majority, and it will obviously be whichever faction has a majority. This is will not work if ties can ocur (e.g., k is even), but will suffice for any problems I give you on a test.] In order for a party to get w candidates elected, its number of voters x must satisfy min(x, xv/w) > min((n-x), (n-x)v/(k-w+1). The quotients in the minimum functions entail spreading their votes equally among the winning candidates, the minimum function (i.e., x and (n-x)) entails that voters can only cast one vote per candidate. cf. the table above.

For example, with n=10,000 voters and k=9 candidates will be elected and v=5 votes per voter, how many voters are needed to elect w=3 candidates? We note first that v/w = 5/3 > 1, so min(x, xv/w) = x. Similarly, v/(k-w+1) = 5/(9-3+1) < 1, so min ((n-x), (n-x)v/(k-w+1)) = (n-x)(5/7). Hence one needs to solve x > (n-x)(5/7) which provides x = 4167.

With n=10,000 voters, k=9 candidates elected, and v=5 votes per voter, how many voters are needed to elect w=6 candidates? v/w = 5/6 < 1, so min(x, xv/w) = xv/w = (5/6)x. Similarly, v/(k-w+1) = 5/4 > 1, so min((n-x), v/(k-w+1)) = (n-x). Hence one needs to solve (5/6)x > (n-x) which provides x = 5455.

Exercise: If there are 25,000 voters and 11 candidates will be elected, how large must a minority be to be assured of getting at least one representative elected if each voter has 1 vote? 6 votes? 11 votes?
to be assured of getting at least 6 representatives elected if each voter has 1 vote? 6 votes? 11 votes?
to be assured of getting at least 11 representatives elected if each voter has 1 vote? 6 votes? 11 votes?

If there are 15,500 majority faction voters and 9500 minority faction voters and 11 candidates are elected, how many candidates can the minority faction get elected if each voter casts 1 vote? 6 votes? 11 votes?