Electoral System Analysis | 选举制度分析

📚 Electoral System Analysis | 选举制度分析

Electoral system analysis applies mathematical tools to understand how votes are converted into seats, how power is distributed, and whether an election outcome is fair. This topic draws on arithmetic, algebra, proportional reasoning and discrete optimisation, making it a valuable extension for A-Level Edexcel Mathematics and Further Mathematics students.

选举制度分析运用数学工具来理解选票如何转化为席位、权力如何分配，以及选举结果是否公平。这一主题涉及算术、代数、比例推理和离散优化，因此对 A-Level Edexcel 数学及进阶数学的学生来说是很有价值的拓展内容。

In the UK and many other countries, different voting systems can produce very different results from the same set of votes. By modelling seat allocation and voting power, we can compare systems such as first-past-the-post, proportional representation and apportionment methods used in legislatures.

在英国和许多其他国家，不同的投票制度可能从同一组选票中产生截然不同的结果。通过对席位分配和投票权力进行建模，我们可以比较简单多数制、比例代表制以及立法机构中使用的席位分配方法。

1. What Is Electoral System Analysis? | 什么是选举制度分析？

Electoral system analysis is the study of how votes are aggregated to elect representatives or decide outcomes. It uses mathematical models to examine fairness, efficiency, proportionality and strategic behaviour in voting.

选举制度分析是研究如何将选票汇总以选出代表或决定结果的方法。它使用数学模型来考察投票中的公平性、效率、比例性以及策略行为。

Typical questions include: How many seats should each party receive? How can we measure wasted votes? When can a voting rule produce a paradox? Answering these questions requires careful definitions and numerical methods.

典型问题包括：每个政党应获得多少席位？如何衡量废票？投票规则何时会产生悖论？回答这些问题需要精确定义和数值方法。

For A-Level students, this topic connects directly to percentages, ratios, rounding, inequalities, weighted averages and iterative algorithms. It also provides a real-world context for critical thinking and mathematical modelling.

对于 A-Level 学生来说，这一主题与百分数、比率、四舍五入、不等式、加权平均数和迭代算法直接相关。它还为批判性思维和数学建模提供了现实背景。

2. Votes, Seats and the Standard Quota | 选票、席位与标准配额

Suppose there are n parties or states, and party i receives Vᵢ votes. The total number of votes is V = Σ Vᵢ, and the total number of seats to be allocated is S. The standard quota for party i is defined as:

假设有 n 个政党或州，政党 i 获得 Vᵢ 张选票。总选票数为 V = Σ Vᵢ，待分配的总席位数为 S。政党 i 的标准配额定义为：

qᵢ = (Vᵢ / V) × S

For example, if a party wins 4,200 out of 10,000 votes and there are 10 seats, its standard quota is (4,200 / 10,000) × 10 = 4.2 seats. Since seats must be integers, we need a method to round or adjust quotas.

例如，某政党在 10,000 张选票中赢得 4,200 张，且总席位为 10，则其标准配额为 (4,200 / 10,000) × 10 = 4.2 个席位。由于席位必须是整数，我们需要一种方法对配额进行取整或调整。

The lower quota of a party is the greatest integer less than or equal to its standard quota, written as ⌊qᵢ⌋. The upper quota is the least integer greater than or equal to qᵢ, written as ⌈qᵢ⌉. Any fair apportionment method should usually give each party either its lower or upper quota.

政党的下配额是小于或等于其标准配额的最大整数，记作 ⌊qᵢ⌋。上配额是大于或等于 qᵢ 的最小整数，记作 ⌈qᵢ⌉。任何公平的分配方法通常应给予每个政党其下配额或上配额之一。

3. Hamilton’s Method | 汉密尔顿法

Hamilton’s method, also called the largest remainder method, begins by awarding each party its lower quota. If the sum of lower quotas is less than S, the remaining seats are allocated one by one to the parties with the largest fractional parts of their standard quotas.

汉密尔顿法，又称最大余数法，首先给每个政党分配下配额。如果下配额之和小于 S，则剩余席位逐一分配给标准配额小数部分最大的政党。

Suppose three parties have standard quotas 4.2, 3.3 and 2.5 with S = 10. The lower quotas are 4, 3 and 2, which sum to 9. One seat remains, and the largest fractional part is 0.5 from the third party, so the final allocation is 4, 3 and 3 seats.

假设三个政党的标准配额分别为 4.2、3.3 和 2.5，且 S = 10。下配额为 4、3 和 2，总和为 9。剩余一个席位，最大的小数部分是第三党的 0.5，因此最终分配为 4、3 和 3 个席位。

Hamilton’s method is easy to apply and satisfies the quota rule, but it can produce counter-intuitive results when the total number of seats changes, known as the Alabama paradox.

汉密尔顿法易于应用且满足配额规则，但当总席位数变化时可能产生违反直觉的结果，即所谓的阿拉巴马悖论。

4. Divisor Methods: Jefferson, Webster and Huntington-Hill | 除数法：杰斐逊、韦伯斯特与亨廷顿-希尔

Divisor methods allocate seats by dividing each party’s votes by a common divisor d, then rounding the resulting modified quotas to integers so that the sum equals S. The choice of rounding rule defines the method.

除数法通过用一个公共除数 d 去除各政党的选票，然后将所得的修正配额取整为整数，使总和等于 S 来分配席位。取整规则的选择定义了不同的方法。

Jefferson’s method rounds every modified quota down to the nearest integer. This tends to favour larger parties because it allows smaller parties to lose fractional parts more easily. Webster’s method rounds to the nearest integer using standard rounding, which is generally more balanced.

杰斐逊法将每个修正配额向下取整到最近的整数。这往往有利于大党，因为它使小党更容易失去小数部分。韦伯斯特法使用标准四舍五入取整到最近的整数，通常更为平衡。

Huntington-Hill method, used to allocate seats in the US House of Representatives, uses a geometric-mean rounding rule. A modified quota between integers n and n + 1 is rounded up if it exceeds √(n(n + 1)), otherwise rounded down.

亨廷顿-希尔法用于美国众议院席位分配，采用几何平均取整规则。介于整数 n 和 n + 1 之间的修正配额如果超过 √(n(n + 1)) 则向上取整，否则向下取整。

In practice, we adjust the divisor d until the rounded modified quotas sum exactly to S. This is an iterative numerical method and can be carried out with a spreadsheet or calculator.

在实际应用中，我们调整除数 d，直到取整后的修正配额总和恰好等于 S。这是一种迭代数值方法，可以用电子表格或计算器实现。

5. Fairness Criteria and Paradoxes | 公平性标准与悖论

A fair apportionment method is expected to satisfy several mathematical criteria. The quota rule states that each party must receive either its lower quota or its upper quota. Hamilton’s method satisfies this rule, but many divisor methods can violate it.

一种公平的分配方法应满足若干数学标准。配额规则规定每个政党必须获得其下配额或上配额之一。汉密尔顿法满足该规则，但许多除数法可能违反它。

The Alabama paradox occurs when increasing the total number of seats causes a party to lose a seat. For example, with Hamilton’s method, changing S from 299 to 300 in historical US data caused Alabama to lose a seat despite no change in population shares.

阿拉巴马悖谬指增加总席位数反而导致某个政党失去一个席位。例如，在历史美国数据中，使用汉密尔顿法将 S 从 299 增加到 300 时，阿拉巴马州在人口比例未变的情况下失去了一个席位。

The population paradox occurs when a party with a higher population growth rate loses a seat to a party with a lower growth rate. These paradoxes highlight why no simple rounding method can satisfy all intuitive fairness properties at once.

人口悖论指人口增长率较高的政党反而将席位输给增长率较低的政党。这些悖论说明，没有一种简单的取整方法能同时满足所有直观的公平性性质。

6. Measuring Disproportionality | 衡量不比例性

Disproportionality measures the gap between vote share and seat share. Let vᵢ be the proportion of votes won by party i and sᵢ be the proportion of seats won. The Loosemore-Hanby index is defined as:

不比例性衡量选票份额与席位份额之间的差距。设 vᵢ 为政党 i 赢得的选票比例，sᵢ 为赢得的席位比例。Loosemore-Hanby 指数定义为：

D = ½ × Σ |vᵢ − sᵢ|

The index ranges from 0 to 1, where 0 means perfect proportionality and 1 means total disproportionality. For example, if a party wins 42% of votes but only 30% of seats, its contribution is |0.42 − 0.30| = 0.12.

该指数范围为 0 到 1，其中 0 表示完全比例，1 表示完全不比例。例如，某政党赢得 42% 的选票但只获得 30% 的席位，其贡献为 |0.42 − 0.30| = 0.12。

Another common measure is the Gallagher index, given by the square root of half the sum of squared differences. It places less weight on small deviations than the Loosemore-Hanby index.

另一种常用指标是 Gallagher 指数，其值为差额平方和的一半的平方根。与 Loosemore-Hanby 指数相比，它对较小偏差赋予的权重更低。

7. The Efficiency Gap and Gerrymandering | 效率差距与选区划分不公

In single-member district systems, gerrymandering occurs when district boundaries are drawn to give one political party an unfair advantage. The efficiency gap is a quantitative measure of this advantage.

在单席位选区制度中，当选区边界被划定以使某一政党获得不公平优势时，就产生了选区划分不公（gerrymandering）。效率差距是衡量这种优势的量化指标。

In a two-party contest, wasted votes are ballots that do not contribute to a candidate’s victory. For a winning candidate, wasted votes are those above the simple majority threshold. For a losing candidate, all votes are wasted.

在两党竞争中，废票是指对候选人获胜没有贡献的选票。对于获胜候选人，废票是超过简单多数门槛的选票。对于失败候选人，所有选票都是废票。

The efficiency gap is calculated as:

EG = (Wasted votes for party A − Wasted votes for party B) / Total votes

A large positive or negative efficiency gap indicates that one party wastes far fewer votes than the other, often due to the shape of district boundaries. This is used in legal and political analysis to detect possible gerrymandering.

效率差距的正值或负值较大，表明一个政党浪费的选票远少于另一个政党，这通常是由于选区边界的形状造成的。这被用于法律和政治分析中，以检测可能存在的选区划分不公。

8. Voting Power Indices | 投票权力指数

In weighted voting systems, such as international organisations or corporate boards, different voters may have different numbers of votes. A voter’s power is not always proportional to their weight, so power indices are used to measure actual influence.

在加权投票系统中，如国际组织或公司董事会，不同投票者可能拥有不同数量的票。投票者的权力并不总是与其权重成正比，因此使用权力指数来衡量实际影响力。

The Banzhaf index counts the number of winning coalitions in which a voter is critical, meaning that their removal would make the coalition losing. A voter’s Banzhaf power is the proportion of all critical defections attributable to that voter.

班扎夫指数计算一个投票者在获胜联盟中起关键作用的次数，所谓关键作用是指若该投票者退出，联盟就会失败。一个投票者的班扎夫权力是其关键退出占所有关键退出总数的比例。

The Shapley-Shubik index considers all possible orders in which voters join a coalition. A voter is pivotal if their addition turns a losing coalition into a winning coalition. The power index is the proportion of orderings in which the voter is pivotal.

沙普利-舒比克指数考虑投票者加入联盟的所有可能顺序。如果某投票者的加入使失败联盟变为获胜联盟，则该投票者是关键者。权力指数是该投票者成为关键者的排序数占总排序数的比例。

These indices often differ from simple vote weights. For instance, in a three-member body with weights 4, 3 and 2 and a quota of 5, the smallest member has no power because no winning coalition needs their vote.

这些指数通常与简单票权不同。例如，在一个权重为 4、3、2，法定票数为 5 的三成员机构中，最小成员没有任何权力，因为没有任何获胜联盟需要其选票。

9. Arrow’s Impossibility Theorem | 阿罗不可能定理

Arrow’s impossibility theorem states that no ranked voting system can simultaneously satisfy all of a set of basic fairness conditions when there are at least three candidates. This is a central result in social choice theory.

阿罗不可能定理指出，当候选人数至少为三个时，没有任何排序投票制度能同时满足一组基本公平条件。这是社会选择理论的核心结论。

The conditions include unrestricted domain, non-dictatorship, Pareto efficiency and independence of irrelevant alternatives. Arrow proved that these four conditions are mutually inconsistent.

这些条件包括无限制域、非独裁、帕累托效率和无关备选方案独立性。阿罗证明了这四个条件相互矛盾。

For students, this shows that there is no perfect voting system. Each method involves trade-offs, and mathematical analysis helps make those trade-offs explicit.

对于学生来说，这表明不存在完美的投票制度。每种方法都涉及权衡，而数学分析有助于将这些权衡明确表达出来。

10. Worked Example: Apportioning 10 Seats | 实例解析：分配 10 个席位

A small council election has three parties: A, B and C. The votes are A = 4,200, B = 3,300 and C = 2,500, giving a total of 10,000 votes. Ten seats are available. Calculate the seat allocation using Hamilton’s method.

某小型议会选举有三个政党：A、B 和 C。得票分别为 A = 4,200，B = 3,300，C = 2,500，总票数为 10,000。有 10 个席位待分配。使用汉密尔顿法计算席位分配。

First compute the standard quotas: qₐ = (4,200 / 10,000) × 10 = 4.2, q_b = 3.3 and q_c = 2.5. The lower quotas are 4, 3 and 2, which sum to 9. One seat remains.

首先计算标准配额：qₐ = (4,200 / 10,000) × 10 = 4.2，q_b = 3.3，q_c = 2.5。下配额为 4、3 和 2，总和为 9。还剩余一个席位。

Compare fractional parts: Party A has 0.2, Party B has 0.3 and Party C has 0.5. The largest fractional part is 0.5, so Party C receives the extra seat.

比较小数部分：A 党为 0.2，B 党为 0.3，C 党为 0.5。最大的小数部分是 0.5，因此 C 党获得额外席位。

Final allocation: A = 4, B = 3, C = 3

Now apply Webster’s method with a suitable divisor. If we choose d = 950, the modified quotas are A = 4,200 / 950 ≈ 4.42, B = 3,300 / 950 ≈ 3.47 and C = 2,500 / 950 ≈ 2.63. Rounding to the nearest integer gives 4 + 3 + 3 = 10, so the same allocation results in this case.

现在使用韦伯斯特法并选择合适的除数。若选择 d = 950，修正配额为 A = 4,200 / 950 ≈ 4.42，B = 3,300 / 950 ≈ 3.47，C = 2,500 / 950 ≈ 2.63。四舍五入后得到 4 + 3 + 3 = 10，因此该情况下结果相同。

11. Exam-Style Practice | 考试风格练习

Question: Four regions have populations 5,200, 3,800, 2,600 and 1,400 (in thousands). A council has 12 seats to allocate by Hamilton’s method. Find the standard quota for each region and the final allocation.

问题：四个地区的人口分别为 5,200、3,800、2,600 和 1,400（单位：千人）。某委员会有 12 个席位需要用汉密尔顿法分配。求每个地区的标准配额和最终分配。

Total population = 5,200 + 3,800 + 2,600 + 1,400 = 13,000. Standard quotas are (5,200 / 13,000) × 12 ≈ 4.80, (3,800 / 13,000) × 12 ≈ 3.51, (2,600 / 13,000) × 12 = 2.40, and (1,400 / 13,000) × 12 ≈ 1.29.

总人口 = 5,200 + 3,800 + 2,600 + 1,400 = 13,000。标准配额分别为 (5,200 / 13,000) × 12 ≈ 4.80，(3,800 / 13,000) × 12 ≈ 3.51，(2,600 / 13,000) × 12 = 2.40，(1,400 / 13,000) × 12 ≈ 1.29。

Lower quotas are 4, 3, 2 and 1, which sum to 10. Two seats remain. The fractional parts are 0.80, 0.51, 0.40 and 0.29. The two largest are 0.80 and 0.51, so the first and second regions each receive an extra seat.

下配额为 4、3、2 和 1，总和为 10。还剩余两个席位。小数部分为 0.80、0.51、0.40 和 0.29。最大的两个是 0.80 和 0.51，因此第一和第二个地区各获得一个额外席位。

Final allocation: 5, 4, 2, 1

Check that the total is 12 and that each region receives a whole number of seats. This allocation satisfies the quota rule because each region’s seat count is equal to either its lower or upper quota.

检查总和为 12 且每个地区获得整数席位。该分配满足配额规则，因为每个地区的席位数等于其下配额或上配额之一。

12. Summary and Key Takeaways | 总结与要点

Electoral system analysis uses standard quotas, divisor methods and rounding rules to allocate seats proportionally. Hamilton’s method is simple but can produce paradoxes, while divisor methods such as Jefferson, Webster and Huntington-Hill offer different trade-offs.

选举制度分析使用标准配额、除数法和取整规则来按比例分配席位。汉密尔顿法简单但可能产生悖论，而杰斐逊、韦伯斯特和亨廷顿-希尔等除数法提供了不同的权衡。

Fairness is measured by criteria such as the quota rule, as well as by disproportionality indices like the Loosemore-Hanby index. The efficiency gap quantifies wasted-vote asymmetry in district-based elections.

公平性通过配额规则等标准以及 Loosemore-Hanby 指数等不比例性指标来衡量。效率差距量化了选区制选举中废票的不对称性。

Power indices like Banzhaf and Shapley-Shubik reveal that voting weight is not the same as voting power. Arrow’s theorem shows that no perfect ranked voting system exists for three or more candidates.

班扎夫和沙普利-舒比克等权力指数揭示投票权重并不等同于投票权力。阿罗定理表明，对于三个或更多候选人，不存在完美的排序投票制度。

For A-Level Edexcel Mathematics, the most important skills are computing standard quotas, carrying out apportionment algorithms and interpreting fairness measures in context. Always check that the final seat allocation sums to the required total.

对于 A-Level Edexcel 数学来说，最重要的技能是计算标准配额、执行分配算法以及在具体情境中解释公平性指标。务必检查最终席位分配之和等于所需总数。

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Electoral System Analysis | 选举制度分析