First Past the Post | 简单多数制

📚 First Past the Post | 简单多数制

The ‘First Past the Post’ (FPTP) electoral system is not only a political concept but also a rich mathematical problem. In Edexcel A-Level Mathematics, particularly in statistics and decision mathematics, the analysis of FPTP offers practical applications of probability, combinatorics, and game theory.

“简单多数制”（FPTP）选举制度不仅是一个政治概念，更是一个丰富的数学问题。在Edexcel A-Level数学中，尤其是在统计学和决策数学领域，对FPTP的分析提供了概率论、组合学和博弈论的实践应用。

1. What is First Past the Post? | 什么是简单多数制

In an FPTP system, each voter casts a single vote for a single candidate. The candidate who receives the highest number of votes wins, even if that number is less than 50% of the total. Mathematically, we define a set of candidates C = {c₁, c₂, … , cₙ} and a set of voters V = {v₁, v₂, … , vₘ}. Each voter vⱼ assigns their vote to exactly one candidate.

在简单多数制中，每位选民只投一票给一名候选人。获得最高票数的候选人获胜，即使该票数不足总票数的50%。在数学上，我们定义候选人集合 C = {c₁, c₂, … , cₙ} 和选民集合 V = {v₁, v₂, … , vₘ}。每位选民 vⱼ 将其选票投给且仅投给一名候选人。

The winning condition is simple: let vᵢ denote the number of votes received by candidate cᵢ. Candidate c_w wins if and only if v_w > vⱼ for all j ≠ w. This is known as the plurality rule, distinct from a majority rule which requires more than 50% of all votes.

获胜条件很简单：设 vᵢ 表示候选人 cᵢ 获得的票数。候选人 c_w 获胜当且仅当对所有的 j ≠ w 都有 v_w > vⱼ。这被称为相对多数原则，与要求超过总票数50%的绝对多数原则不同。

2. The Mathematics of Vote Counting | 计票的数学原理

Consider an election with m voters and n candidates. The total number of votes cast satisfies the conservation equation:

考虑一场有 m 名选民和 n 名候选人的选举。总投票数满足守恒方程：

Σᵢ₌₁ⁿ vᵢ = m

Each vote share is calculated as sᵢ = vᵢ / m × 100%. A candidate’s victory margin is defined as Δ = v_w − v_j, where v_j is the second-highest vote count. When Δ is small, the result is described as a ‘marginal seat’.

每名候选人的得票份额计算为 sᵢ = vᵢ / m × 100%。候选人的胜出差距定义为 Δ = v_w − v_j，其中 v_j 是第二高的票数。当 Δ 很小时，该选区被称为”边缘席位”。

For a two-candidate race, the winner must obtain more than half the votes since v₁ + v₂ = m implies that the larger of the two exceeds m/2. This is a direct application of the pigeonhole principle.

对于两人竞选，获胜者必须获得超过一半的选票，因为 v₁ + v₂ = m 意味着两者中较大者必定超过 m/2。这是鸽巢原理的直接应用。

3. Building a Mathematical Model | 建立数学模型

We can model FPTP as an optimisation problem. Suppose each candidate cᵢ has a ‘true support level’ pᵢ, representing the probability that a randomly selected voter will support them. Since each voter votes independently, the vote count for candidate cᵢ follows a Binomial distribution:

我们可以将FPTP建模为一个最优化问题。假设每位候选人 cᵢ 有一个”真实支持水平” pᵢ，代表随机选择一名选民支持他们的概率。由于每位选民独立投票，候选人 cᵢ 的得票数服从二项分布：

Vᵢ ~ Bin(m, pᵢ)

The expected vote count is E[Vᵢ] = m × pᵢ, with variance Var(Vᵢ) = m × pᵢ × (1 − pᵢ). For large m, by the Central Limit Theorem, Vᵢ is approximately normally distributed: Vᵢ ≈ N(mpᵢ, mpᵢ(1 − pᵢ)).

期望得票数为 E[Vᵢ] = m × pᵢ，方差为 Var(Vᵢ) = m × pᵢ × (1 − pᵢ)。当 m 较大时，根据中心极限定理，Vᵢ 近似服从正态分布：Vᵢ ≈ N(mpᵢ, mpᵢ(1 − pᵢ))。

For candidate cᵢ to win, we require Vᵢ > Vⱼ for all j ≠ i. The probability of this event can be estimated using the joint normal distribution, which in practice is computed using numerical integration techniques taught in higher-level statistics.

候选人 cᵢ 要获胜，需要 Vᵢ > Vⱼ 对所有 j ≠ i 成立。该事件的概率可以通过联合正态分布来估计，在实际中需要使用高级统计学中教授的数值积分方法进行计算。

4. The Spoiler Effect and Vote Splitting | 搅局者效应与选票分裂

One of the most important mathematical phenomena in FPTP is the spoiler effect. Consider three candidates A, B, and C, where A and B are ideologically similar. Voters who prefer A or B split their votes between the two, allowing C to win with fewer than 50% of the votes.

FPTP中最重要的数学现象之一是搅局者效应。考虑三名候选人A、B和C，其中A和B在意识形态上相似。倾向A或B的选民将选票分散给两人，使得C以不到50%的得票率获胜。

Formally, suppose the true preference distribution among voters is: 35% prefer A, 30% prefer B, and 35% prefer C. Under sincere voting, A receives 35%, B receives 30%, and C receives 35% — C wins despite only one group truly supporting them.

形式上，假设选民的偏好分布为：35%倾向A，30%倾向B，35%倾向C。在真实投票下，A获得35%，B获得30%，C获得35%——C虽然只有一组选民真正支持，却仍然获胜。

This situation is closely related to Arrow’s Impossibility Theorem, which states that no ranked voting system can simultaneously satisfy all desirable fairness criteria. In FPTP, the ‘independence of irrelevant alternatives’ condition is violated: removing candidate B from the ballot would change the winner from C to A.

这种情况与阿罗不可能定理密切相关，该定理指出没有任何排名投票制度能同时满足所有理想的公平性标准。在FPTP中，”无关替代方案的独立性”条件被违反：从选票中移除候选人B会将获胜者从C变为A。

5. Probability and Election Outcomes | 概率与选举结果

Probability theory provides powerful tools for analysing FPTP elections. The probability that a candidate wins under uncertainty can be computed using the properties of order statistics. If V₁, V₂, … , Vₙ are the vote counts, the winner is the maximum of these values.

概率论为分析FPTP选举提供了强大的工具。在不确定性下候选人获胜的概率可以利用次序统计量的性质来计算。如果 V₁, V₂, … , Vₙ 是票数，则获胜者是这些值中的最大值。

For two candidates with m voters, where each voter chooses candidate A with probability p, the probability that A wins is:

对于有 m 名选民的两人竞选，每位选民以概率 p 选择候选人A，则A获胜的概率为：

P(A wins) = P(V_A > m/2) = Σₖ₌⌊m/2⌋₊₁ᵐ C(m, k) pᵏ(1 − p)ᵐ⁻ᵏ

This is the cumulative binomial probability. In A-Level exams, for large m and moderate p, this is approximated using the normal distribution with continuity correction. For instance, if m = 1000 and p = 0.52, we compute z = (500.5 − 520) / √(1000 × 0.52 × 0.48) ≈ −1.23, giving P(A wins) ≈ 0.89.

这是累积二项概率。在A-Level考试中，当 m 较大且 p 适中时，使用带连续性校正的正态分布来近似。例如，若 m = 1000 且 p = 0.52，我们计算 z = (500.5 − 520) / √(1000 × 0.52 × 0.48) ≈ −1.23，得到 P(A获胜) ≈ 0.89。

6. Measuring Disproportionality | 度量不成比例性

A key mathematical criticism of FPTP is the disconnect between vote shares and seat shares. The Gallagher Index, also known as the least squares index, quantifies this disproportionality:

对FPTP的一个关键数学批评是得票份额与议席份额之间的脱节。加拉格尔指数，又称最小二乘指数，量化了这种不成比例性：

G = √(½ Σᵢ(vᵢ − sᵢ)²)

where vᵢ is the percentage of votes received by party i and sᵢ is the percentage of seats won by party i. A value of G = 0 indicates perfect proportionality; values above 15 are considered highly disproportional.

其中 vᵢ 是政党i获得的选票百分比，sᵢ 是政党i赢得的议席百分比。G = 0 表示完全成比例；超过15的值被认为是高度不成比例。

Consider a worked example. In a 100-seat parliament, Party X wins 40% of votes and 55 seats. Party Y wins 45% of votes and 40 seats. Party Z wins 15% of votes and 5 seats. Then G = √(½[(40−55)² + (45−40)² + (15−5)²]) = √(½[225 + 25 + 100]) = √175 ≈ 13.2, indicating substantial disproportionality.

考虑一个计算实例。在一个100席的议会中，X党赢得40%选票和55个席位。Y党赢得45%选票和40席。Z党赢得15%选票和5席。则 G = √(½[(40−55)² + (45−40)² + (15−5)²]) = √(½[225 + 25 + 100]) = √175 ≈ 13.2，表明存在显著的不成比例性。

7. Game Theory and Strategic Voting | 博弈论与策略性投票

FPTP creates incentives for strategic (or tactical) voting. A rational voter whose preferred candidate has little chance of winning may instead vote for a ‘lesser evil’ — their most preferred candidate among those who are actually competitive. This is a direct application of decision theory.

FPTP产生了策略性（或战术性）投票的激励。理性选民如果发现自己偏好的候选人胜选希望渺茫，可能会转而投给”两害相权取其轻”——即在真正有竞争力的候选人中他们最偏好的一位。这是决策理论的直接应用。

We can model this as a game with incomplete information. Each voter has a utility uᵢ for each candidate. Under expected utility theory, a voter chooses candidate cⱼ that maximises:

我们可以将此建模为一个不完全信息博弈。每位选民对每名候选人有一个效用 uᵢ。根据期望效用理论，选民选择最大化以下值的候选人 cⱼ：

E[Utility] = Σₖ P(cⱼ wins | voter supports cⱼ) × u(cⱼ)

The probability P(cⱼ wins) depends on the voter’s own action, creating a feedback loop. In equilibrium, voters in safe seats may vote sincerely while those in marginal seats vote strategically. This phenomenon has been empirically confirmed in UK general elections.

概率 P(cⱼ获胜) 取决于选民自身的行动，形成了一个反馈循环。在均衡状态下，安全选区的选民可能会进行真实投票，而边缘选区的选民则进行策略性投票。这一现象已在英国大选中得到实证确认。

8. Statistical Hypothesis Testing in Elections | 选举中的统计假设检验

Statistical methods from the A-Level curriculum can be applied to election data. For example, a χ² goodness-of-fit test can determine whether survey polls before an election are consistent with the actual FPTP results.

A-Level课程中的统计方法可以应用于选举数据。例如，χ² 拟合优度检验可以判断选举前的民调是否与实际FPTP结果一致。

Suppose a poll predicts vote shares of 35% for Party A, 30% for Party B, and 35% for Party C from a sample of 500 voters. The actual election across 10,000 voters yields 3,800 for A, 2,900 for B, and 3,300 for C. The expected counts under the poll are 3,500, 3,000, and 3,500 respectively. The test statistic is:

假设民调从500名选民样本中预测A党得票率35%，B党30%，C党35%。实际选举中10,000名选民的结果是：A获3,800票，B获2,900票，C获3,300票。民调下的期望人数分别为3,500、3,000和3,500。检验统计量为：

χ² = (3800−3500)²/3500 + (2900−3000)²/3000 + (3300−3500)²/3500 = 25.71 + 3.33 + 11.43 = 40.47

With 2 degrees of freedom and α = 0.05, the critical value is 5.991. Since 40.47 > 5.991, we reject the null hypothesis, concluding that the actual result differs significantly from the poll prediction.

自由度为2，显著性水平 α = 0.05 时，临界值为5.991。由于40.47 > 5.991，我们拒绝原假设，得出实际结果与民调预测存在显著差异的结论。

9. Comparing Voting Systems | 投票制度的比较

The mathematics of apportionment provides a rigorous framework for comparing FPTP with alternatives. The d’Hondt method, used in proportional systems, allocates seats by iteratively dividing each party’s vote count by successive divisors 1, 2, 3, and so on.

席位分配的数学为比较FPTP与替代方案提供了严谨的框架。比例制中使用的顿特法，通过每次将各政党得票数依次除以除数1、2、3……来分配席位。

For example, with 5 seats and vote counts A = 100, B = 80, C = 30, the d’Hondt quotients are A: 100, 50, 33.3; B: 80, 40, 26.7; C: 30, 15, 10. The five highest quotients are 100 (A), 80 (B), 50 (A), 40 (B), 33.3 (A), giving A 3 seats, B 2 seats, and C 0 seats.

例如，有5个席位，得票数 A = 100, B = 80, C = 30，顿特商数分别为 A: 100, 50, 33.3；B: 80, 40, 26.7；C: 30, 15, 10。最高的五个商数依次为100(A)、80(B)、50(A)、40(B)、33.3(A)，因此A获得3席，B获得2席，C获得0席。

Under FPTP with single-member constituencies, if these were three candidates in one constituency, A would win with 47.6% of the vote. The mathematical trade-off is clear: FPTP produces decisive outcomes but can distort representation; proportional methods align seats with votes but may produce fragmented parliaments.

在单议席选区FPTP制度下，如果这三名候选人在同一选区竞争，A将以47.6%的得票率获胜。数学上的权衡是明确的：FPTP产生决定性的结果但可能扭曲代表性；比例制使议席与选票对齐，但可能产生碎片化的议会。

10. Exam-Style Applications | 考试型应用

A-Level Mathematics exams may ask students to analyse FPTP through probability trees, binomial distributions, or chi-squared tests. Here is a typical problem: In a two-candidate election, 40% of voters are known to support candidate A and 60% support candidate B. A random sample of 10 voters is selected.

A-Level数学考试可能要求学生通过概率树、二项分布或卡方检验来分析FPTP。以下是一个典型问题：在两候选人选举中，已知40%的选民支持候选人A，60%支持候选人B。随机抽取10名选民。

(a) Find the probability that exactly 4 voters support A. Using the binomial distribution, P(X = 4) = C(10,4) × 0.4⁴ × 0.6⁶ = 210 × 0.0256 × 0.0467 ≈ 0.251. (b) Find the probability that A wins the sample, i.e., at least 6 of 10 voters support A. P(X ≥ 6) = 1 − P(X ≤ 5) ≈ 0.167.

(a) 求恰好4名选民支持A的概率。使用二项分布，P(X = 4) = C(10,4) × 0.4⁴ × 0.6⁶ = 210 × 0.0256 × 0.0467 ≈ 0.251。(b) 求A在样本中获胜的概率，即10人中至少6人支持A。P(X ≥ 6) = 1 − P(X ≤ 5) ≈ 0.167。

These calculations demonstrate how the Binomial(N, p) model underpins quantitative analysis of electoral systems. In your exam, remember to state the distribution of X clearly, use correct probability notation, and check whether a continuity correction is required.

这些计算展示了二项分布模型如何支撑选举制度的定量分析。在考试中，请记得清晰陈述X的分布，使用正确的概率符号，并检查是否需要连续性校正。

11. Conclusion | 结语

First Past the Post is a deceptively simple electoral rule with profound mathematical implications. From Binomial distributions and the Central Limit Theorem to the Gallagher Index and game-theoretic strategic voting, the analysis of FPTP touches on many core topics in the Edexcel A-Level Mathematics syllabus.

简单多数制是一个看似简单却具有深远数学含义的选举规则。从二项分布和中心极限定理到加拉格尔指数和博弈论中的策略性投票，对FPTP的分析涉及Edexcel A-Level数学大纲中的许多核心主题。

By mastering these mathematical tools, you not only gain the skills to analyse electoral systems rigorously but also strengthen your broader statistical and decision-making abilities. Practice applying each formula in context, and you will be well prepared for any election-related question in your examinations.

掌握这些数学工具，你不仅获得了严谨分析选举制度的能力，还能强化更广泛的统计和决策能力。练习在具体情境中应用每个公式，你就能在考试中从容应对任何与选举相关的问题。

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First Past the Post | 简单多数制