📚 Factors Helping Bismarck: An A-Level Mathematical Modelling Study | 帮助俾斯麦的因素：A-Level 数学建模研究
This article uses A-Level Edexcel Mathematics tools - probability, decision theory, game theory, weighted scoring and regression - to quantify the main factors that helped Otto von Bismarck unify Germany between 1862 and 1871.
本文运用 A-Level Edexcel 数学工具——概率、决策论、博弈论、加权评分与回归分析——来量化 1862 至 1871 年间帮助俾斯麦统一德国的主要因素。
The aim is not to replace history with arithmetic, but to show how the same quantitative reasoning tested in Mechanics, Statistics and Decision Mathematics can be applied to a genuine historical problem with defined inputs, defined outputs and measurable uncertainty.
目的不是用算术取代历史，而是展示在力学、统计与决策数学中被考核的同一套定量推理，如何应用于一个输入明确、输出明确、不确定性可度量的真实历史问题。
1. From History to Algebra: Setting Up the Variables | 从历史到代数：设定变量
an military strength, x₂ = diplomatic isolation of rivals, x₃ = railway and industrial capacity, x₄ = nationalist sentiment, and x₅ = mistakes by Austria and France.
我们将结果 Y 定义为俾斯麦实现以普鲁士为主导的德国统一的概率。解释变量为 x₁ = 普鲁士军事实力，x₂ = 对竞争对手的外交孤立，x₃ = 铁路与工业能力，x₄ = 民族主义情绪，x₅ = 奥地利和法国的失误。
These variables can be treated as inputs in a mathematical model, allowing us to compare their relative contribution rather than relying on narrative history alone.
这些变量可以作为数学模型中的输入，使我们能够比较它们的相对贡献，而不是仅仅依赖叙述性历史。
- Y: probability of Prussian-led unification, from 0 to 1
- Y：普鲁士主导统一的概率，范围从 0 到 1
- x₁ to x₅: scored on a 0 to 10 scale
- x₁ 至 x₅：按 0 到 10 分制打分
Each variable needs an operational definition, otherwise the model cannot be tested. Military strength x₁ includes the reformed Prussian general staff under Moltke, the Dreyse breech-loading needle gun, and railway-based mobilisation. Diplomatic isolation x₂ measures how far Russia, Britain, Italy and France were kept from forming a coalition against Prussia.
每个变量都需要可操作的定义，否则模型无法检验。军事实力 x₁ 包括毛奇领导下改革后的普鲁士总参谋部、德莱塞后装针发枪以及基于铁路的动员体系。外交孤立 x₂ 衡量俄国、英国、意大利与法国在多大程度上被阻止组成反普联盟。
| Variable | Meaning | Score 1862 | Score 1870 |
| x₁ | Military strength | 6 | 9 |
| x₂ | Diplomatic isolation | 5 | 9 |
| x₃ | Railways and industry | 5 | 8 |
| x₄ | Nationalist sentiment | 4 | 6 |
| x₅ | Rival mistakes | 4 | 8 |
Once every input sits on the same scale, comparisons become arithmetic rather than rhetorical.
一旦所有输入落在同一量纲上，比较就从修辞变成了算术。
2. Probability Trees and the Unification Scenarios | 概率树与统一情景
Bismarck had to win three limited wars: against Denmark (1864), Austria (1866) and France (1870-71). A probability tree can link these three separate conflicts into a single compound event.
俾斯麦必须赢下三场有限战争：对丹麦（1864 年）、对奥地利（1866 年）以及对法国（1870–71 年）。概率树可以把这三场彼此独立的冲突连接成一个复合事件。
dependence, the probability of winning all three is the product of the three branch probabilities.
设 p₁ = P(1864 年获胜) = 0.92，p₂ = P(1866 年获胜) = 0.72，p₃ = P(1870–71 年获胜) = 0.80。假设条件独立，则三场全胜的概率是三支分支概率的乘积。
P(WWW) = p₁ × p₂ × p₃ = 0.92 × 0.72 × 0.80 = 0.52992 ≈ 0.530
The full tree contains 2³ = 8 outcomes. Listing them shows how thin the margin really was.
完整树形图包含 2³ = 8 种结果。逐一列出可以看到余量其实很薄。
| Path | Probability | Meaning |
| WWW | 0.52992 | Full unification |
| WWL | 0.13248 | North German core only |
| WLW | 0.20608 | Delayed unification |
| LWW | 0.04608 | Risky recovery |
| WLL | 0.05152 | Partial gain |
| LWL | 0.01152 | Failure |
| LLW | 0.01792 | Failure |
| LLL | 0.00448 | Collapse |
5. So the tree says something precise: the compound plan was fragile at the top (only a 53% chance of a clean sweep) but robust at the middle.
三场中至少赢两场的概率为 0.52992 + 0.13248 + 0.20608 + 0.04608 = 0.91456 ≈ 0.915。因此概率树给出了一个精确结论：整个计划在顶端是脆弱的（全胜只有 53% 机会），但在中段是稳健的。
3. Conditional Probability and the Three Wars | 条件概率与三场战争
The independence assumption in Section 2 is convenient but questionable. Prussia's victory over Austria in 1866 changed the strategic environment, so the 1870 branch is not really independent of the 1866 branch.
第 2 节的独立性假设方便但可质疑。普鲁士 1866 年战胜奥地利改变了战略环境，因此 1870 年的分支实际上并不独立于 1866 年的分支。
The conditional probability formula handles this directly.
条件概率公式可以直接处理这一点。
P(A|B) = P(A ∩ B) ÷ P(B)
a defeats Austria, with P(B) = 0.72. Let A be the event that France declares war, with P(A ∩ B) = 0.58. Then P(A|B) = 0.58 ÷ 0.72 ≈ 0.806.
设 B 为普鲁士击败奥地利的事件，P(B) = 0.72。设 A 为法国宣战的事件，P(A ∩ B) = 0.58。则 P(A|B) = 0.58 ÷ 0.
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