A few questions to warm up and check the basics, before the exam-type questions below. They are reshuffled each time, so you can keep retrying.
The following are exam-type questions in the style of the examination paper, with marks at the rates used in the papers. A question totalling fewer than 25 marks would, in the examination, be combined with further parts, often one of the examinable proofs, to make a full 25-mark question. Attempt them in full before reading the worked solutions.
In class we voted on which topic to revise, and compared methods. Nine students rank three topics \(A, B, C\): four rank \(A \succ B \succ C\), three rank \(B \succ C \succ A\), and two rank \(C \succ B \succ A\).
(a) Rank the topics by number of first choice votes and determine the winner. [2]
(b) Compute the Borda scores and the Borda winner. [4]
(c) Determine whether there is a Condorcet winner. [4]
(d) The methods disagree; explain why ranking by first choice votes picks a different winner, with reference to which preferences each method uses, and state which method you would recommend for the class vote, justifying your choice. [5]
(a) Provide definitions for the following terms:
(b) Seven voters rank three candidates \(A, B, C\):
(i) Determine the winner by number of first choice votes. [2]
(ii) Determine the Borda winner. [3]
(iii) Determine the Condorcet winner, if one exists. [3]
(iv) Comment on the disagreement between the methods. [2]
(c) Explain what a Condorcet winner is and why one may fail to exist. State what the existence of a Condorcet winner for this profile tells you about majority rule here. [4]
(a) Define a Condorcet winner and a Borda winner. [4]
(b) Five voters rank three candidates \(A, B, C\):
(i) Compute all three pairwise majority comparisons. [3]
(ii) Show that there is no Condorcet winner. [2]
(iii) Explain the Condorcet cycle: why the collective majority preference can be intransitive even when every individual ranking is transitive. [4]
(c) Apply Borda's method to the same profile and give the resulting ranking. Explain why Borda's method can never produce a cycle, even though pairwise majority does here, referring to positional versus pairwise information. [6]
An election over three alternatives \(A\), \(B\) and \(C\) uses Borda's method. Twelve voters have preferences
(a) Define what it means for a voting rule to be manipulable, and what it means for a rule to be strategy-proof. [4]
(b) Compute the Borda score of each alternative and determine the sincere outcome. [6]
(c) Consider one of the five voters whose sincere preference is \(A \succ B \succ C\). Show that by submitting an insincere ballot this voter can change the outcome to one they strictly prefer, stating the ballot they submit and the resulting scores. [9]
(d) The manipulating voter altered only the position of \(C\), the alternative they care least about, yet the collective ranking of \(A\) and \(B\) changed. Explain how this violates independence of irrelevant alternatives, and relate the example to the Gibbard-Satterthwaite theorem. [6]
These exercises are entirely optional: they are extra practice, not extra content. The best way to use an AI assistant is to be its examiner, not its student: checking work is exactly the skill the examination rewards, and the skill you will need to use these tools well. Each exercise gives a transcript in the style of a widely used free AI chatbot answering one of the example questions above. Every transcript contains at least one error. Mark it as an examiner would: find each error, explain what goes wrong, and award a mark out of the marks of the original question. A marker's report for each transcript is on the solutions page.
A student pastes Question 1(b) into a free AI chatbot, which replies:
Borda's method gives 3 points for a first place, 2 for a second and 1
for a third.
A: 4(3) + 3(1) + 2(1) = 17
B: 4(2) + 3(3) + 2(2) = 21
C: 4(1) + 3(2) + 2(3) = 16
The Borda winner is B. 🏆 Note that the choice of points does not
matter: any decreasing point scheme gives the same Borda ranking.
A student pastes Question 4(b) and (c) into a free AI chatbot, which replies:
Sincere scores: A: 5(2) + 3(1) = 13, B: 4(2) + 5(1) = 13,
C: 3(2) + 4(1) = 10. So A and B tie on 13 points. ✅
The manipulation: one of the A > B > C voters submits A > C > B
instead. Only B's score changes: it drops by one point. The new scores
are A: 13, B: 12, C: 10, so A now wins outright, which the manipulator
strictly prefers to the tie.
This possibility violates the Gibbard-Satterthwaite theorem, which
states that a fair voting rule cannot be manipulated.
Worked solutions to the example questions →
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