Solutions to the example questions on the Social Choice page.
(a) First choice votes. [2]
First choice votes: \(A = 4\), \(B = 3\), \(C = 2\). The winner by first choice votes is \(A\).
(b) Borda. [4]
Each alternative scores the number of alternatives a voter ranks it above, so here a first place is worth 2 points, a second 1 and a third 0:
The Borda winner is \(B\).
(c) Condorcet winner. [4]
\(B\) beats both \(A\) and \(C\), so \(B\) is the Condorcet winner.
(d) Why the methods disagree, and a recommendation. [5]
Counting first choice votes picks \(A\) because it looks only at top choices, and \(A\) has the most. But a majority (\(5\) of \(9\)) rank \(A\) last, which the first choice count ignores. Borda and Condorcet use the full rankings, the lower preferences and the pairwise comparisons, and both select \(B\), who is preferred to each rival by a majority. The methods disagree because counting first choice votes discards everything except each voter's top choice. For the class vote, \(B\) is the better outcome: it is the Condorcet winner, beating each alternative head-to-head, so it commands majority support against any rival, whereas \(A\), the winner on first choice votes, would lose a straight vote to either of the others. A Condorcet (or Borda) method is therefore preferable here, as it reflects the whole class's preferences rather than only first choices.
(a) Definitions. (Bookwork.) [4]
(b)(i) First choice votes. [2]
First choice votes: \(A = 3\), \(B = 2\), \(C = 2\); the winner by first choice votes is \(A\).
(b)(ii) Borda. [3]
The Borda winner is \(B\).
(b)(iii) Condorcet. [3]
\(A\) vs \(B\): \(B\) wins \(4\)-\(3\). \(B\) vs \(C\): \(B\) wins \(5\)-\(2\). So \(B\) beats both and is the Condorcet winner.
(b)(iv) Comment. [2]
Counting first choice votes elects \(A\) on top choices alone, but \(A\) loses to both \(B\) and \(C\) head-to-head. Borda and Condorcet both elect \(B\), which is preferred by a majority against each rival. The methods disagree because counting first choice votes ignores lower preferences.
(c) Condorcet winner. [4]
A Condorcet winner beats every other candidate in pairwise majority votes. It may fail to exist because pairwise majorities can be cyclic, in which case no candidate beats all others. For this profile, however, a Condorcet winner does exist, namely \(B\): so for these seven voters majority rule is well behaved, the pairwise relation is transitive (\(B \succ A\), \(B \succ C\), and \(C \succ A\)), and there is a single alternative that a majority prefers to every other. The existence of a Condorcet winner tells us there is no majority cycle here.
(a) Definitions. (Bookwork.) [4]
(b)(i) Pairwise majorities. [3]
(b)(ii) No Condorcet winner. [2]
The majorities cycle \(A \succ B \succ C \succ A\), so no candidate beats all the others; there is no Condorcet winner.
(b)(iii) Condorcet cycle. [4]
Every individual ranking is transitive, yet the collective majority relation is cyclic. This is a Condorcet cycle: aggregating transitive individual preferences by pairwise majority can produce an intransitive collective preference, so majority rule need not yield a consistent ranking.
(c) Borda's method. [6]
Each alternative scores the number of alternatives a voter ranks it above (2 points for a first place, 1 for a second, 0 for a third):
The Borda ranking is \(B \succ A \succ C\): a complete, transitive ranking, even though pairwise majority cycles on the very same profile. The reason is the information each method uses: pairwise majority looks only at isolated head-to-head contests, in each of which the losers' margins are discarded, so the separate verdicts can disagree and form a cycle. Borda instead uses positional information, each alternative's rank in every voter's full list, and sums a single number per alternative; ordering alternatives by their scores can never produce a cycle. The price of escaping the cycle is the pairwise viewpoint itself: here the Borda winner \(B\) loses its head-to-head contest with \(A\) by \(3\) votes to \(2\). Borda avoids intransitivity precisely by giving up pairwise (independence) information in favour of positional scores.
(a) Manipulability and strategy-proofness. [4]
A voter manipulates a voting rule when, by submitting a ballot that does not reflect their true preferences, they obtain a collective outcome they strictly prefer to the one their sincere ballot would produce, with the other voters' ballots held fixed. A rule is manipulable if some preference profile admits such a voter, and strategy-proof if no voter can ever gain by voting insincerely, so that sincere voting is always a best response.
(b) Sincere Borda scores. [6]
With 2, 1 and 0 points for first, second and third,
So \(A\) and \(B\) tie on 13 points, ahead of \(C\) on 10: the sincere outcome is a tie between \(A\) and \(B\).
(c) A profitable manipulation. [9]
Take one of the five voters whose sincere preference is \(A \succ B \succ C\). They rank \(A\) first, so they would rather \(A\) win outright than share first place with \(B\). Suppose this voter instead submits \(A \succ C \succ B\), keeping \(A\) top but demoting \(B\) below \(C\). Only their ballot changes: \(B\) loses the point it received from this voter and \(C\) gains one, while \(A\) is unaffected. The profile becomes
with scores
Now \(A\) wins outright with 13 points. The manipulating voter, who ranks \(A\) first, strictly prefers this to the sincere \(A\)-\(B\) tie, so the misreport is profitable and Borda's method is manipulable on this profile.
(d) Independence of irrelevant alternatives and Gibbard-Satterthwaite. [6]
The voter changed only where \(C\) sits relative to \(B\); they still rank \(A\) above \(B\). Independence of irrelevant alternatives requires the collective ranking of \(A\) and \(B\) to depend only on how the voters rank \(A\) against \(B\), so moving the irrelevant alternative \(C\) should leave that verdict untouched. Under Borda it does not: promoting \(C\) drains a point from \(B\) and breaks the tie in \(A\)'s favour. This failure of independence of irrelevant alternatives is exactly what makes the manipulation possible. The Gibbard-Satterthwaite theorem shows the phenomenon is unavoidable: every non-dictatorial voting rule over three or more alternatives is manipulable, so no reasonable rule, Borda included, can be strategy-proof.
Marking exercise 1 (Question 1(b)).
The winner is right but the scores are not Borda scores as defined in this course: an alternative scores the number of alternatives a voter ranks it above, so 2, 1 and 0 points here, giving \(A: 8\), \(B: 12\), \(C: 7\) as in the solution above. The \((3, 2, 1)\) scheme adds one point per voter to every alternative, which is why the ranking survives; an answer using it has not answered the question asked.
The closing claim is the serious error: it is not true that any decreasing point scheme gives the same ranking. Only schemes obtained from \((2, 1, 0)\) by a positive affine transformation do. With points \((10, 1, 0)\), for example, this same profile gives \(A: 40 + 3 = 43\), \(B: 4 + 30 + 2 = 36\), \(C: 3 + 20 = 23\), and the winner switches to \(A\). A fair mark is [2] of [4].
Marking exercise 2 (Question 4(b) and (c)).
Part (b) is correct: the sincere outcome is the \(A\)-\(B\) tie on 13, as in the solution above, so [6] of [6].
In part (c) the chosen ballot and the conclusion are right, but the scores are wrong: moving \(C\) above \(B\) does not only take a point from \(B\), it gives one to \(C\), whose score becomes \(11\), not \(10\). The slip is detectable without recomputing anything: each of the 12 voters hands out \(2 + 1 + 0 = 3\) points, so the scores must total \(36\), and \(13 + 12 + 10 = 35\). A fair mark is [7] of [9].
The final sentence gets the Gibbard-Satterthwaite theorem backwards, twice. The theorem states that every non-dictatorial rule over three or more alternatives is manipulable, so the example confirms it, violating nothing; and it is a theorem about all voting rules, not a definition of fairness. That sentence belongs to part (d), where it would cost marks.