This topic runs over two classes. The first class is the Traitors activity and its debrief; the second class is the formal discussion of the chapter and the worked exam question. The activity is the spine of both, so we set it out in full.
Goal. Show that a plan of action can look like an equilibrium at the start of a game yet stop being optimal once a particular subgame is reached. This is the gap between a Nash equilibrium of the whole game and one that is optimal in every subgame, and it motivates subgame perfection.
We play the game three times: first with no guidance, then with a prescribed voting rule, and finally with the rule in place but players free to break it for fun. The contrast between the runs is the whole point, so leave time for all three.
Setup and numbers. Deal a hidden role to every student: a small number are Traitors, the rest are Faithful. Only the Traitors learn who the other Traitors are. Seat the group in a fixed circle and keep that ordering for the whole game. A clean default is ten players with two Traitors. The reason for this particular count is worth knowing. Each round removes two players, one banished by the day vote and one murdered at night, so the state \((n, m)\) (with \(n\) players and \(m\) Traitors) loses two players a round. The Traitors win the moment they reach parity, \(2m \ge n\); the Faithful win once every Traitor has been banished. Starting from \((10, 2)\) the game runs
reaching parity in three rounds if no Traitor is banished along the way, and on average a little under four rounds once the random tie-break is taken into account. That is short enough to play twice in a class. For a larger room, keep two compliance rounds before the endgame by choosing \(n \approx 2m + 6\), so three Traitors wants twelve players and four Traitors wants fourteen. This uses more Traitors per head than the television show, which is deliberate: with the show's ratio the interesting endgame is never reached inside a class.
Each round has two phases:
First game: no instruction. Deal the roles, explain only the two phases and the win conditions, and say nothing about how to vote. Let the group vote however it likes. With free voting the Traitors collude covertly, and that collusion is statistically indistinguishable from ordinary disagreement, so it usually goes unpunished and the Traitors tend to win. Keep this run short, one or two rounds is enough to make the point: the Faithful have no way to detect coordination.
Second game: the Vote-Left rule. Reset and deal fresh roles. Now prescribe a rule: every player votes for the next surviving player to their left in the circle. Two properties are worth drawing out by hand.
Pair the rule with a punishment: a player seen to deviate is banished next. While there are enough Faithful to carry out that punishment, complying with Vote-Left is a best response for everyone, Faithful and Traitor alike, since a detected deviation leads to certain banishment and a winning probability of zero.
The endgame, and the deviation. Play the second game on towards its end and watch for the state \((6, 2)\), where \(n = 2m + 2\). From here a Traitor who deviates is detected, but after the following night murder there are no longer enough Faithful to outvote and banish them. The punishment threat has stopped being credible, and the Traitors' best response flips: they should now collude openly, banish a Faithful, and reach parity. This is the moment to pause. The Traitors' colluding move is the optimal action in that subgame, and it is a well-defined best response there whether or not any particular game actually reaches it. That is exactly what subgame perfection asks of a strategy: it must prescribe an optimal action at every subgame, including ones never reached in a given play.
Third game: when you are not sure. Reset once more and keep the Vote-Left rule, but now tell the group that anyone may break it now and then for fun, with no warning. Reintroducing this noise undoes what Vote-Left bought us. A vote that breaks the rule no longer points to a Traitor, since a Faithful might simply be playing around, so the Faithful can no longer be sure who, if anyone, is colluding. Let \(q\) be the Faithful's belief that a flagged suspect really is the Traitor. Banishing a suspect becomes a gamble: with probability \(q\) it removes the Traitor, and with probability \(1 - q\) it removes one of their own. When \(q\) is low the threat to banish on suspicion is empty, because carrying it out is as likely to cost the Faithful a friend as to catch a Traitor, and the Traitors exploit exactly this. This is the game written up as the marked exam question below, where the threshold on \(q\) is worked out in full.
Debrief. Three threads to pull together.
First, a rule that makes deviation visible is what turns the Faithful's threat into a credible one. With free voting the Traitors' collusion is indistinguishable from disagreement, so the threat to punish is empty; under Vote-Left a deviation is unmistakable, so the threat bites and the Traitors comply.
Second, even Vote-Left has a limit. The punishment is credible early and not credible late: with many Faithful left, "deviate and you will be banished" is a real deterrent, but in the endgame the Faithful can no longer carry it out. Draw the partial game tree for the last couple of rounds and circle the subgame at \((6, 2)\) where the Traitors' best response changes.
Third, the rule's bite depends on being sure. Once players break Vote-Left for fun, a flagged suspect is the Traitor only with some probability \(q\), and a threat to banish on suspicion is non-credible when \(q\) is low. This is the same picture as the marked exam question below: a plan can be a Nash equilibrium of the whole game while relying on an action, banishing a suspect, that is not a best response once its subgame is reached.
A note for the curious. The Vote-Left rule, the threshold \(n > 2m + 2\) for its credibility, and the Traitors' optimal endgame deviation are worked out in full in Knight, The Vote-Left Equilibrium: A Deterministic Coordination Strategy for the Faithful in The Traitors, arXiv:2605.10233. The paper shows that Vote-Left is a Perfect Bayesian Equilibrium for every state with \(n > 2m + 2\), and that it roughly triples the Faithful's winning probability over random voting when the Traitors collude. None of this is examinable; it is here for anyone who wants to see where the activity comes from.
The second class formalises the activity. Discuss the Subgame Perfection chapter.
Work through the centipede game as the formal worked example of backwards induction:
Solve it by backwards induction, then contrast the subgame perfect equilibrium with the strategy that passes at the first two nodes and takes at the last two.
Discussion point: After the definition of backwards induction, ask what it leads to for the centipede game.
Discussion point: After the subgame perfection definition, ask which equilibrium is subgame perfect, and relate it to the Traitors punishment threat that was credible early and failed in the endgame subgame.
The third game above is written up as a marked exam question: Question 1 (the in-class activity) on the Subgame Perfection page, with a full worked solution. Closing the loop here is the step that helps students who find exams hard: work through that question together, or set it as the immediate follow-up, so they see the game they just played turned into a full-mark answer.
General email templates to send before and after this class. Fill in the bracketed placeholders before sending.
Hi all,
A reminder that our next Game Theory class covers Subgame Perfection.
All of the course materials, including the relevant chapter, are available
at https://vknight.org/gt/. It is worth skimming the chapter beforehand.
See you in class,
Vince
Dear all,
Thanks for your work in today's class on Subgame Perfection.
A recording is available here [RECORDING LINK] and on Learning Central.
All class resources are available at https://vknight.org/gt/.
Thanks,
Vince