Past exam of the computer science course of the University of Oxford / 2018 / Computational game theory / 2 / d / Solution

Yes: let the order be voter 2, then voter 3, then voter 1. When voter 1 moves last, it makes $A$ win whenever that is possible; otherwise it prefers $B$ to $C$. For the first two votes in this new order, its optimal continuation winners areRows now record voter 2's initial choice and columns voter 3's choice. In a three-way split, the specified tie-break produces $A$, which explains, for example, the off-diagonal $B,C$ entry.

Voter 3 optimizes its ranking $B\succ C\succ A$ in each row, giving the continuation winners $A,B,C$. Voter 2 then chooses $C$, its favourite of the three. The winner is therefore $C$ in every subgame perfect equilibrium, again by unique outcome under strict outcome preferences. In particular, if voter 2 starts with $C$, voter 3 must also vote $C$: either other vote lets the last voter force $A$.