We provide a game-theoretic analysis of consensus, assuming that processes are controlled by rational agents and may fail by crashing. We consider agents that \emph{care only about consensus}: that is, (a) an agent's utility depends only on the consensus value achieved (and not, for example, on the number of messages the agent sends) and (b) agents strictly prefer reaching consensus to not reaching consensus. We show that, under these assumptions, there is no \emph{ex post Nash Equilibrium}, even with only one failure. Roughly speaking, this means that there must always exist a \emph{failure pattern} (a description of who fails, when they fail, and which agents they do not send messages to in the round that they fail) and initial preferences for which an agent can gain by deviating. On the other hand, if we assume that there is a distribution $π$ on the failure patterns and initial preferences, then under minimal assumptions on $π$, there is a Nash equilibrium that tolerates $f$ failures (i.e., $π$ puts probability 1 on there being at most $f$ failures) if $f+1 < n$ (where $n$ is the total number of agents). Moreover, we show that a slight extension of the Nash equilibrium strategy is also a \emph{sequential} equilibrium (under the same assumptions about the distribution $π$).