 The Hawk-Dove game has no Nash equilibria in pure strategies and exactly one Nash equilibrium in mixed strategies. The mixed strategy Nash equilibrium has both individuals playing Hawk with probability \(\frac{V}{C}\) and Dove with probability \(1-\frac{V}{C}\). Denote this strategy by \(\sigma\). According to the fundamental theorem of mixed-strategy Nash equilibria (see Gintis, 2009), it is the case that \[ \pi(\text{Hawk} \mid \sigma) = \pi(\text{Dove} \mid \sigma) = \pi(\sigma \mid \sigma), \] where “\(\pi( x \mid y)\)” denotes the payoff obtained when playing strategy \(x\) against someone using the strategy \(y\). From this, it follows that for any other mixed strategy \(\mu\) it is the case that \(\pi( \mu \mid \sigma ) = \pi(\sigma \mid \sigma)\), and so the Nash equilibrium is not strict. Yet a population where everyone follows the strategy \(\sigma\) is still nevertheless able to resist invasion, for it can be shown that \(\pi(\sigma \mid \mu) &gt; \pi(\mu \mid \mu)\). That is, the incumbent strategy \(\sigma\) receives a higher payoff when played against any mutant strategy \(\mu\), than the mutant strategy receives when played against itself.