# Nash Equilibria of Quantum Games in the Special Two-Parameter Strategy Space
## 4 Games without Pure Nash Equilibria
Without loss of generality, we can assume that \(\theta_{b}^{*}=0\). However, then, player \(b\) obtains \(\alpha\) by choosing \((\theta_{a},\varphi_{a})=(\pi,\phi_{b}^{*})\); i.e.,  
\[\Pi_{a}(\pi,\phi_{b}^{*},0,\phi_{b}^{*})=\alpha>\gamma=\Pi_{a}(\theta_{a}^{*}, \phi_{a}^{*},\theta_{b}^{*},\phi_{b}^{*}). \tag{47}\]  
This contradicts our assumption that \((\theta_{a}^{*},\phi_{a}^{*},\theta_{b}^{*},\phi_{b}^{*})\) is a Nash equilibrium. An immediate consequence of Proposition 3 is the following corollary. **Corollary 1**. : _There are no pure Nash equilibria of the game_  
\[ \begin{pmatrix}(\gamma,\gamma)&(\beta,\alpha)\\ (\alpha,\beta)&(\delta,\delta)\end{pmatrix} ,\quad\alpha>\gamma>\delta>\beta, \;2\gamma>\alpha+\beta. \tag{48}\]  
The game (48) is a well known prisoner's dilemma. The condition that \((\gamma,\gamma)\) is beyond the convex hull of the remaining payoffs (41) is equivalent to the last inequality in (48). Note that it is enough to replace the first inequality \(\alpha>\gamma\) with an equality \(\alpha=\gamma\) for the game (48) to have a dominant pair of payoffs and Nash equilibria in pure strategies. Such a case is presented in Example 2, where \(\alpha=3,\gamma=3,\delta=1,\beta=0\). The first inequality \(\alpha>\gamma\) in Proposition 3 is therefore crucial for the non-existence of Nash equilibria. It is also worth noting that the assumptions of Proposition 3 cover a wider class of games than just the Prisoner's Dilemma. Assuming that \(\beta>\delta\) in (41), we get the next corollary. **Corollary 2**. : _There are no pure Nash equilibria of the game_  
\[ \begin{pmatrix}(\gamma,\gamma)&(\beta,\alpha)\\ (\alpha,\beta)&(\delta,\delta)\end{pmatrix} ,\quad\alpha>\gamma>\beta>\delta,\;2 \gamma>\alpha+\beta. \tag{49}\]  
The game (49) is a special case of the game of Chicken. Examples for both corollaries can be given if we assume that \(\alpha=4\), \(\gamma=3\), \(\delta=1\) and \(\beta=0\) (for the prisoner's dilemma) or \(\beta=1\) and \(\delta=0\) (for the game of chicken). **Example 3**. : _According to Corollaries 1 and 2, the following games do not have Nash equilibria in pure strategies._  
\[ \begin{pmatrix}(3,3)&(0,4)\\ (4,0)&(1,1)\end{pmatrix} \text{(the prisoner's dilemma)}\quad\text{and}\quad \begin{pmatrix}(3,3)&(1,4)\\ (4,1)&(0,0)\end{pmatrix} \text{(the game of chicken)}. \tag{50}\]  
_The payoff polygons of both games are shown in Figure 2. Here, one can also see the equivalence of conditions for \((\gamma,\gamma)\) from Equations (41), (48) and (49), where gray payoffs are outside the convex hull of the set of red payoffs._  
Figure 2: Payoff polygons for games (50). In what follows, we examine another class of games that have no Nash equilibria. Let us now consider the following game. \[ \begin{pmatrix}(\alpha,\beta)&(\gamma,\gamma)\\ (\gamma,\gamma)&(\beta,\alpha)\end{pmatrix} ,\quad\alpha>\beta,\quad(\alpha+\beta )/2>\gamma. \tag{51}\]  
This game is more general than the battle of the sexes, for which \(\alpha>\beta>\gamma\). For games of this type, the following proposition holds:  
Proposition 4: _There are no pure Nash equilibria of the game (51) in parameterization (3)._  
Proof: Using the expressions (11) and (10), the player's \(p=1,2\) payoff function of the game (51) is  
\[\Pi_{p}(\theta_{a},\phi_{a},\theta_{b},\phi_{b})= \frac{\alpha+\beta}{4}(1+\cos\theta_{a}\cos\theta_{b})+\frac{ \gamma}{2}(1-\cos\theta_{a}\cos\theta_{b})+\] \[\pm\frac{\beta-\alpha}{4}(\cos(\phi_{a}+\phi_{b})(\cos\theta_{a} +\cos\theta_{b})+\sin(\phi_{a}+\phi_{b})\sin\theta_{a}\sin\theta_{b}),\]  
where \"\(\pm\)\" means \"\(+\)\" for player \(b\) and \"\(-\)\" for player \(a\). Let us assume that the game has a Nash equilibrium at \((\theta_{a}^{*},\phi_{a}^{*},\theta_{b}^{*},\phi_{b}^{*})\).