## 4 Computation of \(\epsilon\) Proper Equilibria
### Analytical Computation  
We now use the optimization results in Tables 2 and 3 to analytically generate sequences of \(\epsilon\)-proper equilibria from extreme Nash equilibrium 3. In the extreme Nash equilibrium 3 strategy profile, player I randomizes on \(x_{11}\) and \(x_{13}\). A sequence of \(\epsilon\)-proper equilibria should then take into account that player II is indifferent between his first and third strategies. One should think that if player I is not indifferent between his first and third strategies, he would have to assign to the less interesting strategy a probability less than \(\epsilon\) times the probability of the other strategy.  
Therefore, \(3x_{21}+4x_{22}+x_{23}+3x_{31}+2x_{32}=x_{21}+x_{22}+3x_{23}+x_{31}+x_{32}\).  
Thus, we have  
\[2x_{23}=2x_{21}+3x_{22}+2x_{31}+x_{32}\]  
\[\Leftrightarrow\]  
\[x_{23}=\frac{1}{2}+x_{21}+\frac{3}{2}x_{22}+\frac{1}{2}x_{31}. \tag{3}\]  
One should also note that this is satisfied by equilibrium 3. Hence, player II would have to assign a probability larger than \(\frac{1}{2}\) to his third strategy.  
Simultaneously, a sequence of \(\epsilon\)-proper equilibria should take into account that player II is indifferent between his first and third strategies. Therefore, \(5x_{11}+2x_{12}+2x_{13}+3x_{14}+2x_{31}+5x_{32}=x_{11}+4x_{12}+2x_{13}+x_{14} +4x_{31}+x_{32}\).  
Thus, we have  
\[2x_{11}+x_{14}+3x_{32}=x_{12}+1. \tag{4}\]  
This condition is satisfied by equilibrium 3. Taking into account that \(x_{12}\) and \(x_{14}\) would receive very small but strictly positive probabilities, player I would have to assign a probability less than \(\frac{1}{2}\) to his first strategy, and player III would have to assign a probability less than \(\frac{1}{3}\) to his second strategy.  
Finally, a sequence of \(\epsilon\)-proper equilibria would take into account that player III is indifferent between his first and second strategies. Therefore, \(x_{11}+x_{12}+2x_{13}+3x_{14}+5x_{21}+4x_{22}+3x_{23}=2x_{11}+x_{12}+2x_{13}+ x_{14}+2x_{21}+x_{22}+3x_{23}\).  
Thus, we have  
\[2x_{14}+3x_{21}+2x_{22}=x_{11}. \tag{5}\]  
This condition is also satisfied by equilibrium 3. Player II would have to assign a probability less than \(\frac{1}{6}\) to his first strategy.  
We here detail some of the steps to generate a sequence of \(\epsilon\)-proper equilibria converging to the extreme Nash equilibrium 3.  
We first set \(x_{31}=\frac{7}{9}-\alpha\epsilon\), \(x_{32}=\frac{2}{9}+\alpha\epsilon\), \(x_{21}=\frac{1}{18}+\lambda\epsilon\), \(x_{22}=\frac{1}{36}\epsilon\) and \(x_{23}=\frac{17}{18}+\rho\epsilon\), where \(\lambda+\rho+\frac{1}{36}=0\).  
With conditions (3), we obtain \(\lambda=\frac{36\alpha-5}{144}\) and \(\rho=\frac{1-36\alpha}{144}\).  
Since \(x_{32}\) needs to be less than \(\frac{1}{3}\), we have  
\[\alpha\epsilon\leq\frac{1}{9}. \tag{6}\]  
For simplification purposes, we consider that player I would be indifferent between his second and fourth strategies, which yields \(x_{14}=x_{12}\). This assumption does not eliminate the possibility for player I to have a marked preference for one of these two strategies.  
We then set \(x_{14}=x_{12}\) and substitute condition (5) in (4).  
\[x_{12}=x_{14}=\frac{7-324\alpha}{284}\times\epsilon. \tag{7}\]Since \(x_{12}\leq\frac{1}{6}\epsilon\), because player I prefers his first strategy to his second, we obtain a lower bound on \(\alpha\):  
\[\alpha\geq-\frac{121}{972}.\]  
Then, by (5) we have \(x_{11}=\frac{1}{6}+(2\beta+3\lambda+\frac{1}{8})\epsilon\) and \(x_{13}=\frac{5}{6}-(4\beta+3\lambda+\frac{1}{18})\epsilon\), where  
\[\beta=\frac{7-324\alpha}{284}.\]  
It has now been made clear that we obtained a sequence of \(\epsilon\)-proper equilibria converging to the extreme Nash equilibrium 3 when \(\epsilon\) converges to 0.