Appendix
Proof of Proposition 3.1:
Part (1)
\(\ 0\leqslant x<75\): By inspection, \((T,M)\) and \((B,L)\) are pure strategy Nash equilibria. Let the probability of Player 1 choosing \(T\) (resp. \(B)\) be \(q\) (resp. \((1-q))\), and the probability of Player 2 choosing \(L\) (resp. \(M)\) be \(p\) (resp. \((1-p))\). For \(x\) in this range, \(R\) is dominated by \(\frac{1}{4}\cdot L+\frac{3}{4}\cdot M\), which yields an expected payoff of \(75\). Thus, \(R\) cannot be played in Nash equilibrium. There are 3 equilibria: \((T,M), (B,L)\) and \(\left( \frac{3}{4}\cdot T+\frac{1}{4} \cdot B,\frac{1}{4}\cdot L+\frac{3}{4}\cdot M\right) \).
Part (2)
\(75<x\leqslant 100:\) For \(x\) in this range, \((T,M)\) and \((B,L)\) remain pure strategy Nash equilibria. Player 2 is indifferent between \(M\) and \(R\) when \(100q=x\) or \(q=x/100\). Player 1 is indifferent between \(T\) and \(B\) when \(300p\,+\,50(1-p)=55(1-p)\), or \(p=1/61\). There are 3 equilibria: \((T,M), (B,L)\) and \(\left( \frac{x}{100}\cdot T+\frac{ 100-x}{100}\cdot B,\frac{1}{61}\cdot M+\frac{60}{61}\cdot R\right) \).Footnote 25
Part (3)
\(100<x<300:\) For this range, \(M\) is dominated for Player 2 by \(R\). Once \(M\) is eliminated, Player 1 will never play \(T\), which is now a dominated strategy. He thus plays \(B\). The best response for Player 2 is to play \(L\). In this case, there is a unique Nash equilibrium: \((B,L)\), which is also an iterated dominance equilibrium. \(\square \)
Proof of Proposition 3.2:
Part 1. \(0\leqslant x\leqslant \left( 1-\delta _{2}\right) 75\): In this range, there are two EUA in pure strategies and one in mixed strategies. In the pure equilibria, the supports of the equilibrium beliefs are given by \((T,M)\) and \((B,L)\). Consider the first of these. Define a neo-additive capacity, \(\nu _{1}\), by \(\nu _{1}=\left( 1-\delta _{1}\right) \pi _{M}\left( A\right) \), where \(\pi _{M}\) is the additive probability on \( S_{2}\) defined by \(\pi _{M}\left( A\right) =1\) if \(M\in A\), \(\pi _{M}\left( A\right) =0\) otherwise. Similarly, define Player \(2\)’s beliefs \(\nu _{2}\) by \( \nu _{2}=\left( 1-\delta _{2}\right) \pi _{T}\left( A\right) .\) By definition, \(\hbox {supp}\nu _{1}=M\) and \(\hbox {supp}\nu _{2}=T.\) With these beliefs the (Choquet) expected payoffs of Players 1 and \(2\) are, respectively, \(V_{1}\left( T\right) =\left( 1-\delta _{1}\right) 300,V_{1}\left( B\right) =0,V_{2}\left( L\right) =0,V_{2}\left( M\right) =\left( 1-\delta _{2}\right) 100,V_{2}\left( R\right) =75.\) It can be been that the support of both players’ beliefs consists only of best responses. Denote this equilibrium by \(\left\langle T,M\right\rangle \) . By similar reasoning, we may show that there exists a pure equilibrium where \(\hbox { supp}\nu _{1}=L\) and \(\hbox {supp}\nu _{2}=B,\) which we denote by \( \left\langle B,L\right\rangle \) .
Now consider the mixed equilibria. Denote the equilibrium beliefs of Players \(1\) and \(2\), respectively, by \(\tilde{\nu }_{1}=\left( 1-\tilde{\delta } _{1}\right) \tilde{\pi }_{1}\) and \(\tilde{\nu }_{2}=\left( 1-\tilde{\delta } _{2}\right) \tilde{\pi }_{2}\). Player \(2\)’s Choquet expected payoffs are given by, \(V_{2}\left( L\right) =300\left( 1-\tilde{\delta }_{2}\right) \tilde{\pi }_{2}\left( B\right) , V_{2}\left( M\right) =100\left( 1-\tilde{ \delta }_{2}\right) \tilde{\pi }_{2}\left( T\right) \) and \(V_{2}\left( R\right) =x\). If \(V_{2}\left( L\right) <x\leqslant (1-\tilde{\delta }_{2})75\), then \(\tilde{\pi }_{2}\left( B\right) <\frac{1}{4}\), which implies \(\tilde{\pi }_{2}\left( T\right) >\frac{3}{4}\). Hence \(V_{2}\left( M\right) =100\left( 1- \tilde{\delta }_{2}\right) \tilde{\pi }_{2}\left( T\right) >(1-\tilde{\delta } _{2})75\geqslant x\). Thus, \(R\) cannot be a best response for Player 2 and hence \( \tilde{\pi }_{1}\left( R\right) =0\). Consequently, in any mixed equilibrium, \(2\) ’s strategies are \(L\) and \(M\).
In a mixed equilibrium, Player \(2\) must be indifferent between \(L\) and \(M\) and hence
$$\begin{aligned} V_{2}\left( L\right)&= V_{2}\left( M\right) \Leftrightarrow 300\left( 1- \tilde{\delta }_{2}\right) \tilde{\pi }_{2}\left( B\right) =100\left( 1-\tilde{ \delta }_{2}\right) \tilde{\pi }_{2}\left( T\right) \\&\Leftrightarrow \tilde{\pi }_{2}\left( T\right) =\frac{3}{4}. \end{aligned}$$
In this equilibrium, \(V_{2}\left( L\right) =V_{2}\left( M\right) =75\left( 1- \tilde{\delta }_{2}\right) .\) Similarly, we may show that for Player \(1\) to be indifferent between \(T\) and \(B\), we must have \(\tilde{\pi }_{1}\left( L\right) =\frac{1}{4}\) and \(\tilde{\pi }_{1}\left( M\right) =\frac{3}{4}\).
Thus, in the mixed equilibrium, \(\tilde{\nu }_{1}=\left( 1-\tilde{\delta } _{1}\right) \tilde{\pi }_{1}\) with \(\tilde{\pi }_{1}\left( L\right) =\frac{1}{4 }\) and \(\tilde{\pi }_{1}\left( M\right) =\frac{3}{4}\) and \(\hbox {supp} \tilde{\nu }_{1}=\left\{ L,M\right\} \), while \(\tilde{\nu }_{2}=\left( 1-\tilde{ \delta }_{2}\right) \tilde{\pi }_{2}\) with \(\tilde{\pi }_{2}\left( T\right) = \frac{3}{4}\) and \(\tilde{\pi }_{2}\left( B\right) =\frac{1}{4}\), with support \(\left\{ T,B\right\} \) . In this equilibrium, \(V_{2}\left( L\right) =V_{2}\left( M\right) =75\left( 1-\tilde{\delta }_{2}\right) \) . We shall denote this equilibrium by \(\left\langle \frac{3}{4}\cdot T+\frac{1}{4}\cdot B,\frac{1}{4}\cdot L+\frac{3}{4}\cdot M\right\rangle \) .
Part 2. \((1-\delta _{2})75<x<(1-\delta _{2})100:\) In this range, there are two EUA in pure strategies: \( (T,M) \) and \((B,L)\). The reasoning is similar to that used in Part 1 above.
In addition, there is a mixed strategy equilibrium. Denote the equilibrium beliefs of Players \(1\) and \(2\), respectively, by \(\tilde{\nu }_{1}=\left( 1- \tilde{\delta }_{1}\right) \tilde{\pi }_{1}\) and \(\tilde{\nu }_{2}=\left( 1- \tilde{\delta }_{2}\right) \tilde{\pi }_{2}\). Player \(2\)’s Choquet expected payoffs are given by, \(V_{2}\left( L\right) =300\left( 1-\tilde{\delta } _{2}\right) \tilde{\pi }_{2}\left( B\right) , V_{2}\left( M\right) =100\left( 1-\tilde{\delta }_{2}\right) \tilde{\pi }_{2}\left( T\right) \) and \( V_{2}\left( R\right) =x\).
First note that there is no longer equilibrium where Player \(2\) mixes between \(L\) and \(M\). Such an equilibrium would require Player \(1\) to be indifferent between \(T\) and \(B\). As in part 1, this would imply \(\tilde{\pi }_{2}\left( T\right) =\frac{3}{4}\) and \(\tilde{\pi } _{2}\left( B\right) =\frac{1}{4}\). Hence, \(V_{2}\left( M\right) =75\left( 1- \tilde{\delta }_{2}\right) <V_{2}\left( R\right) \), which implies that \(M\) is not a best response. A similar argument demonstrates that there is no equilibrium where player \(2\) randomises between all three of his strategies. We can also eliminate the possibility that Player 2 mixes between \(L\) and \( R\), as follows. If \(2\) never plays \(M\), then \(B\) is a best response for Player \(1\). However, \(B\) is not compatible with Player \(2\) being indifferent between \(L\) and \(R\).
The remaining possibility is that \(2\) mixes between \(M\) and \(R\). Player 2 is indifferent between \(M\) and \(R\) when
$$\begin{aligned} V_{2}\left( M\right) =V_{2}\left( R\right) \Leftrightarrow 100\left( 1- \tilde{\delta }_{2}\right) \tilde{\pi }_{2}\left( T\right) =x\Leftrightarrow \tilde{\pi }_{2}\left( T\right) =\frac{x}{\left( 1-\tilde{\delta }_{2}\right) 100}. \end{aligned}$$
Similarly, Player 1’s Choquet expected payoff is given by \(V_{1}(T){\,=\,}300\left( \!1-\tilde{\delta }_{1}\!\right) \tilde{\pi }_{1}(M) +50\left( 1-\tilde{\delta }_{1}\right) \tilde{\pi }_{1}\left( R\right) \) and \(V_{1}\left( B\right) =55(1-\tilde{\delta }_{1})\tilde{\pi } _{1}\left( R\right) \) . Player 1 is indifferent between \(T\) and \(B\) when
$$\begin{aligned} V_{1}\left( T\right) =V_{1}\left( B\right) \Leftrightarrow \tilde{\pi } _{1}\left( M\right) =\frac{1}{61}. \end{aligned}$$
Thus, in the mixed equilibrium, \(\tilde{\nu }_{1}=\left( 1-\tilde{\delta } _{1}\right) \tilde{\pi }_{1},\) with \(\tilde{\pi }_{1}\left( M\right) =\frac{1}{ 61}\) and \(\tilde{\pi }_{1}\left( R\right) =\frac{60}{61}\) and \(\hbox {supp} \tilde{\nu }_{1}=\left\{ M,R\right\} \), while \(\tilde{\nu }_{2}=\left( 1- \tilde{\delta }_{2}\right) \tilde{\pi }^{2}\) with \(\tilde{\pi }_{2}\left( T\right) =\frac{x}{\left( 1-\tilde{\delta }_{2}\right) 100}\) and \(\tilde{\pi } _{2}\left( B\right) =\frac{(1-\tilde{\delta }_{2})100\text { }-\text { }x}{(1- \tilde{\delta }_{2})100},\) with support \(\left\{ T,B\right\} .\) In this equilibrium \(V_{2}\left( M\right) =V_{2}\left( R\right) =x\). The mixed strategy equilibrium is \(\left\langle \frac{x}{(1-\tilde{\delta }_{2})100} \cdot T+\frac{(1-\tilde{\delta }_{2})100\text { }-\text { }x}{(1-\tilde{\delta } _{2})100}\cdot B,\frac{1}{61}\cdot M+\frac{60}{61}\cdot R\right\rangle \).
Part 3. \(\ (1-\delta _{2})100<x<(1-\delta _{2})300:\) Denote the equilibrium beliefs of Players \(1\) and \(2\), respectively, by \(\tilde{\nu }_{1}=\left( 1-\tilde{\delta }_{1}\right) \tilde{\pi }_{1}\) and \( \tilde{\nu }_{2}=\left( 1-\tilde{\delta }_{2}\right) \tilde{\pi }_{2}.\) Player \( 2\)’s Choquet expected payoffs are given by \(V_{2}\left( L\right) =300\left( 1-\tilde{\delta }_{2}\right) \tilde{\pi }_{2}\left( B\right) , V_{2}\left( M\right) =100\left( 1-\tilde{\delta }_{2}\right) \tilde{\pi } _{2}\left( T\right) \) and \(V_{2}\left( R\right) =x\). For \(x\) in this range, \( V_{2}\left( R\right) >V_{2}\left( M\right) \) for any beliefs of Player 2 and hence \(\tilde{\pi }_{1}\left( M\right) =0\). Player \(1\)’s Choquet expected payoffs are given by \(V_{1}\left( T\right) =50\left( 1-\tilde{\delta } _{1}\right) \tilde{\pi }_{1}\left( R\right) \) and \(V_{1}\left( B\right) =100\left( 1-\tilde{\delta }_{1}\right) \tilde{\pi }_{1}\left( L\right) +55\left( 1-\tilde{\delta }_{1}\right) \tilde{\pi }_{1}\left( R\right) \) . Strategy \(B\) yields a higher Choquet expected payoff than \(T\) for any beliefs of Player 1, with support contained in \(\{L,R\}\). For Player 2, \(L\) is the best response to \(B\). In this case, there is a unique EUA: \( \left\langle B,L\right\rangle \).
Part 4. \(x>(1-\delta _{2})300:\) Denote the equilibrium beliefs of Players \(1\) and \(2\), respectively, by \(\tilde{\nu } _{1}=\left( 1-\tilde{\delta }_{1}\right) \tilde{\pi }_{1}\) and \(\tilde{\nu } _{2}=\left( 1-\tilde{\delta }_{2}\right) \tilde{\pi }_{2}\). Player \(2\)’s Choquet expected payoffs are given by \(V_{2}\left( L\right) =300\left( 1- \tilde{\delta }_{2}\right) \tilde{\pi }_{2}\left( B\right) , V_{2}\left( M\right) =100\left( 1-\tilde{\delta }_{2}\right) \tilde{\pi }_{2}\left( T\right) \) and \(V_{2}\left( R\right) =x\), where \(\ x>(1-\tilde{\delta } _{2})300\).
For \(x\) in this range, \(R\) strictly dominates both \(L\) and \(M\) for any beliefs of Player 2 and hence \(\tilde{\pi }_{1}\left( L\right) =\tilde{\pi } _{1}\left( M\right) =0\). Player \(1\)’s best response is to play \(B\), with \( \hbox {supp}\nu _{1}=R\). There is a unique EUA: \(\left\langle B,R\right\rangle \) . \(\square \)