Ellsberg games

Abstract

In the standard formulation of game theory, agents use mixed strategies in the form of objective and probabilistically precise devices to conceal their actions. We introduce the larger set of probabilistically imprecise devices and study the consequences for the basic results on normal form games. While Nash equilibria remain equilibria in the extended game, there arise new Ellsberg equilibria with distinct outcomes, as we illustrate by negotiation games with three players. We characterize Ellsberg equilibria in two-person conflict and coordination games. These equilibria turn out to be related to experimental deviations from Nash equilibrium play.

Appendix

We provide here the proofs of Theorems 1, 2, 5, 6 and 7, and of Propositions 4, 6 and 7 in this order (in which they appear in the text). First we state the definition of an Ellsberg equilibrium, which was stated solely in the text of Sect. 3.

Definition 3

Let $G$ be a normal form game. A profile $(((\varOmega _1^*,\mathcal{F }_1^*,\mathcal{P }_1^*),f^*_1),\ldots ,((\varOmega _n^*, \mathcal{F }_n^*,\mathcal{P }_n^*),f^*_n))$ of Ellsberg strategies is an Ellsberg equilibrium of $G$ if for all players $i\in N$, all Ellsberg urns $(\varOmega _i,\mathcal{F }_i,\mathcal{P }_i)$ and all acts $f_i$ we have

$$\begin{aligned} U_i(((\varOmega ^*,\mathcal{F }^*,\mathcal{P }^*),f^*)) \ge U_i(((\varOmega _i,\mathcal{F }_i,\mathcal{P }_i),f_i),((\varOmega ^*_{-i},\mathcal{F }^*_{-i},\mathcal{P }^*_{-i}),f_{-i}^*)), \text{ that } \text{ is } \end{aligned}$$

$$\begin{aligned}&\min _{P_i\in \mathcal{P }_i^*,P_{-i}\in \mathcal{P }^*_{-i}}\int \limits _{\varOmega _i^*}\int \limits _{\varOmega ^*_{-i}}u_i(f_i^*(\omega _i),f_{-i}^*(\omega _{-i})) \,dP_{-i} dP_{i} \\&\quad \ge \quad \min _{P_i\in \mathcal{P }_i,P_{-i}\in \mathcal{P }^*_{-i}}\int \limits _{\varOmega _i}\int \limits _{\varOmega ^*_{-i}}u_i(f_i(\omega _i),f_{-i}^*(\omega _{-i})) \,dP_{-i} dP_{i}. \end{aligned}$$

Reduced form Ellsberg equilibrium is defined in Definition 1.

Proof of Theorem 1

“$\Leftarrow$” Let $\mathcal{Q }^*$ be a reduced form Ellsberg equilibrium according to Definition 1. We choose the states of the world $\varOmega =S$ to be the set of pure strategy profiles, thereby we see that player $i$ uses the Ellsberg urn $(S_i,\mathcal{F }_i,\mathcal{Q }_i^*)$, where $\mathcal{F }_i$ is the power set of $S_i$. We define the act $f_i^*:(S_i,\mathcal{F }_i)\rightarrow \varDelta S_i$ to be the embedding $f_i^*(s_i)=\delta _{s_i}$ of $S_i$ into $\varDelta S_i$, where $\delta _{s_i}\in \varDelta S_i$ is the degenerate mixed strategy which puts all weight on the pure strategy $s_i$. $f^*_i$ induces an image measure $Q_i^{f_i^*}$ of $Q_i\in \mathcal{Q }_i^*$ on $\varDelta S_i$,

$$\begin{aligned} Q_i^{f_i^*}: \delta _{s_i}\mapsto Q_i(f_i^{*^{-1}}(\delta _{s_i})). \end{aligned}$$

The image measure $Q_i^{f_i^*}$ can be identified with $Q_i\in \mathcal{Q }_i^*$. Thus, the reduced form Ellsberg equilibrium $\mathcal{Q }^*$ can be written as $((S,\mathcal{F },\mathcal{Q }^*),f^*)$. This strategy is an Ellsberg equilibrium according to Definition 3.

“$\Rightarrow$” Let now $((\varOmega ^*,\mathcal{F }^*,\mathcal{P }^*),\;f^*)$ be an Ellsberg equilibrium according to Definition 3. Every $P_i\in \mathcal{P }_i^*$ induces an image measure $P_i^{f_i^{*}}$ on $\varDelta S_i$ that assigns a probability to a distribution $f_i^*(\omega _i)\in \varDelta S_i$ to occur. To describe the probability that a pure strategy $s_i$ is played, given a distribution $P_i$ and an Ellsberg strategy $((\varOmega _i^*,\mathcal{F }_i^*,\mathcal{P }_i^*),\;f_i^*)$, we integrate $(f_i^*(\omega _i))(s_i)$ over all states $\omega _i\in \varOmega _i$. Thus, we can define $Q_i$ to be

$$\begin{aligned} Q_i(s_i):=\int \limits _{\varOmega _i^*} f_i^*(\omega _i)(s_i)\,dP_i. \end{aligned}$$

(5)

Recall that $\mathcal{P }_i$ is a closed and convex set of probability distributions. We call the resulting set of probability measures $\mathcal{Q }_i^*$. $\mathcal{Q }_i^*$ is closed and convex, since $\mathcal{P }_i^*$ is.

A straightforward, if tedious, reasoning shows that the profile $(\mathcal{Q }^*_1,\ldots ,\mathcal{Q }^*_n)$ yields the same payoff as $((\varOmega _i^*,\mathcal{F }_i^*,\mathcal{P }_i^*),f_i^*)_{i=1,\ldots ,n}$.

Now suppose $\mathcal{Q }^*$ is not a reduced form Ellsberg equilibrium. Then for some player $i\in N$ there exists a set $\mathcal{Q }_i$ of probability measures on $S_i$ that yields a higher minimal expected utility. We can then define the Ellsberg strategy ${\tilde{\varOmega }}=S_i,\,{\tilde{\mathcal{F }}}_i = \mathcal{P } S_i$ (the power set of $S_i$), and $f_i(s)=\delta _s$ to obtain a profitable deviation in the original Ellsberg game, a contradiction. $\square$

The main part of the proof of Theorem 2 is that every player can find for every profile of Ellsberg strategies a mixed strategy that gives him at least the same utility as his Ellsberg strategy. We first prove the latter result in the following lemma. We let $P_i^*$ abbreviate the constant act that maps every state of the world to the mixed strategy $P_i^*\in \varDelta S_i$.

Lemma 1

Let $G=\langle N,(S_i),(u_i)\rangle$ be a normal form game. Then for any profile of Ellsberg strategies $(\mathcal{P }_1,\ldots ,\mathcal{P }_n)$ and every $i\in N$ there exists a mixed strategy $P_i^*\in \varDelta S_i$ such that $U_i(\{P_i^*\},\mathcal{P }_{-i})= U_i(\mathcal{P }_i,\mathcal{P }_{-i})$.

Proof

As the set of pure strategies is finite, the set of mixed strategies $\varDelta S_i$ is compact. The linear utility function is continuous, and thus a minimizer in the compact set $\mathcal{P }_i$ exists. $\square$

Proof of Theorem 2

Let $\mathcal{P }^*$ with $\mathcal{P }^*_i=\{{P^*_i}\}$ be an Ellsberg equilibrium of $G$; then $U_i(\{{P^*_i}\},\{{P^*_{-i}}\})\ge U_i(\mathcal{P }_i,\{{P^*_{-i}}\})$ holds for all $i\in N$ for all Ellsberg strategies $\mathcal{P }_i\subseteq \varDelta S_i$. In particular this holds for all singletons $\{{P_i}\}$, so $U_i(\{{P^*_i}\},\{{P^*_{-i}}\})\ge U_i(\{{P_i}\},\{{P^*_{-i}}\})$ and thus $u_i(P^*_i,P^*_{-i})\ge u_i(P_i,P^*_{-i})$ for all $P_i \in \varDelta S_i.$ Therefore the profile $(P^*_1,\ldots ,P^*_n)$ is a Nash equilibrium of $G$.

Next, assume that $(P^*_1,\ldots ,P^*_n)$ is a Nash equilibrium of $G$. Suppose it was not an Ellsberg equilibrium of $G$, that is there exists an Ellsberg strategy $\mathcal{P }_i$ for some player $i$ such that $U_i(\mathcal{P }_i,\{{P^*_{-i}}\})> U_i(\{{P^*_i}\},\{{P^*_{-i}}\})$. By Lemma 1, there exists a $P_i^\prime$ such that $U_i(\{{P_i^\prime }\},\{{P^*_{-i}}\}) = U_i(\mathcal{P }_i,\{{P^*_{-i}}\})> U_i(\{{P^*_i}\},\{{P^*_{-i}}\}).$ This contradicts the assumption that $(P^*_1,\ldots ,P^*_n)$ is a Nash equilibrium of the game $G$. $\square$

To prove Theorem 5 we use the following notation. A two-person normal form game $G$ is described by the payoff matrices $A$ and $B$ of player 1 and player 2, we write $G=(A,B)$. We call $G$ square when $A$ and $B$ are square matrices; in this section we only consider square games. The row vectors of $A\,(B)$ are denoted by subscripts, $a_i \,(b_i)\in \mathbb{R }^n$, and the column vectors by superscripts, $a^j\,(b^j)\in \mathbb{R }^n$. In the following it will be convenient to write the expected utility of a mixed strategy $P$ of player 1 with payoff matrix $A$ as $P A Q$ when player 2 plays mixed strategy $Q$. Transpose signs are suppressed.

A necessary condition for $G$ to have a completely mixed Nash equilibrium is that no player has weakly dominated strategies. That is, no row (column) of $A$ ($B$) is dominated by another row (column) or a convex combination of rows (columns). This condition can be expressed as follows. Let $\tilde{A}$ be the $(n+1)\times n$-matrix consisting of the matrix $A$ with an additional last column $(1,\ldots ,1)$, and $\tilde{B}$ the $n\times (n+1)$-matrix with an additional last row $(1,\ldots ,1)$. Furthermore let $\tilde{k}=(k,\ldots ,k,1)$ and $\tilde{l}=(l,\ldots ,l,1)$ with $k,l\in \mathbb{R }$. Then $G$ has a completely mixed Nash equilibrium $(P,Q)=(P^*,Q^*)$ when it is a nonnegative solution to the two systems of linear equations $P\tilde{B}=\tilde{k}$ and $\tilde{A}Q=\tilde{l}$. The existence of immunization strategies can be expressed analogously: no column (row) of matrix $A$ ($B$) is dominated by another column (row) or a convex combination of columns (rows). Let $\tilde{u}=(u,\ldots ,u,1)$ and $\tilde{v}=(v,\ldots ,v,1)$ with $u,v\in \mathbb{R }$. Player 1 has an immunization strategy $P=\bar{P}$ in $G$ when the vector is a nonnegative solution to the system $P\tilde{A}=\tilde{u}$, and player 2 has an immunization strategy $Q=\bar{Q}$ in $G$ when the vector is a nonnegative solution to the system $\tilde{B}Q=\tilde{v}$. Note that for the existence of a completely mixed Nash equilibrium both solutions $P^*$ and $Q^*$ have to exist, whereas the immunization strategy is defined for a single player. To a square game $G$ we define associated zero-sum games $G^1$ and $G^2$. $G^1$ is the game with payoff matrices $(A,-A),\,G^2$ the game with payoff matrices $(-B,B)$. We first prove the following lemma.

Lemma 2

Let $G$ be a square two-person normal form game, $G^1$ and $G^2$ its associated zero-sum games. If $G$ has a completely mixed Nash equilibrium $(P^*,Q^*)$ and player 1 has an immunization strategy $\bar{P}$, then $G^1$ has a Nash equilibrium $(P_1^*,Q_1^*)$ where $P_1^*=\bar{P}$ and $Q_1^*=Q^*$.

Proof

If $(P^*,Q^*)$ is a completely mixed Nash equilibrium of $G$, then $Q^*$ solves the system $\tilde{A}Q=\tilde{l}$ for some $l\in \mathbb{R }$. Furthermore, if $\bar{P}$ is an immunization strategy of player 1, then $\bar{P}$ solves the system $P\tilde{A}=\tilde{u}$ for some $u\in \mathbb{R }$. Then $\bar{P}$ also solves $P(\tilde{-A})=\tilde{-u}$ and is therefore a Nash equilibrium strategy for player 1 of the game $G^1$. $\square$

Now, we can prove Theorem 5.

Proof of Theorem 5

We prove the result only for player 1. $\bar{P}$ is an immunization strategy of player 1 in the game $G$ if and only if it is a nonnegative solution to the system $P\tilde{A}=\tilde{u}$. Now by assumption there exists a completely mixed Nash equilibrium $(P^*,Q^*)$ of $G$, thus with Lemma 2 there exists a completely mixed Nash equilibrium $(P_1^*,Q_1^*)$ in $G^1$. $P_1^*$ is therefore a nonnegative solution to the system $P(\tilde{-A})=\tilde{k}$ for some $k\in \mathbb{R }$. Then $P_1^*$ also solves the system $P\tilde{A}=(\tilde{-k})$ and we see that $P_1^*$ must be the immunization strategy $\bar{P}$ of player 1.

Now, since $G^1$ is a zero-sum game, $\bar{P}$ is by the classic Minimax Theorem for zero-sum games also the maximin strategy in $G^1$, that means

$$\begin{aligned} \bar{P}\in \arg \max _P\min _Q PAQ. \end{aligned}$$

This condition only depends on the payoff matrix $A$ of player 1, thus $\bar{P}$(=$P_1^*)$ is the maximin strategy in any game where player 1 has the payoff matrix $A$, in particular in the game $G$.

To show the analog statement for player 2 we use the associated zero-sum game $G^2$. $\square$

Lemma 1 in Pruzhansky (2011) proves a result very close to our Theorem 5, but by different means, making use of geometric properties of the various types of strategies. Our approach has an interesting interpretation: a player who is immunizing himself against strategic ambiguity blindfolds himself and drops the strategic element of the game. Instead of playing the original game, he plays the associated zero-sum game which results from his own payoff matrix. Thus, he plays an optimal strategy $\bar{P}$, the immunization strategy, against an imaginary self that receives the negative of his own payoff.

Proof of Theorem 6

Let $\left( \{\bar{P}\},\varDelta S_2\right)$ be an equilibrium with unilateral full ambiguity. Then $\bar{P}$ is a best reply to $\varDelta S_2$, or in other words, $\bar{P}$ is a maximin strategy.

By the indifference principle (Theorem 3), player 2 is indifferent between all $Q \in \varDelta S_2$ when player 1 plays $\bar{P}$. It follows that $(\bar{P}, Q^*)$ is a Nash equilibrium. As we have assumed uniqueness of Nash equilibrium, $\bar{P}=P^*$, and $P^*$ is therefore maximin.

Similarly, if $P^*$ is maximin, then the singleton $\{P^*\}$ is a best reply in the Ellsberg game to $\varDelta S_2$. Also, player 2 is indifferent against $P^*$, so $\varDelta S_2$ is a best reply to $P^*$ for player 2.

Now suppose that $P^*$ is also an immunization strategy. Let $v^*=u_1(P^*,Q^*)$ be the Nash equilibrium payoff, and $\bar{v}=\min _{Q\in \varDelta S_2} u_1 (P^*,Q)$ be the maximin payoff. As $P^*$ immunizes player 1, we have $v^*=\bar{v}$. As $P^*$ is part of a completely mixed Nash equilibrium, any $\mathcal{Q }\subset \varDelta S_2$ is a best reply in the Ellsberg game for player 2. $\square$

Theorem 7 follows from the following three lemmata.

Lemma 3

Let $P$ be a probability distribution on the set $S=\{s_{1},\ldots ,s_{m}\}$. Let $u(P)$ be a continuous function that maps the $m$-dimensional vector $P$ into the real numbers. Then

$$\begin{aligned} \max _{\mathcal{P }\subseteq \varDelta S}\min _{P\in \mathcal{P }}u(P)&= \max _{P\in \varDelta S}u(P)\,,\end{aligned}$$

(6)

$$\begin{aligned} \text{ and } \quad \min _{\mathcal{P }\subseteq \varDelta S}\min _{P\in \mathcal{P }}u(P)&= \min _{P\in \varDelta S}u(P). \end{aligned}$$

(7)

Proof

We start with Eq. (6). The left hand side is clearly less or equal than the right hand side. But we can obtain equality if we take a maximizer $P^* \in \varDelta S$ for the right hand side (which exists by continuity of $u$ and compactness of $\varDelta S$), and use the singleton $\{P^*\}$ on the left hand side. The same argument proves (6). $\square$

Lemma 4

Let $G$ be a zero-sum game with two players and value $v\in \mathbb{R }$. Then

$$\begin{aligned}&\quad \quad \min _{\mathcal{Q }\subseteq \varDelta S_2}\max _{\mathcal{P }\subseteq \varDelta S_1}U_1(\mathcal{P },\mathcal{Q })\overset{(1)}{=}\max _{\mathcal{P }\subseteq \varDelta S_1}\min _{\mathcal{Q }\subseteq \varDelta S_2}U_1(\mathcal{P },\mathcal{Q })\overset{(2)}{=}u_1(P^*,Q^*)=v\,, \end{aligned}$$

(8)

$$\begin{aligned}&\text{ and } \quad \min _{\mathcal{P }\subseteq \varDelta S_1}\max _{\mathcal{Q }\subseteq \varDelta S_2}U_2(\mathcal{P },\mathcal{Q })=\max _{\mathcal{Q }\subseteq \varDelta S_2}\min _{\mathcal{P }\subseteq \varDelta S_1}U_2(\mathcal{P },\mathcal{Q })=-u_1(P^*,Q^*)=-v.\nonumber \\ \end{aligned}$$

(9)

Proof

We start by showing equality (2) of Eq. (8), followed by equality (1). The proof of (9) is analog. For all equalities we use the fact that for all linear functions $f(x,y),\,\min _x\min _y f(x,y)=\min _y\min _x f(x,y)$, and Lemma 3.

$$\begin{aligned}&\max _{\mathcal{P }\subseteq \varDelta S_1}\min _{\mathcal{Q }\subseteq \varDelta S_2}U_1(\mathcal{P },\mathcal{Q })=\max _{\mathcal{P }\subseteq \varDelta S_1}\min _{\mathcal{Q }\subseteq \varDelta S_2}\min _{P\in \mathcal{P }}\min _{Q\in \mathcal{Q }}u_1(P,Q) \\ =&\max _{\mathcal{P }\subseteq \varDelta S_1}\min _{\mathcal{Q }\subseteq \varDelta S_2}\min _{Q\in \mathcal{Q }}\min _{P\in \mathcal{P }}u_1(P,Q)\overset{\tiny \text{ Lemma } \text{3 }}{=}\max _{\mathcal{P }\subseteq \varDelta S_1}\min _{Q\in \varDelta S_2}\min _{P\in \mathcal{P }}u_1(P,Q) \\ =&\max _{\mathcal{P }\subseteq \varDelta S_1}\min _{P\in \mathcal{P }}\min _{Q\in \varDelta S_2}u_1(P,Q)\overset{\tiny \text{ Lemma } \text{3 }}{=}\max _{P\in \varDelta S_1}\min _{Q\in \varDelta S_2}u_1(P,Q)=u_1(P^*,Q^*)=v. \end{aligned}$$

To proof equality (1) we need the Minimax Theorem 1 (Theorem 4). We derive

$$\begin{aligned}&\min _{\mathcal{Q }\subseteq \varDelta S_2}\max _{\mathcal{P }\subseteq \varDelta S_1}U_1(\mathcal{P },\mathcal{Q })=\min _{\mathcal{Q }\subseteq \varDelta S_2}\max _{\mathcal{P }\subseteq \varDelta S_1}\min _{P\in \mathcal{P }}\min _{Q\in \mathcal{Q }}u_1(P,Q) \\ =&\min _{\mathcal{Q }\subseteq \varDelta S_2}\max _{P\in \varDelta S_1}\min _{Q\in \mathcal{Q }}u_1(P,Q)\overset{\tiny \text{ Thm. } \text{4 }}{=}\min _{\mathcal{Q }\subseteq \varDelta S_2}\min _{Q\in \mathcal{Q }}\max _{P\in \varDelta S_1}u_1(P,Q) \\ =&\min _{Q\in \varDelta S_2}\max _{P\in \varDelta S_1}u_1(P,Q)=u_1(P^*,Q^*)=\max _{\mathcal{P }\subseteq \varDelta S_1}\min _{\mathcal{Q }\subseteq \varDelta S_2}U_1(\mathcal{P },\mathcal{Q }). \end{aligned}$$

$\square$

Lemma 5

Let $(\mathcal{P }^*,\mathcal{Q }^*)$ be an Ellsberg equilibrium of the two-person zero-sum game $G$. Then

$$\begin{aligned}&\quad \quad U_1(\mathcal{P }^*,\mathcal{Q }^*)=u_1(P^*,Q^*)=v\,, \end{aligned}$$

(10)

$$\begin{aligned}&\text{ and } \,U_2(\mathcal{P }^*,\mathcal{Q }^*)=-u_1(P^*,Q^*)=-v. \end{aligned}$$

(11)

Proof

We start by showing Eq. (10). Equation (11) follows analogously.

$$\begin{aligned}&\,U_2(\mathcal{P }^*,\mathcal{Q }^*)\ge U_2(\mathcal{P }^*,\mathcal{Q }) \text{ for } \text{ all } \mathcal{Q }\subseteq \varDelta S_2 \\ \Rightarrow&\,U_2(\mathcal{P }^*,\mathcal{Q }^*)=\max _{\mathcal{Q }\subseteq \varDelta S_2} U_2(\mathcal{P }^*,\mathcal{Q }) \\ \Rightarrow&\min _{P\in \mathcal{P }^*}\min _{Q\in \mathcal{Q }^*}u_2(P,Q)=\max _{\mathcal{Q }\subseteq \varDelta S_2}\min _{P\in \mathcal{P }^*}\min _{Q\in \mathcal{Q }^*}u_2(P,Q) \\ \Rightarrow&\min _{P\in \mathcal{P }^*}\min _{Q\in \mathcal{Q }^*}u_2(P,Q)=\max _{Q\in \varDelta S_2}\min _{P\in \mathcal{P }^*}u_2(P,Q) \\ \Rightarrow&\max _{P\in \mathcal{P }^*}\max _{Q\in \mathcal{Q }^*}u_1(P,Q)=\min _{Q\in \varDelta S_2}\max _{P\in \mathcal{P }^*}u_1(P,Q) \\ \Rightarrow&\max _{P\in \mathcal{P }^*}\max _{Q\in \mathcal{Q }^*}u_1(P,Q)\le \min _{Q\in \varDelta S_2}\max _{P\in \varDelta S_1}u_1(P,Q). \end{aligned}$$

Furthermore,

$$\begin{aligned}&\,U_1(\mathcal{P }^*,\mathcal{Q }^*)\ge U_1(\mathcal{P },\mathcal{Q }^*) \text{ for } \text{ all } \mathcal{P }\subseteq \varDelta S_1 \\ \Rightarrow&\,U_1(\mathcal{P }^*,\mathcal{Q }^*)=\max _{\mathcal{P }\subseteq \varDelta S_1} U_1(\mathcal{P },\mathcal{Q }^*) \\ \Rightarrow&\min _{P\in \mathcal{P }^*}\min _{Q\in \mathcal{Q }^*}u_1(P,Q)=\max _{\mathcal{P }\subseteq \varDelta S_1}\min _{P\in \mathcal{P }}\min _{Q\in \mathcal{Q }^*}u_1(P,Q) \\ \Rightarrow&\min _{P\in \mathcal{P }^*}\min _{Q\in \mathcal{Q }^*}u_1(P,Q)=\max _{P\in \varDelta S_1}\min _{Q\in \mathcal{Q }^*}u_1(P,Q) \\ \Rightarrow&\min _{P\in \mathcal{P }^*}\min _{Q\in \mathcal{Q }^*}u_1(P,Q)\ge \max _{P\in \varDelta S_1}\min _{Q\in \varDelta S_2}u_1(P,Q). \end{aligned}$$

From the above relations and the classical Minimax Theorem follows that

$$\begin{aligned}&\max _{P\in \mathcal{P }^*}\max _{Q\in \mathcal{Q }^*}u_1(P,Q)\le \min _{Q\in \varDelta S_2}\max _{P\in \varDelta S_1}u_1(P,Q)=\max _{P\in \varDelta S_1}\min _{Q\in \varDelta S_2}u_1(P,Q)\\&\quad \le \min _{P\in \mathcal{P }^*}\min _{Q\in \mathcal{Q }^*}u_1(P,Q) \end{aligned}$$

and we finally have $U_1(\mathcal{P }^*,\mathcal{Q }^*)=u_1(P^*,Q^*)=v$. $\square$

Lemma 4 and 5 prove Theorem 7.

Proof of Proposition 4

The Nash equilibrium strategies follow from the usual analysis. To calculate the Ellsberg equilibria of the general conflict game (Fig. 7), we first derive the utility functions of player 1 and player 2. Due to the assumption that $a,d>b,c \text{ and } e,h < f,g$, the denominator $a-b-c+d$ is positive, and the denominator $e-f-g+h$ is negative. This reflects the competitiveness of the game in the payoff functions; player 1 uses $Q_0$ as a minimizer when $P>\frac{d-c}{a-b-c+d}$, and on the contrary, player 2 uses $P_0$ as a minimizer when $Q<\frac{h-f}{e-f-g+h}$.

$$\begin{aligned}&U_1(P,[Q_0,Q_1])=\!\!\min _{Q_0\le Q \le Q_1}aPQ+bP(1-Q)+c(1-P)Q+d(1-P)(1\!-\!Q)\\&\qquad =\min _{Q_0\le Q \le Q_1}Q((a-b-c+d)P+c-d)+(b-d)P+d\\&\qquad =\left\{ \begin{array}{cl} Q_0((a-b-c+d)P+c-d)+(b-d)P+d &{} \text{ if } P>\frac{d-c}{a-b-c+d}\,,\\ \frac{(b-d)(c-d)}{a-b-c+d}+d &{} \text{ if } P=\frac{d-c}{a-b-c+d}\,,\\ Q_1((a-b-c+d)P+c-d)+(b-d)P+d &{}\text{ if } P<\frac{d-c}{a-b-c+d}\,, \end{array}\right. \\&U_2([P_0,P_1]\,,Q)=\min _{P_0\le P \le P_1}ePQ+g(1-P)Q+fP(1-Q)+h(1-P)(1\!-\!Q)\\&\qquad =\min _{P_0\le P \le P_1}P((e-f-g+h)Q+f-h)+(g-h)Q+h \\&\qquad =\left\{ \begin{array}{cl} P_0((e-f-g+h)Q+f-h)+(g-h)Q+h &{} \text{ if } Q<\frac{h-f}{e-f-g+h}\,,\\ \frac{(g-h)(h-f)}{e-f-g+h}+h &{} \text{ if } Q=\frac{h-f}{e-f-g+h}\,,\\ P_1((e-f-g+h)Q+f-h)+(g-h)Q+h&{}\text{ if } Q>\frac{h-f}{e-f-g+h}. \end{array}\right. \end{aligned}$$

We see that

$$\begin{aligned} M_1= \frac{d-c}{a-b-c+d},\quad \text{ respectively }\quad M_2=\frac{h-f}{e-f-g+h} \end{aligned}$$

are indeed the immunization strategies of player 1 and 2 respectively. At the boundaries the utility functions become

$$\begin{aligned}&U_1(0\,,[Q_0,Q_1])=(c-d)Q_1+d,\quad U_2([P_0,P_1]\,,0)=(f-h)P_0+h\,, \\&U_1(1\,,[Q_0,Q_1])=(a-b)Q_0+b,\quad U_2([P_0,P_1]\,,1)=(e-g)P_1+g. \end{aligned}$$

The payoff function of player 1 is constant when $Q_0=Q_1=Q^*$, and the payoff function of player 2 is constant when $P_0=P_1=P^*$. The best response correspondences for both players are listed below. We see immediately that, of course, there cannot be an Ellsberg equilibrium in pure strategies: when player 2 plays $L$, player 1 best responds $T$, and when player 1 plays $T$, player 2 best responds $R$. The intersections of the best response correspondences are discussed in detail below.

$$\begin{aligned}&Q_0>Q^*:&\quad B_1([Q_0,Q_1])=1\quad \quad (1)\\&Q_0=Q^*<Q_1:&\quad B_1([Q_0,Q_1])=\{[P_0,P_1]\subseteq [M_1,1]\}\quad \quad (2)\\&Q_0<Q^*<Q_1:&\quad B_1([Q_0,Q_1])=M_1\quad \quad (3)\\&Q_0<Q^*=Q_1:&\quad B_1([Q_0,Q_1])=\{[P_0,P_1]\subseteq [0,M_1]\}\quad \quad (4)\\&Q_1<Q^*:&\quad B_1([Q_0,Q_1])=0\quad \quad (5)\\&Q_0=Q^*=Q_1:&\quad B_1([Q_0,Q_1])=\{[P_0,P_1]\subseteq [0,1]\}\quad \quad (6) \\&P_0>P^*:&\quad B_2([P_0,P_1])=0\\&P_0=P^*<P_1:&\quad B_2([P_0,P_1])=\{[Q_0,Q_1]\subseteq [0,M_2]\}\\&P_0<P^*<P_1:&\quad B_2([P_0,P_1])=M_2\\&P_0<P^*=P_1:&\quad B_2([P_0,P_1])=\{[Q_0,Q_1]\subseteq [M_2,1]\}\\&P_1<P^*:&\quad B_2([P_0,P_1])=1\\&P_0=P^*=P_1:&\quad B_2([P_0,P_1])=\{[Q_0,Q_1]\subseteq [0,1]\} \end{aligned}$$

To find the Ellsberg equilibria, we look at the different cases in turn.

• (1) $Q_0>Q^*:$ player 1 responds $P_0=P_1=1$ and player 2 chooses $Q_0=Q_1=0$, thus this is not an Ellsberg equilibrium.

• (2) $\left[ \frac{d-b}{a-b-c+d},Q_1\right] \Rightarrow [P_0,P_1]\subseteq \left[ \frac{d-c}{a-b-c+d},1\right] ,$ then, depending on the size of $M_1$ in relation to $P^*$, player 1 has the following choices:

  • $\frac{d-c}{a-b-c+d}\le \frac{h-g}{e-f-g+h}:$ player 1 can play either

    • $\left[ \frac{h-g}{e-f-g+h},P_1\right] ,$ then the best response of player 2 is $[Q_0,Q_1]\subseteq \left[ 0,\frac{h-f}{e-f-g+h}\right]$. Now, depending on the size of $M_2$ in relation to $Q^*$, player 2 has the following choices:

      • $\frac{h-f}{e-f-g+h}\ge \frac{d-b}{a-b-c+d}:\,\left[ \frac{d-b}{a-b-c+d},Q_1\right]$ with $Q_1\le \frac{h-f}{e-f-g+h}$, we thus have here the Ellsberg equilibrium

        $$\begin{aligned}&\left( \left[ \frac{h-g}{e-f-g+h}, P_1\right] ,\left[ \frac{d-b}{a-b-c+d},Q_1 \right] \right) ,\\&\quad \text{ where } Q_1\le \frac{h-f}{e-f-g+h}\,, \\&\text{ that } \text{ is } ([P^*,P_1]\,,[Q^*,Q_1]), \text{ where } Q_1\le M_2. \end{aligned}$$

      • $\frac{h-f}{e-f-g+h}\le \frac{d-b}{a-b-c+d}$: for all $[Q_0,Q_1]$ the best response is $P_0=P_1=0$, thus no Ellsberg equilibrium arises.

    • or $\left[ P_0,\frac{h-g}{e-f-g+h}\right]$ with $\frac{d-c}{a-b-c+d}\le P_0\le \frac{h-g}{e-f-g+h}$, then the best response of player 2 is $[Q_0,Q_1]\subseteq \left[ \frac{h-f}{e-f-g+h},1\right]$. Now, depending on the size of $M_2$ in relation to $Q^*$, player 2 has the following choices:

      • $\frac{h-f}{e-f-g+h}\le \frac{d-b}{a-b-c+d}:\,\left[ \frac{d-b}{a-b-c+d},Q_1\right]$, then the best response of player 1 is $[P_0,P_1]\subseteq \left[ \frac{d-c}{a-b-c+d},1\right]$. We thus have here the Ellsberg equilibrium

        $$\begin{aligned}&\left( \left[ P_0,\frac{h-g}{e-f-g+h}\right] ,\left[ \frac{d-b}{a-b-c+d},Q_1 \right] \right) , \\&\quad \text{ where } \frac{d-c}{a-b-c+d}\le P_0\,, \\&\text{ that } \text{ is } ([P_0,P^*]\,,[Q^*,Q_1]), \text{ where } M_1\le P_0. \end{aligned}$$

      • $\frac{h-f}{e-f-g+h}\ge \frac{d-b}{a-b-c+d}:$ for all $[Q_0,Q_1]$ the best response is $P_0=P_1=1$, thus no Ellsberg equilibrium arises.

  • $\frac{d-c}{a-b-c+d}\ge \frac{h-g}{e-f-g+h}$: to any $[P_0,P_1]$ the best response is $Q_0=Q_1=0$, thus no Ellsberg equilibrium arises.

• (3) $Q_0<Q^*<Q_1:$ player 1 responds with $P_0=P_1=M_1$. Only when $M_1=P^*$, player 1 sticks to his strategy and we get the equilibrium $(P^*,[Q_0,Q_1]), \text{ where } Q_0<Q^*<Q_1.$

• (4) $\left[ Q_0,\frac{d-b}{a-b-c+d}\right] \Rightarrow [P_0,P_1]\subseteq \left[ 0,\frac{d-c}{a-b-c+d}\right] ,$ then, depending on the size of $M_1$ in relation to $P^*$, player 1 has the following choices:

  • $\frac{d-c}{a-b-c+d}\ge \frac{h-g}{e-f-g+h}:$ player 1 can play either

    • $\left[ P_0,\frac{h-g}{e-f-g+h}\right] ,$ then the best response of player 2 is $[Q_0,Q_1]\subseteq \left[ \frac{h-f}{e-f-g+h},1\right]$. Now, depending on the size of $M_2$ in relation to $Q^*$, player 2 has the following choices:

      • $\frac{h-f}{e-f-g+h}\le \frac{d-b}{a-b-c+d}:\,\left[ Q_0,\frac{d-b}{a-b-c+d}\right]$ with $\frac{h-f}{e-f-g+h}\le Q_0$, we thus have here the Ellsberg equilibrium

        $$\begin{aligned}&\left( \left[ P_0, \frac{h-g}{e-f-g+h}\right] ,\left[ Q_0,\frac{d-b}{a-b-c+d}\right] \right) , \\&\quad \text{ where } \frac{h-f}{e-f-g+h}\le Q_0\,, \\&\text{ that } \text{ is } ([P_0,P^*]\,,[Q_0,Q^*]), \text{ where } M_2\le Q_0. \end{aligned}$$

      • $\frac{h-f}{e-f-g+h}\ge \frac{d-b}{a-b-c+d}:$ for all $[Q_0,Q_1]$ the best response is $P_0=P_1=1$, thus no Ellsberg equilibrium arises.

    • or $\left[ \frac{h-g}{e-f-g+h},P_1\right]$ with $\frac{h-g}{e-f-g+h}\le P_1\le \frac{d-c}{a-b-c+d}$, then the best response of player 2 is $[Q_0,Q_1]\subseteq \left[ 0,\frac{h-f}{e-f-g+h}\right]$. Now, depending on the size of $M_2$ in relation to $Q^*$, player 2 has the following choices:

      • $\frac{d-b}{a-b-c+d}\le \frac{h-f}{e-f-g+h}:\,\left[ Q_0,\frac{d-b}{a-b-c+d}\right]$, then the best response of player 1 is $[P_0,P_1]\subseteq \left[ 0,\frac{d-c}{a-b-c+d}\right]$. We thus have here the Ellsberg equilibrium

        $$\begin{aligned}&\left( \left[ \frac{h-g}{e-f-g+h},P_1\right] ,\left[ Q_0,\frac{d-b}{a-b-c+d}\right] \right) , \\&\quad \text{ where } P_1\le \frac{d-c}{a-b-c+d}\,, \\&\text{ that } \text{ is } ([P^*,P_1]\,,[Q_0,Q^*]), \text{ where } P_1\le M_1. \end{aligned}$$

      • $\frac{d-b}{a-b-c+d}\ge \frac{h-f}{e-f-g+h}:$ for all $[Q_0,Q_1]$ the best response is $P_0=P_1=0$, thus no Ellsberg equilibrium arises.

  • $\frac{d-c}{a-b-c+d}\le \frac{h-g}{e-f-g+h}$: to any $[P_0,P_1]$ the best response is $Q_0=Q_1=0$, thus no Ellsberg equilibrium arises.

• (5) $Q_1<Q^*:$ player 1 responds $P_0=P_1=0$ and player 2 chooses $Q_0=Q_1=1$ thereafter, thus this is not an Ellsberg equilibrium.

• (6) $Q_0=Q^*=Q_1:$ player 1 responds with $[P_0,P_1]\subseteq [0,1]$. Only when $M_2=Q^*$, player 2 sticks to his strategy and we get the equilibrium $([P_0,P_1]\,,Q^*), \text{ where } P_0<P^*<P_1.$ $\square$

Proof of Proposition 6

We start by calculating the minimal expected utility functions of player 1 and 2. Player 1 minimizes over $Q_1$ and $Q_2$. $Q_3$ is expressed as $1-Q_1-Q_2$. The lower and upper bounds of $Q_1$ and $Q_2$ are $w_1,z_1$ and $w_2,z_2$, respectively.

$$\begin{aligned}&U_1(P_1,P_2,Q_1,Q_2)\\&\quad =\min _{Q_1,Q_2}2P_1Q_2-P_1(1-Q_1-Q_2)-P_2Q_1\\&\quad \quad \,+P_2(1-Q_1-Q_2)+(1-P_1-P_2)Q_1-(1-P_1-P_2)Q_2\\&\quad =\min _{Q_2}\left\{ \begin{array}{cl} w_1(1-3P_2)+4P_1Q_2-P_1+P_2-Q_2 &{}\text{ if } P_2<1/3\,,\\ 4P_1Q_2-P_1-Q_2+1/3 &{} \text{ if } P_2=1/3\,,\\ z_1(1-3P_2)+4P_1Q_2-P_1+P_2-Q_2 &{}\text{ if } P_2>1/3 \end{array}\right. \\&\quad =\left\{ \begin{array}{r} \left. \begin{array}{cl} w_2 (4P_1-1)-3P_2w_1 +w_1 -P_1+P_2 &{} \text{ if } P_1>1/4\,,\\ w_1 -3P_2w_1 -1/4+P_2&{} \text{ if } P_1=1/4\,,\\ z_2 (4P_1-1)-3P_2w_1 +w_1 -P_1+P_2 &{}\text{ if } P_1<1/4\,, \end{array}\right\} P_2<1/3\,,\\ \\ \left. \begin{array}{cl} w_2 (4P_1-1)+1/3-P_1 &{} \text{ if } P_1>1/4\,,\\ 1/12&{} \text{ if } P_1=1/4\,,\\ z_2 (4P_1-1)+1/3-P_1 &{}\text{ if } P_1<1/4\,, \end{array}\right\} P_2=1/3\,,\\ \\ \left. \begin{array}{cl} w_2 (4P_1-1)-3P_2z_1 +z_1 -P_1+P_2 &{} \text{ if } P_1>1/4\,,\\ z_1 -3P_2z_1 -1/4+P_2&{} \text{ if } P_1=1/4\,,\\ z_2 (4P_1-1)-3P_2z_1 +z_1 -P_1+P_2 &{}\text{ if } P_1<1/4\,, \end{array}\right\} P_2>1/3.\end{array}\right. \end{aligned}$$

Player 2 minimizes over $P_1$ and $P_2,\,P_3$ is expressed as $1-P_1-P_2$. The lower and upper bounds of $P_1$ and $P_2$ are $x_1,y_1$ and $x_2,y_2$, respectively.

$$\begin{aligned}&U_2(P_1,P_2,Q_1,Q_2)\\&\quad =\min _{P_1,P_2}-P_1Q_2+P_1(1-Q_1-Q_2)+P_2Q_1\\&\quad \quad \,-P_2(1-Q_1-Q_2)-(1- P_1-P_2)Q_1+(1-P_1-P_2)Q_2\\&\quad =\min _{P_2}\left\{ \begin{array}{cl} x_1(1-3Q_2)+3P_2Q_1-P_2-Q_1+Q_2 &{}\text{ if } Q_2<1/3\,,\\ 3P_2Q_1-P_2-Q_1+1/3 &{} \text{ if } Q_2=1/3\,,\\ y_1(1-3Q_2)+3P_2Q_1-P_2-Q_1+Q_2 &{}\text{ if } Q_2>1/3 \end{array}\right. \\&\quad =\left\{ \begin{array}{r} \left. \begin{array}{cl} x_2 (3Q_1-1)-3Q_2x_1 +x_1 -Q_1+Q_2 &{} \text{ if } Q_1>1/3\,,\\ x_1 -3Q_2x_1 -1/3+Q_2&{} \text{ if } Q_1=1/3\,,\\ y_2 (3Q_1-1)-3Q_2x_1 +x_1 -Q_1+Q_2 &{}\text{ if } Q_1<1/3\,, \end{array}\right\} Q_2<1/3\,,\\ \\ \left. \begin{array}{cl} x_2 (3Q_1-1)+1/3-Q_1 &{} \text{ if } Q_1>1/3\,,\\ 0 &{} \text{ if } Q_1=1/3\,,\\ y_2 (3Q_1-1)+1/3-Q_1 &{}\text{ if } Q_1<1/3\,, \end{array}\right\} Q_2=1/3\,,\\ \\ \left. \begin{array}{cl} x_2 (3Q_1-1)-3Q_2y_1 +y_1 -Q_1+Q_2 &{} \text{ if } Q_1>1/3\,,\\ y_1 -3Q_2y_1 -1/3+Q_2&{} \text{ if } Q_1=1/3\,,\\ y_2 (3Q_1-1)-3Q_2y_1 +y_1 -Q_1+Q_2 &{}\text{ if } Q_1<1/3\,, \end{array}\right\} Q_2>1/3. \end{array}\right. \end{aligned}$$

Now we proceed as follows to derive the Ellsberg equilibria. Recall the Nash equilibria and immunization strategies of the game: $(P^*,Q^*)=((1/3,1/3,1/3),(1/3,1/4,5/ 12))$ and $(M_1,M_2)=((1/4,1/3,5/12),(1/3,1/3,1/3))$. Player 1 has to use his Nash equilibrium strategies at the boundary of his Ellsberg equilibrium strategy, except for the component $P_2$, where the Nash equilibrium probability is the same as the immunization strategy. Furthermore, the set of probability distributions in the Ellsberg equilibrium may not extend across the immunization strategy.

Hence, for the first component for player 1, we consider either $\{1/4\le x_1 \le P_1\le 1/3\},$ or $\{1/3\le P_1\}$. In Ellsberg equilibrium, player 2 uses $1/3$ as his worst case measure, thus in the first case we only consider those elements of $U_2$ which use $y_1$ as worst case measure, in the second case only those which use $x_1$. $y_1$ is only used when $Q_2\ge 1/3$. Since the range of $Q_2$ must contain $1/4$ at the boundary, which is impossible when $Q_2\ge 1/3,\,\{1/4\le x_1 \le P_1\le 1/3\}$ cannot be an Ellsberg equilibrium strategy. We proceed with the second case: $x_1$ is only used for $Q_2\le 1/3$, this is compatible with the Nash equilibrium strategy and yields $\{1/4\le Q_2 \le z_2 \le 1/3\}$. Finally we check if $\{1/3\le P_1\}$ is a best response to $\{1/4\le Q_2 \le z_2 \le 1/3\}$. Player 1 must use $w_2$ in his utility evaluation $U_1$, this is true whenever $P_1\ge 1/4$ which is compatible with the strategy $\{1/3\le P_1\}$. Thence we have found a best response pair.

In the second component, player 1 can play $\{P_2=1/3\}$, which is his Nash equilibrium probability and therefore makes player 2 indifferent between all $\{w_1\le Q_1\le z_1\}$, and at the same time his best response to player 2 playing any probability $\{w_1\le Q_1\le z_1\}$ with $0\le w_1\le 1/3\le z_1\le 1$. Or, conversely, player 2 can play $\{Q_1=1/3\}$ and player 1 $\{x_2\le P_2\le y_2\}$ with $0\le x_2\le 1/3\le y_2\le 1$. The latter case collapses to Nash equilibrium. The restrictions on $w_1,z_1$, which are $0\le w_1\le 1/3=z_1$, follow from the third component and the fact that each element of the Ellsberg equilibrium strategy must be a probability distribution.

We cannot determine the range for $P_3$ and $Q_3$ from the two utility functions, since $P_3$ and $Q_3$ are only implicitly given. Thus, we derive the utility functions again, now using $(P_1, 1-P_1-P_3,P_3)$ and $(1-Q_2-Q_3,Q_2,Q_3)$ as probabilities. Since we already know that only the boundaries $x_1$ and $w_2$ are relevant for the equilibrium, we get

$$\begin{aligned}&U_1(P_1,P_3,Q_2,Q_3)\\&\quad =\min _{Q_2,Q_3}2P_1Q_2-P_1Q_3-(1-P_1-P_3)(1-Q_2-Q_3)\\&\quad \quad \,+(1-P_1-P_3)Q_3+P_3(1-Q_2-Q_3)-P_3Q_2\\&\quad =\left\{ \begin{array}{cl} w_3(2-3P_1-3P_3)+w_2(P_1-3P_3+1)+2P_3+P_1-1 &{}\text{ if } P_3<2/3\!-\!P_1\,,\\ w_2(5/3-4P_3)+P_3-1/3 &{} \text{ if } P_3=2/3\!-\!P_1\,,\\ z_3(2-3P_1-3P_3)+w_2(P_1-3P_3+1)+2P_3+P_1-1 &{}\text{ if } P_3>2/3\!-\!P_1\,, \end{array}\right. \end{aligned}$$

when $P_1>3P_3-1$. On the other hand,

$$\begin{aligned}&U_2(P_1,P_3,Q_2,Q_3)\\&\quad =\min _{P_1,P_3}-P_1Q_2+P_1Q_3+(1-P_1-P_3)(1-Q_2-Q_3)\\&\quad \quad \,-(1-P_1-P_3)Q_3-P_3(1-Q_2-Q_3)+P_3Q_2\\&\quad =\left\{ \begin{array}{cl} x_3(3Q_2+3Q_3-2)+x_1(3Q_3-1)-Q_2-2Q_3+1 &{}\text{ if } Q_2>2/3-Q_3\,,\\ x_1(3Q_3-1)-Q_3+1/3 &{} \text{ if } Q_2=2/3-Q_3\,,\\ y_3(3Q_2+3Q_3-2)+x_1(3Q_3-1)-Q_2-2Q_3+1 &{}\text{ if } Q_2<2/3-Q_3\,, \end{array}\right. \end{aligned}$$

when $Q_3>1/3$.

Now we use the following reasoning: we see from $U_2$ that $Q_3$ must be greater than (or in fact equal to, this is suppressed in the shortened statement of $U_2$) $1/3$. Since $5/12$ is the Nash equilibrium probability of $Q_3$, we get the two following possible probability sets for $Q_3$:

$$\begin{aligned} \text{ either } \, \{1/3\le w_3\le Q_3\le 5/12\}\,, \, \text{ or } \, \{5/12\le Q_3\le z_3\}. \end{aligned}$$

In the first case, player 1 must use $z_3$ in his utility evaluation, according to $U_1$ this is the case only when $P_3\ge 2/3-P_1$. We know that $P_1\ge 1/3$, therefore $P_3\ge 1/3$. This leads to the set of probabilities $\{1/3\le P_3\le y_3\le 5/12\}$ for $P_3$, since $1/3$ as the Nash equilibrium probability must be part of the set. Here $x_3$ is used by player 2, hence from $U_2$ we see this is only the case if $Q_2\ge 2/3-Q_3$. $Q_3\ge 1/3$ and therefore $Q_2$ must be greater than $1/3$. This is not possible, since we have already found $Q_2$ to be played with the set of probabilities $\{1/4\le Q_2\le z_2\le 1/3\}$.

Consider the second possible set for $Q_2$. Here $w_3$ is used by player 1, and hence $P_2\le 1/3$. This leads to $\{x_3\le P_3\le 1/3\}$ for $P_3$. Player 2 uses $y_3$ and we conclude that $Q_2$ must be less than or equal to $1/3$. This is exactly what we found to be true for $Q_2$ before. Therefore we have found the Ellsberg equilibrium of RSP to be

$$\begin{aligned} ((\{1/3&\le P_1\le y_1\le 2/3\},\{P_2= 1/3\},\{x_3\le P_3\le 1/3\}),\\&(\{w_1\le Q_1\le z_1\},\{1/4\le Q_2 \le z_2\le 1/3\},\{5/12\le Q_3 \le z_3\le 3/4\}))\,,\\&\text{ where } 0\le w_1\le 1/3 =z_1. \square \end{aligned}$$

Proof of Proposition 7

The Nash equilibrium strategies follow from the usual analysis. To calculate the Ellsberg equilibria of the general coordination game (Fig. 7), we first derive the utility functions of player 1 and player 2. Due to the assumption that $a,d\ge b,c$ and $e,h \ge f,g$, with $a-b-c+d\ne 0$ and $e-f-g+h\ne 0$, the denominators $a-b-c+d$ and $e-f-g+h$ are positive. This reflects in the payoff functions that the game is a coordination game; player 1 uses $Q_0$ as a minimizer when $P>\frac{d-c}{a-b-c+d}$, and, the same way, player 2 uses $P_0$ as a minimizer when $Q>\frac{h-f}{e-f-g+h}$.

$$\begin{aligned}&U_1(P,[Q_0,Q_1])\!=\!\min _{Q_0\le Q \le Q_1}aPQ+bP(1-Q)+c(1-P)Q+d(1-P)(1-Q) \\&\quad \quad =\min _{Q_0\le Q \le Q_1}Q((a-b-c+d)P+c-d)+(b-d)P+d \\&\quad \quad =\left\{ \begin{array}{cl} Q_0((a-b-c+d)P+c-d)+(b-d)P+d &{} \text{ if } P>\frac{d-c}{a-b-c+d}\,,\\ \frac{(b-d)(c-d)}{a-b-c+d}+d &{} \text{ if } P=\frac{d-c}{a-b-c+d}\,,\\ Q_1((a-b-c+d)P+c-d)+(b-d)P+d &{}\text{ if } P<\frac{d-c}{a-b-c+d}\,, \end{array}\right. \\&U_2([P_0,P_1]\,,Q)\!=\!\min _{P_0\le P \le P_1}ePQ+g(1-P)Q+fP(1-Q)+h(1-P)(1-Q) \\&\quad \quad =\min _{P_0\le P \le P_1}P((e-f-g+h)Q+f-h)+(g-h)Q+h\\&\quad \quad =\left\{ \begin{array}{cl} P_0((e-f-g+h)Q+f-h)+(g-h)Q+h &{} \text{ if } Q>\frac{h-f}{e-f-g+h}\,,\\ \frac{(g-h)(h-f)}{e-f-g+h}+h &{} \text{ if } Q=\frac{h-f}{e-f-g+h}\,,\\ P_1((e-f-g+h)Q+f-h)+(g-h)Q+h&{}\text{ if } Q<\frac{h-f}{e-f-g+h}. \end{array}\right. \end{aligned}$$

We see that

$$\begin{aligned} M_1= \frac{d-c}{a-b-c+d}\,,\quad \text{ respectively } \quad M_2=\frac{h-f}{e-f-g+h} \end{aligned}$$

are the immunization strategies of player 1 and 2, respectively. At the boundaries the utility functions become

$$\begin{aligned}&U_1(0\,,[Q_0,Q_1])=(c-d)Q_1+d\,,&U_2([P_0,P_1]\,,0)=(f-h)P_1+h\,,\\&U_1(1\,,[Q_0,Q_1])=(a-b)Q_0+b\,,&U_2([P_0,P_1]\,,1)=(e-g)P_0+g. \end{aligned}$$

The payoff function of player 1 is constant for all $P$, when $Q_1=Q_0=Q^*$, and the payoff function of player 2 is constant for all $Q$, when $P_1=P_0=P^*$. We calculate the best response correspondences:

$$\begin{aligned}&Q_0>Q^*:&\quad B_1([Q_0,Q_1])=1\quad \quad (1)\\&Q_0=Q^*<Q_1:&\quad B_1([Q_0,Q_1])=\{[P_0,P_1]\subseteq \left[ M_1,1\right] \}\quad \quad (2)\\&Q_0<Q^*<Q_1:&\quad B_1([Q_0,Q_1])=M_1\quad \quad (3)\\&Q_0<Q^*=Q_1:&\quad B_1([Q_0,Q_1])=\{[P_0,P_1]\subseteq \left[ 0,M_1\right] \}\quad \quad (4)\\&Q_1<Q^*:&\quad B_1([Q_0,Q_1])=0\quad \quad (5)\\&Q_0=Q^*=Q_1:&\quad B_1([Q_0,Q_1])=\{[P_0,P_1]\subseteq [0,1]\}\quad \quad (6)\\&P_0>P^*:&\quad B_2([P_0,P_1])=1\\&P_0=P^*<P_1:&\quad B_2([P_0,P_1])=\{[Q_0,Q_1]\subseteq \left[ M_2,1\right] \}\\&P_0<P^*<P_1:&\quad B_2([P_0,P_1])=M_2\\&P_0<P^*=P_1:&\quad B_2([P_0,P_1])=\{[Q_0,Q_1]\subseteq \left[ 0,M_2\right] \}\\&P_1<P^*:&\quad B_2([P_0,P_1])=0\\&P_0=P^*=P_1:&\quad B_2([P_0,P_1])=\{[Q_0,Q_1]\subseteq [0,1]\} \end{aligned}$$

To find the Ellsberg equilibria, we look at the different cases in turn.

• (1) $Q_0>Q^*:$ player 1 responds $P_0=P_1=1$. If also player 2 chooses $Q_0=Q_1=1$, this is the Ellsberg equilibrium that is identical to the pure Nash equilibrium $(T,L)$.

• (2) $\left[ \frac{d-b}{a-b-c+d},Q_1\right] \Rightarrow [P_0,P_1]\subseteq \left[ \frac{d-c}{a-b-c+d},1\right] ,$ then, depending on the size of $M_1$ in relation to $P^*$, player 1 has the following choices:

  • $\frac{d-c}{a-b-c+d}\le \frac{h-g}{e-f-g+h}:$ player 1 can play either

    • $\left[ \frac{h-g}{e-f-g+h},P_1\right] ,$ then the best response of player 2 is $[Q_0,Q_1]\subseteq \left[ \frac{h-f}{e-f-g+h},1\right]$. Now, depending on the size of $M_2$ in relation to $Q^*$, player 2 has the following choices:

      • $\frac{h-f}{e-f-g+h}\le \frac{d-b}{a-b-c+d}:\,\left[ \frac{d-b}{a-b-c+d},Q_1\right]$, we thus have here the Ellsberg equilibrium

        $$\begin{aligned}&\left( \left[ \frac{h-g}{e-f-g+h}, P_1\right] ,\left[ \frac{d-b}{a-b-c+d},Q_1 \right] \right) , \\&\text{ that } \text{ is } ([P^*,P_1]\,,[Q^*,Q_1]). \end{aligned}$$

      • $\frac{h-f}{e-f-g+h}\ge \frac{d-b}{a-b-c+d}:$ for all $[Q_0,Q_1]$ the best response is $P_0=P_1=1$, thus no Ellsberg equilibrium arises.

    • or $\left[ P_0,\frac{h-g}{e-f-g+h}\right]$ with $\frac{d-c}{a-b-c+d}\le P_0\le \frac{h-g}{e-f-g+h}$, then the best response of player 2 is $[Q_0,Q_1]\subseteq \left[ 0,\frac{h-f}{e-f-g+h}\right]$. Now, depending on the size of $M_2$ in relation to $Q^*$, player 2 has the following choices:

      • $\frac{h-f}{e-f-g+h}\ge \frac{d-b}{a-b-c+d}:\,\left[ \frac{d-b}{a-b-c+d},Q_1\right]$ with $Q_1\le \frac{h-f}{e-f-g+h}$, then the best response of player 1 is $[P_0,P_1]\subseteq \left[ \frac{d-c}{a-b-c+d},1\right]$. We thus have here the Ellsberg equilibrium

        $$\begin{aligned}&\left( \left[ P_0,\frac{h-g}{e-f-g+h}\right] ,\left[ \frac{d-b}{a-b-c+d},Q_1 \right] \right) , \\&\text{ where } \frac{d-c}{a-b-c+d}\le P_0 \text{ and } Q_1\le \frac{h-f}{e-f-g+h}\,, \\&\text{ that } \text{ is } ([P_0,P^*]\,,[Q^*,Q_1]), \text{ where } M_1\le P_0 \text{ and } Q_1\le M_2. \end{aligned}$$

      • $\frac{h-f}{e-f-g+h}\le \frac{d-b}{a-b-c+d}:$ for all $[Q_0,Q_1]$ the best response is $P_0=P_1=0$, thus no Ellsberg equilibrium arises.

  • $\frac{d-c}{a-b-c+d}\le \frac{h-g}{e-f-g+h}$: to any $[P_0,P_1]$ the best response is $Q_0=Q_1=1$, thus no Ellsberg equilibrium arises.

• (3) $Q_0<Q^*<Q_1:$ player 1 responds with $P_0=P_1=M_1$. Only when $M_1=P^*$, player 1 sticks to his strategy and we get the equilibrium $(P^*,[Q_0,Q_1]), \text{ where } Q_0<Q^*<Q_1.$

• (4) $\left[ Q_0,\frac{d-b}{a-b-c+d}\right] \Rightarrow [P_0,P_1]\subseteq \left[ 0,\frac{d-c}{a-b-c+d}\right] ,$ then, depending on the size of $M_1$ in relation to $P^*$, player 1 has the following choices:

  • $\frac{d-c}{a-b-c+d}\ge \frac{h-g}{e-f-g+h}:$ player 1 can play either

    • $\left[ P_0,\frac{h-g}{e-f-g+h}\right] ,$ then the best response of player 2 is $[Q_0,Q_1]\subseteq \left[ 0,\frac{h-f}{e-f-g+h}\right]$. Now, depending on the size of $M_2$ in relation to $Q^*$, player 2 has the following choices:

      • $\frac{h-f}{e-f-g+h}\ge \frac{d-b}{a-b-c+d}:\,\left[ Q_0,\frac{d-b}{a-b-c+d}\right]$, we thus have here the Ellsberg equilibrium

        $$\begin{aligned}&\left( \left[ P_0, \frac{h-g}{e-f-g+h}\right] ,\left[ Q_0,\frac{d-b}{a-b-c+d}\right] \right) , \\&\text{ that } \text{ is } ([P_0,P^*]\,,[Q_0,Q^*]). \end{aligned}$$

      • $\frac{h-f}{e-f-g+h}\le \frac{d-b}{a-b-c+d}:$ for all $[Q_0,Q_1]$ the best response is $P_0=P_1=0$, thus no Ellsberg equilibrium arises.

    • or $\left[ \frac{h-g}{e-f-g+h},P_1\right]$ with $\frac{h-g}{e-f-g+h}\le P_1\le \frac{d-c}{a-b-c+d}$, then the best response of player 2 is $[Q_0,Q_1]\subseteq \left[ \frac{h-f}{e-f-g+h},1\right]$. Now, depending on the size of $M_2$ in relation to $Q^*$, player 2 has the following choices:

      • $\frac{d-b}{a-b-c+d}\ge \frac{h-f}{e-f-g+h}:\,\left[ Q_0,\frac{d-b}{a-b-c+d}\right]$ with $\frac{h-f}{e-f-g+h}\le Q_0$, then the best response of player 1 is $[P_0,P_1]\subseteq \left[ 0,\frac{d-c}{a-b-c+d}\right]$. We thus have here the Ellsberg equilibrium

        $$\begin{aligned}&\left( \left[ \frac{h-g}{e-f-g+h},P_1\right] ,\left[ Q_0,\frac{d-b}{a-b-c+d}\right] \right) , \\&\text{ where } P_1\le \frac{d-c}{a-b-c+d} \text{ and } \frac{h-f}{e-f-g+h}\le Q_0\,, \\&\text{ that } \text{ is } ([P^*,P_1]\,,[Q_0,Q^*]), \text{ where } P_1\le M_1 \text{ and } M_2\le Q_0. \end{aligned}$$

      • $\frac{d-b}{a-b-c+d}\le \frac{h-f}{e-f-g+h}:$ for all $[Q_0,Q_1]$ the best response is $P_0=P_1=1$, thus no Ellsberg equilibrium arises.

  • $\frac{d-c}{a-b-c+d}\le \frac{h-g}{e-f-g+h}$: to any $[P_0,P_1]$ the best response is $Q_0=Q_1=1$, thus no Ellsberg equilibrium arises.

• (5) $Q_1<Q^*:$ player 1 responds $P_0=P_1=0$, if player 2 chooses $Q_0=Q_1=0$ this is the Ellsberg equilibrium that is identical to the pure Nash equilibrium $(B,R)$.

• (6) $Q_0=Q^*=Q_1:$ player 1 responds with $[P_0,P_1]\subseteq [0,1]$. Only when $Q^*=M_2$, player 2 sticks to his strategy and we get the equilibrium $([P_0,P_1]\,,Q^*), \text{ where } P_0<P^*<P_1.$ $\square$