Games with the total bandwagon property meet the Quint–Shubik conjecture

Abstract

This paper revisits the total bandwagon property (TBP) introduced by Kandori and Rob (Games Econ Behav 22:30–60, 1998). With this property, we characterize the class of two-player symmetric $n\times n$ games, showing that a game has TBP if and only if the game has $2^{n}-1$ symmetric Nash equilibria. We extend this result to bimatrix games by generalizing TBP. This sheds light on the (wrong) conjecture of Quint and Shubik (Int J Game Theory 26:353–359, 1997) that any nondegenerate $n\times n$ bimatrix game has at most $2^{n}-1$ Nash equilibria. We also provide an equilibrium selection criterion to two subclasses of games with TBP.

Appendix: proofs and examples

Consider any given finite set of strategies A with $|A|=n$. Any game $\mathbf {g}=(A,g)$ with TBP is nondegenerate. From Shapley (1974, Theorem 2) and Quint and Shubik (1997, Lemma 2.2), we know that the number of NE is both finite and odd if a game is nondegenerate, and also that for any given subset $S \subseteq A$ there exists at most one symmetric NE with support S, otherwise all convex combinations of two NE with identical support would constitute a continuum of NE, which violates the condition that the game has a finite number of NE. Therefore, game $\mathbf {g}=(A,g)$ with TBP has at most symmetric $2^{n}-1$ NE. To prove Proposition 1 and Theorem 1, we will apply these associated facts with TBP:

• (f1) the finiteness property (the number of NE is finite);

• (f2) the oddness property;

• (f3) the upper bound of the number of symmetric NE is $2^{n}-1$.

1.1 Proof of Proposition 1

Proof

Consider any given symmetric game $\mathbf {g}=(A,g)$ with TBP. Assume that the game has an asymmetric NE, $(x^{1},x^{2})\in \varDelta ^{2}$ with $x^{1}\ne x^{2}$. There are two cases to consider: (i) $\text {supp}(x^{1})=\text {supp}(x^{2})$ and (ii) $\text {supp}(x^{1})\ne \text {supp}(x^{2})$. We show below that neither case can hold.

Consider case (i) and take $S = \text {supp}(x^{1})=\text {supp}(x^{2}) \subseteq A$ for $x^{1}\ne x^{2}$. Since $(x^{1},x^{2})\in \varDelta ^{2}$ is in equilibrium, $\text {br}(x^{1})=\text {br}(x^{2}) = S$. With symmetry, $(x^{2},x^{1})\in \varDelta ^{2}$ is also an equilibrium with support S. Let $\hat{\varDelta }(S) \equiv \{ x \in A \mid \forall \lambda \in [0,1], x = \lambda x^1 + (1-\lambda )x^2 \}$ be the set of all convex combinations of two points $x^1$ and $x^2$. Since the set of best responses is convex, $\text {br}(x) = S$ must hold for all $x \in \hat{\varDelta }(S)$, implying that any pair of two points in the line segment of $\hat{\varDelta }(S)$ constitutes a continuum of NE. Since the game with TBP is nondegenerate, though, the number of NE must be finite due to the associated fact (f1) with TBP, which contradicts the existence of the continuum of NE.

In case (ii), without loss of generality, there must be some pure strategy $k \in \text {supp}(x^{1}) \backslash \text {supp}(x^{2})$. By TBP, player 1 takes a non-best response k to the opponent’s strategy $x^{2}$, which is a contradiction because all pure strategies in $\text {supp}(x^{1})$ must be the best response to $x^{2}$ in equilibrium.

1.2 Proof of Theorem 1

We will prove the if part and the only if part of Theorem 1 in order. For the only if part, we will use the associated facts (f2) and (f3) with TBP mentioned above.

1.2.1 Proof of the if part

To begin with, I first provide a lemma, which is subsequently used in the proof of the if part of Theorem 1. For this, consider a symmetric game $\mathbf {g}=(A,g)$ with $|A| = n$. Suppose that $\mathbf {g}$ has exactly $2^{n}-1$ symmetric NE. For any given subset of strategies $S\subseteq A$ we denote by $x^{*}|_{S} \in \varDelta$ the strategy in the symmetric NE with support S. Those NE must have different supports from one another, that is, they must be regular (Harsanyi 1973), and include all symmetric pure strategy profiles, otherwise there are at least two distinct symmetric (mixed strategy) NE with identical support and all convex combinations of them will be NE, and so the number of NE would be infinite, leading to a contradiction. By the same token, we can show:

Lemma 2

Suppose that a symmetric game $\mathbf {g}=(A,g)$ with $|A| = n$ has $2^n -1$ symmetric Nash equilibria. For any given $S \subseteq A$, $x^{*}|_{S}$ satisfies

$$\begin{aligned} \text {br}(x^{*}|_{S})= & {} \text {supp}(x^{*}|_{S}). \end{aligned}$$

(A.1)

Proof

By assumption, there are $2^n -1$ symmetric NE for all distinct supports $S \subseteq A$. Take one of them for any given $S \subseteq A$. Denote its equilibrium strategy by $x^{*}|_{S} \in \mathring{\varDelta }(S)$. By the property of NE, $\text {supp}(x^{*}|_{S}) \subseteq \text {br}(x^{*}|_{S})$ holds true. Below we show that $\text {supp}(x^{*}|_{S})$ cannot be a proper subset of $\text {br}(x^{*}|_{S})$, and therefore (A.1) must hold instead.

Consider first the case of $S = A$. In this case, $\text {br}(x^{*}|_{S}) \subseteq A = \text {supp}(x^{*}|_{S})$ holds. Together with $\text {supp}(x^{*}|_{S}) \subseteq \text {br}(x^{*}|_{S})$, this implies that (A.1) must hold.

Next consider any given $S \subsetneq A$ with $|S| =1, \dots ,n-1$ and the respective symmetric NE strategy of $x^{*}|_{S}$.

For this, suppose that there exists a pure strategy $j \in A$ such that $j \in \text {br}(x^{*}|_{S}) \backslash \text {supp}(x^{*}|_{S})$. Let $\tilde{S} = S \cup \{j\} \subseteq A$. By assumption, we know that there is a symmetric NE strategy, $x^{*}|_{\tilde{S}}$, such that $\tilde{S} \subseteq \text {br}(x^{*}|_{\tilde{S}})$.

With this, consider now all convex combinaitons connecting $x^{*}|_{S}$ to $x^{*}|_{\tilde{S}}$, and denote them by

$$\begin{aligned} X= & {} \{ x \in \mathring{\varDelta }(\tilde{S}) \mid \forall t \in [0,1], x = t x^{*}|_{S} + (1 - t) x^{*}|_{\tilde{S}} \}. \end{aligned}$$

Taking any point $x \in X$ as given, $\tilde{S} \subseteq \text {br}(x)$ must hold as x is a convex combination of two strategies to which $\tilde{S}$ is the set of best responses. This implies that there is a continuum of equilibria along with X, which contradicts the fact that the game has a finite number of NE.

Thus, for any given $S \subseteq A$, the symmetric NE strategy of $x^{*}|_{S}$ satisfies (A.1).

With this lemma, we show the if part of Theorem 1 as follows.

Proof

Consider a symmetric game $\mathbf {g}=(A,g)$ with $|A| = n$ where there are $2^{n}-1$ symmetric NE.

In the case of $|\text {supp}(x)| = 1$, the strategy $x \in \varDelta$ represents a pure strategy $i \in A$. Since any symmetric pure strategy profile is a NE, Lemma 2 implies that $\text {br}(x) = \text {supp}(x)$ must hold.

In the case of $|\text {supp}(x)| = n$, both $\text {br}(x) \subseteq A$ and $\text {supp}(x) = A$ hold true, which leads to $\text {br}(x) \subseteq \text {supp}(x)$.

So, what we have to show is that $\text {br}(x) \subseteq \text {supp}(x)$ holds for any other $x \in \varDelta$ with $|\text {supp}(x)| \in \{2, \dots , n-1\}$ when $n \ge 3$.

Suppose $n \ge 3$. For any given $x \in \varDelta$ with $|\text {supp}(x)| \in \{2,\dots , n-1\}$, consider the sequence of tuples composed of the subsets of strategies and their associated subsets and constants $(S^k, T^k,c^k)_{k=0}^{m}$ with $S^k \subseteq A$, $T^k \subseteq S^k$, and $c^k>0$ for some $m \in \mathbb {N}_{+}$ and $k=0,\dots ,m$ such that

$$\begin{aligned}&S^0 = \text {supp}(x), \quad T^0 = \arg \min _{i \in S^0} \frac{x_i}{x_i^{*}|_{S^0}}, \quad c^0 = \min _{i \in S^0} \frac{x_i}{x_i^{*}|_{S^0}}, \\&S^{l+1} = S^l \backslash T^l, \quad T^{l+1} = \arg \min _{i \in S^{l+1}} \frac{x_i - \sum _{j< l}c^j x_{i}^{*}|_{S^j}}{x_i^{*}|_{S^{l+1}}}, \\&c^{l+1} = \min _{i \in S^{l+1}} \frac{x_i - \sum _{j < l}c^j x_{i}^{*}|_{S^j}}{x_i^{*}|_{S^{l+1}}} \end{aligned}$$

for $l = 0,\dots ,m-1$ and $\emptyset = S^{m+1} \subsetneq S^m \subsetneq \cdots \subsetneq S^1 \subsetneq S^0 \subseteq A$. With this sequence, we seek to exclude all non-best responses in $\text {supp}(x)$ on the basis of the equilibria step-by-step so as to extract the best responses from $\text {supp}(x)$.

We can now decompose x into $(m+1)$ parts as

$$\begin{aligned} x= & {} \sum _{k=0}^m c^k x^{*}|_{S^k} \end{aligned}$$

where $x^*|_{S^k}$ satisfies, by Lemma 2, $\text {br}(x^*|_{S^k}) = \text {supp}(x^*|_{S^k}) = S^k$ for any given $k = 0,\dots ,m$. Let

$$\begin{aligned} \pi _i^k= & {} \sum _{j \in S^k} g_{ij} x_j^{*}|_{S^k} \end{aligned}$$

be the expected payoff of strategy i given that the opponent plays $x^{*}|_{S^k}$ for any given $k = 0,\dots ,m$. With this, the condition of $\text {br}(x^*|_{S^k}) = \text {supp}(x^*|_{S^k}) = S^k$ implies that the strict inequality of $\pi _i^k > \pi _h^k$ holds true for all $i \in S^k$ and $h \in A \backslash S^k$, whereas the equality of $\pi _i^k = \pi _h^k$ for any given $i,h \in S^k$. Furthermore, let

$$\begin{aligned} \pi _i(x)= & {} \sum _{k = 0}^m c^k \pi _i^k \end{aligned}$$

be the payoff of strategy i given that the opponent plays x. By construction, $\pi _i(x) > \pi _h(x)$ holds for all $i \in S^m$ and $h \in A \backslash S^m$ while $\pi _i(x) = \pi _h(x)$ for any given $i,h \in S^m$, thus $\text {br}(x) = S^m$ holds true. Together with $S^m \subseteq S^0 = \text {supp}(x)$, this leads to $\text {br}(x) \subseteq \text {supp}(x)$.

From above, we have shown that $\text {br}(x) \subseteq \text {supp}(x)$ holds for all $x \in \varDelta$.

1.2.2 Proof of the only if part

We show the only if part of Theorem 1 by induction. In doing so, we use Proposition 1 to ensure that there are no asymmetric NE in any given symmetric game with TBP, by which we can restrict attention to symmetric NE. For any given two-player symmetric game ${\mathbf{g}}=(A,g)$ and any given subset $S \subset A$, define by ${\mathbf{g}}|_{S}$ a restricted game in which both players choose their strategies only from S.

First, we show:

Lemma 3

Let ${\mathbf{g}}=(A,g)$ be any symmetric $n \times n$ game with TBP for any integer $n \ge 2$. Then any restricted game ${\mathbf{g}}|_{S}$ with $|S|\le k(=2,3,\dots ,n)$ has a unique symmetric Nash equilibrium with support S.

Proof

We show Lemma 3 by induction as follows.

• (I) It is obvious for $k=2$.

• (II) Suppose that every restricted game ${\mathbf{g}}|_{S}$ with the support size |S| that is smaller than or equal to $k(=2,3,\dots ,n-1)$ has a unique symmetric NE with support S. Taking into account that any restricted game ${\mathbf{g}}|_{S}$ has TBP if ${\mathbf{g}}$ has TBP, this implies that the restricted game ${\mathbf{g}}|_{S^{\prime }}$ with $|S^{\prime }|=k^{\prime }(=1,\dots ,k)$ has $2^{k^{\prime }}-1$ symmetric NE. Take now any (restricted) game ${\mathbf{g}}|_{S^{\prime \prime }}$ with $|S^{\prime \prime }|=k+1$, and consider all restricted games of ${\mathbf{g}}|_{S^{\prime \prime }}$. Each of those restricted games has at least one strategy in $S^{\prime \prime }$ and at most $k+1$ strategies. The number of those restricted games except for ${\mathbf{g}}|_{S^{\prime \prime }}$ is $2^{k+1}-2$. Again by TBP, ${\mathbf{g}}|_{S^{\prime \prime }}$ should have at least $2^{k+1}-2$ symmetric NE. Since any game with TBP is nondegenate, using the fact (f3) we know that ${\mathbf{g}}|_{S^{\prime \prime }}$ has at most $2^{k+1}-1$ symmetric NE. Together with the fact (f2), ${\mathbf{g}}|_{S^{\prime \prime }}$ must have a unique NE with support $S^{\prime \prime }$, and this NE must be symmetric due to Proposition 1.

Using Lemma 3, we can show the only if part of Theorem 1.

Proof

Any restricted game ${\mathbf{g}}|_{S}$ with $|S|\le n-1$ has a unique symmetric NE with support S due to Lemma 3, and the number of all restricted games ${\mathbf{g}}|_{S}$ with $|S|\le n-1$ is $2^{n}-2$. Since any restricted game ${\mathbf{g}}|_{S}$ has TBP if ${\mathbf{g}}$ has TBP, the game ${\mathbf{g}}=(A,g)$ has at least $2^{n}-2$ symmetric NE, which do not include an interior NE in $\varDelta$. Following Lemma 3, ${\mathbf{g}}$ must have a unique interior (symmetric) NE and therefore it has $(2^{n}-2)+1=2^{n}-1$ symmetric NE.

1.3 Proof of Theorem 2

1.3.1 Proof of the if part

Proof

Fix any given $i,j=1,2$ with $i \ne j$ and $x \in \varDelta$. In the same way as in the proof of the if part of Theorem 1, we can show that $\text {br}_i(x) \subseteq \text {supp}(x)$ holds for player i by constructing a sequence of tuples composed of subsets of strategies and their associated subsets and constants on the basis of player j’s equilibrium strategies that have different supports from one another.

1.3.2 Proof of the only if part

The main point of the proof is to use each player’s payoff function for finding out a symmetric NE in a “fictitious” symmetric game, which gives two (possibly different) symmetric NE with the identical support, and to combine those two NE strategies to construct a NE in the original game.

Proof

Fix any game ${\mathbf{g}}=(\{1,2\},A,(g^{i})_{i=1,2})$ where $A=\{1,2,\dots ,n\}$ and GTBP holds. For any given player $i=1,2$, construct the symmetric two-player game by using the player i’s payoff function $g^{i}$. Denote it by ${\mathbf{g}}^{i}=(A,g^{i})$. Since ${\mathbf{g}}^{i}$ has the same set of strategies A for both two players and the symmetric payoff function $g^{i}$, the game ${\mathbf{g}}^{i}$ satisfies TBP by GTBP. From Theorem 1, if ${\mathbf{g}}^{i}$ has TBP, it has $2^{n}-1$ symmetric NE that are all different in terms of supports. Consider the symmetric NE $(x^{i}|_{S},x^{i}|_{S})\in \varDelta$ with $\text {supp}(x^{i}|_{S})=S \subseteq A$ in ${\mathbf{g}}^{i}$. The strategy profile $(x^{2}|_{S},x^{1}|_{S})$ is a NE of the game ${\mathbf{g}}$ such that $\text {supp}(x^{1}|_{S})=\text {supp}(x^{2}|_{S})=S$. Since ${\mathbf{g}}^{i}$ has $2^{n}-1$ symmetric NE for every $i=1,2$, the bimatrix game ${\mathbf{g}}=(\{1,2\},A,(g^{i})_{i=1,2})$ also has $2^{n}-1$ NE, and any two of those NE are described by $(x^{2}|_{S},x^{1}|_{S})$ and $(x^{2}|_{S^{\prime }},x^{1}|_{S^{\prime }})$ for $S,S^{\prime } \subseteq A$ with $S \ne S^{\prime }$, respectively.

1.3.3 Examples

In the following, we give an example regarding Theorem 2.

Example 1

(Asymmetric pure-coordination game) Let us consider the $4\times 4$ pure-coordination game where $A=\{1,2,3,4\}$ and for any player $i=1,2$, $g_{kh}^i>0$ for $k=h$, otherwise $g_{kh}^i=0$. With player i’s payoffs, we define 11 different mixed strategies $x^j \in \varDelta$ of player $j \ne i$ based on these support sizes |S|: 6 different strategies in the case of $|S|=2$, for any two different strategies $h,l \in A$ and any other strategy $k \in A\backslash \{h,l\}$,

$$\begin{aligned} (x_h^j,x_l^j) =\frac{1}{g_{hh}^i+g_{ll}^i}(g_{ll}^i ,g_{hh}^i), \quad x_k^j=0; \end{aligned}$$

4 different strategies in the case of $|S|=3$, for any three different strategies $h,l,k \in A$ and any other strategy $m \in A\backslash \{h,l,k\}$,

$$\begin{aligned} (x_h,x_l,x_k) = \frac{1}{g_{hh}^i g_{ll}^i + g_{hh}^i g_{kk}^i+g_{ll}^i g_{kk}^i } (g_{ll}^ig_{kk}^i,g_{hh}^ig_{kk}^i,g_{hh}^ig_{ll}^i), \quad x_m=0; \end{aligned}$$

1 strategy in the case of $|S|=4$,

$$\begin{aligned}&\frac{1}{g_{11}^i g_{22}^i g_{33}^i + g_{11}^i g_{22}^i g_{44}^i + g_{11}^i g_{33}^i g_{44}^i+ g_{22}^i g_{33}^i g_{44}^i }\\&\quad (g_{22}^ig_{33}^i g_{44}^i, g_{11}^i g_{33}^i g_{44}^i, g_{11}^i g_{22}^i g_{44}^i, g_{22}^i g_{33}^i g_{44}^i). \end{aligned}$$

The game has $2^{4}-1(=15)$ NE consisting of the pair of the eleven mixed strategies described above and also of the four symmetric pure-strategy profiles. One can easily observe that the game satisfies GTBP.

To illustrate how the proof of the if part of Theorem 2 works out, consider, for instance, the payoffs of $g_{11}^1=g_{44}^2=1$, $g_{22}^1=g_{33}^2=2$, $g_{33}^1=g_{22}^2=3$, $g_{44}^1=g_{11}^2=4$, and then take $i=1$ and a mixed strategy $x =(4/7 , 2/7 ,1/7 , 0) \in \varDelta$ in order to demonstrate that $\text {br}_1(x) \subseteq \text {supp}(x)$ holds. From above, we can easily compute all NE of the game. Denote by $x^{*}|_S =(x_h^{*}|_S)_{h=1}^4 \in \varDelta$ the player 2’s strategy in the NE with support $S \subseteq A$. With these equilibrium strategies of player 2, we construct the sequence of tuples composed of subsets of strategies and their associated subsets and constants by $(S^k,T^k,c^k)_{k}$ such that

$$\begin{aligned} S^0= & {} \text {supp}(x) = \{1,2,3\}, \quad T^0 = \arg \min _{h \in S^0} \frac{x_h}{x_h^{*}|_{S^0}} = \{ 3\}, \\ c^0= & {} \min _{h \in S^0} \frac{x_h}{x_h^{*}|_{S^0}} = \frac{1/7}{ 2/11} = \frac{11}{14}, \\ S^1= & {} S^0 \backslash T^0 = \{1,2\}, \quad T^1 = \arg \min _{h \in S^1} \frac{x_h - c^0 x_{h}^{*}|_{S^0}}{x_h^{*}|_{S^1}} = \{1,2\}, \\ c^1= & {} \min _{h \in S^1} \frac{x_h - c^0 x_{h}^{*}|_{S^0}}{x_h^{*}|_{S^1}} = \left. \frac{4/7 - (11/14)(6/11)}{2/3}\right| _{h=1} \\= & {} \left. \frac{2/7 - (11/14)(3/11)}{1/3}\right| _{h=2} = \frac{3}{14}, \end{aligned}$$

and $S^{2} = S^1 \backslash T^1 = \emptyset$, where $x^{*}|_{S^0} = (6/11,3/11,2/11,0)$ and $x^{*}|_{S^1} = (2/3,1/3,0,0)$. This sequence leads to $\text {br}_1(x) \subseteq \text {supp}(x)$, because $\text {br}_1(x) = \text {br}_1(x^*|_{S^0}) \cap \text {br}_1(x^*|_{S^1}) = \{1,2\} = \text {supp}(x^*|_{S^0}) \cap \text {supp}(x^*|_{S^1}) = S^0 \cap S^1 \subseteq \{1,2,3\}= S^0 = \text {supp}(x)$.

Note that GTBP holds for all $n\times n$ pure coordination games and those slightly perturbed ones (see footnote 9 for the definition of a slightly perturbed pure coordination game).

1.3.4 Games with $2^{n}-1$ NE without the restriction on the support of NE

GTBP does not necessarily hold for a given asymmetric game ${\mathbf{g}}$ with $2^{n}-1$ NE. To ensure GTBP, ${\mathbf{g}}$ must be nondegenerate and also satisfy the restriction on the support of NE imposed in Theorem 2. Below we will provide those corresponding counterexamples.

Example 2

(Degenerate $3\times 3$ coordination game) Consider the asymmetric $3\times 3$ coordination game given by Table 1. The game has 3 symmetric pure-strategy NE as well as 4 other mixed-strategy NE:

$$\begin{aligned}&((2/3,1/3,0),(1/2,1/2,0)),((1/2,0,1/2),(1/2,0,1/2)),\\&\quad ((0,3/5,2/5),(0,1/2,1/2)),((2/7,3/7,2/7),(1/2,0,1/2)). \end{aligned}$$

While the game has $7(=2^{3}-1)$ NE, the strategy profile $(x^{1},x^{2})=((2/7,3/7,2/7),(1/2,0,1/2))$ violates GTBP as the game is degenerate with $|\text {br}_{1}(x^{2})|>|\text {supp}(x^{2})|$.

Table 1 An asymmetric $3\times 3$ coordination game

Example 3

(Nondegenerate asymmetric $4\times 4$ game) Consider the asymmetric $4\times 4$ game given by Table 2.¹⁷ This game has the following 15 NE.¹⁸ $|\{(x^{1},x^{2})\in \text {NE}\mid \forall i=1,2,|\text {supp}(x^{i})|=1\}|=2:$

$$\begin{aligned} \left( (0,1,0,0),(0,0,0,1)),((0,0,1,0),(1,0,0,0) \right) . \end{aligned}$$

$|\{(x^{1},x^{2})\in \text {NE}\mid \forall i=1,2,|\text {supp}(x^{i})|=2\}|=10:$

$$\begin{aligned}&\left( (11/15,4/15,0,0),(4/15,11/15,0,0) \right) , \end{aligned}$$

(A.2)

$$\begin{aligned}&\left( (0,0,1/2,1/2),(0,0,1/2,1/2) \right) ,\\&\left( (51/70,19/70,0,0),(0,23/27,4/27,0) \right) , \nonumber \\&\left( (23/27,0,0,4/27) (19/70,51/70,0,0) \right) ,\nonumber \\&\left( (2/7,5/7,0,0),(0,0,19/42,23/42) \right) , \nonumber \\&\left( (93/112,0,0,19/112),(0,93/112,19/112,0) \right) ,\nonumber \\&\left( (0,0,23/42,19/42),(2/7,5/7,0,0) \right) , \nonumber \\&\left( (0,0,31/59,28/59),(0,15/19,4/19,0) \right) ,\nonumber \\&\left( (15/19,0,0,4/19),(0,0,28/59,31/59) \right) , \nonumber \\&\left( (0,33/53,20/53,0),(33/53,0,0,20/53)\right) .\nonumber \end{aligned}$$

(A.3)

$|\{(x^{1},x^{2})\in \text {NE}\mid \forall i=1,2,|\text {supp}(x^{i})|=3\}|=2:$

$$\begin{aligned}&\left( (60/109,39/109,10/109,0),(31/74,0,17/111,95/222) \right) ,\\&\left( (0,31/74,95/222,17/111),(39/109,60/109,0,10/109) \right) . \end{aligned}$$

$|\{(x^{1},x^{2})\in \text {NE}\mid \forall i=1,2,|\text {supp}(x^{i})|=4\}|=1:$

$$\begin{aligned} \left( (5/11,4/11,5/33,1/33),(4/11,5/11,1/33,5/33) \right) . \end{aligned}$$

(A.4)

The game is nondegenerate and has $15(=2^{4}-1)$ NE but does not satisfy GTBP because it is not a coordination game (even if we can allow for all kinds of permutations of the strategies). The condition that both players have an identical support in every NE holds only for 3 NE of (A.2)–(A.4).

Table 2 An asymmetric $4\times 4$ game

Example 3 illustrates that a nondegenerate $n \times n$ game with $2^{n}-1$ NE does not need to be a coordination game. To show that a nongedenerate $n \times n$ game with $2^{n}-1$ NE has GTBP, the game must be a coordination game, but this alone does not suffice. Especially when there are many actions, we must impose the restriction on the support of NE.

1.4 Proof of Lemma 1

Proof

Suppose that a game with TBP has a GPRD–equilibrium $(i,i)\in A^2$. Then, br$(x) = \{i \}$ must hold for $x \in \mathring{\varDelta }(\{i,j\})$ with $x_i=x_j=1/2$ for any given pure strategy $j \in A\backslash \{i\}$. Since all pure-strategy best-response sets are convex in $\varDelta$, this implies that br$(x) = \{i \}$ holds for all $x \in \varDelta$ with $x_i \ge 1/2$, thus (i, i) is a 1 / 2–dominant equilibrium.

In the following we provide an example to highlight the difference between Lemma 1 in this paper and Lemma 2 in Kandori and Rob (1998).

Example 4

(Symmetric $3\times 3$ game with/out TBP) Kandori and Rob (1998, p.44–45) consider a symmetric $3\times 3$ game as shown in the left panel of Table 3, where only the row player’s payoffs are given due to symmetry. The associated best response region is depicted as the left panel of Fig. 1, where the numbers in parentheses represent the best response for each region. The figure indicates that TBP fails as there are mixed strategies $x \in \varDelta (\{1,3\})$ such that $\text {br}(x) = 2 \notin \{1,3\} = \text {supp}(x)$, and also that the strategy profile (3, 3) is a GPRD-equilibrium as well as a 1 / 2-dominant equilibrium. From Proposition 2 and the subsequent argument for Figures 1–3 in Kandori and Rob (1998), MBP is shown to hold in this game.

Table 3 A symmetric $3 \times 3$ game where MBP holds but TBP fails (left panel); TBP holds but MBP fails (right panel)

Fig. 1

The best response regions of the games on the left-hand-side (left panel) and on the right-hand-side (right panel) in Table 3

Consider now a symmetric $3\times 3$ game as shown in the right panel of Table 3. The associated best response region is described as the right panel of Fig. 1. The figure indicates that TBP holds true as well as (3, 3) is a 1 / 2-dominant equilibrium, while showing that MBP fails as the borderline of the best response regions of strategies 1 and 2 should lie in the gray-colored region to meet MBP (see Kandori and Rob 1998, for the detail).

If a symmetric $3 \times 3$ game satisfies both TBP and $g_{23} - g_{13} > g_{22} - g_{12}$, MBP should fail (Kandori and Rob 1998, Proposition 2). The game shown in the right panel of Table 3 is supermodular and meets both, therefore violating MBP.

1.5 Proof of Proposition 2

We first introduce the following notations. For $x,x^{\prime }\in \varDelta$, we write $x\succ x^{\prime }$ if x stochastically dominates $x^{\prime }$, that is, for any $i \in A$, $\sum _{i \le j \le n}x_{j} \ge \sum _{i \le j \le n} x_{j}^{\prime }$ holds with strict inequality for at least some i. In particular when considering a mixed strategy $x \in \mathring{\varDelta }(\{i,j\})$ for $i,j\in A$ with $i \ne j$, we write $x^{i j}$ instead of x for clarity. Let $\min \text {br}(x)$ be the lowest pure-strategy best-response against the opponent’s strategy x and $\text {max}\text {br}(x)$ the highest. With these notations, we prove Proposition 2 by using the property on supermodular games ${\mathbf{g}}=(A,g)$ that the best-response correspondence for any given player is non-decreasing in the opponent’s strategies, that is, for any $x,x^{\prime }\in \varDelta$ with $x\succ x^{\prime }$, both $\min \text {br}(x)\ge \min \text {br}(x^{\prime })$ and $\max \text {br}(x)\ge \max \text {br}(x^{\prime })$ hold.¹⁹

Proof

By TBP, it follows that for $x^{1n} \in \mathring{\varDelta }(\{1,n\})$ with $x_{1}^{1n}=x_{n}^{1n}=1/2$,

$$\begin{aligned} \text {br}(x^{1n})\subseteq \{1,n\}. \end{aligned}$$

By the assumption of $g_{11}-g_{n1}\ne g_{nn}-g_{1n}$, . Suppose that holds, so does . Now take with . Since $x^{1n}\succ x^{1i}$ holds for , by supermodularity, it follows that

$$\begin{aligned} \{1\}=\max \text {br}(x^{1n})\ge \max \text {br}(x^{1i}), \end{aligned}$$

which implies that

$$\begin{aligned} \text {br}(x^{1i})&=\{1\} \end{aligned}$$

holds for all $i \in A$ and the belief $x^{1i} \in \mathring{\varDelta }(\{1,i\})$ with $x_{1}^{1i} = x_{i}^{1i}=1/2$. This condition implies that the strategy profile (1,1) is a GPRD-equilibrium, and by Lemma 1, it is a 1 / 2–dominant equilibrium.

To show the uniqueness, suppose that there are two 1 / 2–dominant equilibria, (i, i) and (j, j). This gives the following two inequalities,

$$\begin{aligned} \dfrac{1}{2}g_{ii}+\dfrac{1}{2}g_{ij}>\dfrac{1}{2}g_{jj}+\dfrac{1}{2}g_{ji},\\ \dfrac{1}{2}g_{jj}+\dfrac{1}{2}g_{ji}>\dfrac{1}{2}g_{ii}+\dfrac{1}{2}g_{ij}, \end{aligned}$$

which leads to a contradiction.

Similarly, in the case of $g_{11}-g_{n1} < g_{nn}-g_{1n}$, we can show that (n, n) is the unique 1 / 2–dominant equilibrium.

1.6 Proof of Proposition 3

Proof

Suppose that there is a potential function v such that (i, i) is a unique potential maximizer. Then, both $v_{ii}>v_{jj}$ and $v_{ij}=v_{ji}$ must hold for any other strategy $j \in A \backslash \{ i\}$, which leads to

$$\begin{aligned} v_{ii} - v_{jj} > 0 = v_{ji} - v_{ij}. \nonumber \end{aligned}$$

Rewrite this as $v_{ii} - v_{ji} > v_{jj} - v_{ij}$. By definition of the potential maximizer, $g_{ii} - g_{ji} > g_{jj} - g_{ij}$ should hold, which implies that

$$\begin{aligned} \frac{1}{2}( g_{ii} + g_{ij} ) > \frac{1}{2} ( g_{ji} + g_{jj} ) \nonumber \end{aligned}$$

holds for all $j\in A \backslash \{i\}$, and therefore (i, i) is a GPRD–equilibrium. Together with Lemma 1, it is a 1 / 2–dominant equilibrium.