Nash equilibria in random games with right fat-tailed distributions

Abstract

We study the distribution of the number of mixed strategy Nash equilibria in two-player games where each player’s payoffs are independently drawn from an identical distribution. When the payoff distributions are sufficiently right fat-tailed, we characterize the Nash equilibria by best reply cycles of pure strategies, and we show that the expected number of Nash equilibria is approximately $\sqrt{\pi mn/\left( m+n\right) }$ in a random $m\times n$ asymmetric game and approximately n/2 in a random $n\times n$ symmetric game. We also provide new lower bounds for the expected number of Nash equilibria in a random game with any type of payoff distribution.

Appendix

1.1 Proof of Proposition 1

Without loss of generality, we assume that $m\ge n$. The proof of Proposition 1 consists of two parts: the derivation of the expected number of Nash equilibria, ${{\,\textrm{E}\,}}$, and an asymptotic analysis. By Theorem 1, the expectation of ${{\,\textrm{E}\,}}$ can be derived from the distribution of the number of ${{\,\textrm{PR}\,}}$, the number of point rationalizable strategies, and the conditional expectation of ${{\,\textrm{E}\,}}$ given ${{\,\textrm{PR}\,}}$. The distribution of ${{\,\textrm{PR}\,}}$ in the following lemma is given by Theorems 3 and 10 in Pei and Takahashi (2019).

Lemma 3

In a random $m\times n$ asymmetric game, we have for any $1\le k\le \min \left( {m,n}\right)$

$$\begin{aligned} {\mathbb {P}}_{m,n}\left( {{\,\textrm{PR}\,}} \ge k\right)&=\frac{m!}{\left( m-k\right) !m^{k}}\frac{n!}{\left( n-k\right) !n^{k}}. \end{aligned}$$

In a random $n\times n$ symmetric game, for any $1\le k\le n$, we have

$$\begin{aligned} {\mathbb {P}}_{n}\left( {{\,\textrm{PR}\,}} =k\right)&=k!\left( {\begin{array}{c}n\\ k\end{array}}\right) kn^{-k-1},\\ {\mathbb {P}}_{n}\left( {{\,\textrm{PR}\,}} \ge k\right)&=\frac{n!}{\left( n-k\right) !n^{k}}. \end{aligned}$$

In the rest of this proof, we first show the conditional expectation of E given PR, which is the Permutation Cycle Index. We then complete the proof with an asymptotic analysis.

1.1.1 Permutation cycle index

Denote the expectation under ${\mathbb {Q}}^{k}$ as ${\mathbb {E}}^{k}$. We compute the expectation of $2^{ BC }-1$ conditional on each player having exactly k point rationalizable strategies, i.e., ${\mathbb {E}}^{k}\left[ 2^{{{\,\textrm{PC}\,}}_{o}+{{\,\textrm{PC}\,}}_{e}}-1\right]$ for asymmetric case and ${\mathbb {E}}^{k}\left[ 2^{{{\,\textrm{PC}\,}}_{o}+2{{\,\textrm{PC}\,}}_{e}}-1\right]$ for symmetric case.

Lemma 4

${\mathbb {E}}^{k}\left[ 2^{{{\,\textrm{PC}\,}}_{o}+{{\,\textrm{PC}\,}}_{e}}-1\right] =k$ and ${\mathbb {E}}^{k}\left[ 2^{{{\,\textrm{PC}\,}}_{o}+2{{\,\textrm{PC}\,}}_{e}}-1\right] =k^{2}/4+k+\left( -1\right) ^{k}/8-1/8$.

Proof

Let ${\mathfrak {p}}\left( k,l;c_{1},c_{2},\ldots \right)$ be the number of k-permutations that have l permutation cycles, where $c_{1},c_{2},\ldots$ represent the number of permutation cycles with length $1,2,\ldots$ Clearly, $c_{1}+c_{2}+\cdots =l$ and $c_{1}+2c_{2}+\cdots =k$. By Comtet (1974, Theorem C in Section 6.2), we have the generating function:

$$\begin{aligned} \sum _{k,l,c_{1},c_{2},\ldots \ge 0}{\mathfrak {p}}\left( k,l;c_{1},c_{2},\ldots \right) \frac{t^{k}}{k!}x_{1}^{c_{1}}x_{2}^{c_{2}}\ldots&=\exp \left( x_{1}t+x_{2}\frac{t^{2}}{2}+x_{3}\frac{t^{3}}{3}+\cdots \right) . \end{aligned}$$

Note that

$$\begin{aligned} {\mathbb {E}}^{k}\left[ 2^{PC_{o}+PC_{e}}\right]&=\frac{1}{k!}\sum _{l,c_{1},c_{2},\ldots \ge 0}{\mathfrak {p}}\left( k,l;c_{1},c_{2},\ldots \right) 2^{l}, \end{aligned}$$

which is the coefficient of $t^{k}$ of the generating function with $x_{1}=x_{2}=\cdots =2$, i.e., $\exp \left( 2t+\frac{2t^{2}}{2}+\frac{2t^{3}}{3}+\cdots \right) =\left( 1-t\right) ^{-2}$. Hence,

$$\begin{aligned} {\mathbb {E}}^{k}\left[ 2^{PC_{o}+PC_{e}}-1\right]&=\frac{1}{k!}\frac{\partial ^{k}}{\partial t^{k}}\left( 1-t\right) ^{-2}|_{t=0}-1=k+1-1=k. \end{aligned}$$

Similarly,

$$\begin{aligned} {\mathbb {E}}^{k}\left[ 2^{PC_{o}+2PC_{e}}\right]&=\frac{1}{k!}\sum _{l,c_{1},c_{2},\ldots \ge 0}{\mathfrak {p}}\left( k,l;c_{1},c_{2},\ldots \right) 2{}^{c_{1}}2^{2c_{2}}\ldots , \end{aligned}$$

which is the coefficient of $t^{k}$ of the generating function with $x_{1}=x_{3}=\cdots =2$ and $x_{2}=x_{4}=\cdots =4$, i.e., $\exp \left( 2t+\frac{4t^{2}}{2}+\frac{2t^{3}}{3}+\frac{4t^{4}}{4}+\cdots \right) =\left( 1-t\right) ^{-3}\left( 1+t\right) ^{-1}$. Hence,

$$\begin{aligned} {\mathbb {E}}^{k}\left[ 2^{{{\,\textrm{PC}\,}}_{o}+2{{\,\textrm{PC}\,}}_{e}}-1\right]&=\frac{1}{k!}\left. \frac{\partial ^{k}}{\partial t^{k}}\left( 1-t\right) ^{-3}\left( 1+t\right) ^{-1}\right| _{t=0}-1\\&=\frac{1}{k!}\left. \sum _{i=0}^{k}\left( {\begin{array}{c}k\\ i\end{array}}\right) \frac{\partial ^{i}\left( 1-t\right) ^{-3}}{\partial t^{i}}\frac{\partial ^{k-i}\left( 1+t\right) ^{-1}}{\partial t^{k-i}}\right| _{t=0}-1\\&=\frac{1}{k!}\sum _{i=0}^{k}\frac{k!}{\left( k-i\right) !i!}\frac{\left( i+2\right) !}{2}\left( -1\right) ^{k-i}\left( k-i\right) !-1\\&=k^{2}/4+k+\left( -1\right) ^{k}/8-1/8. \end{aligned}$$

$\square$

1.1.2 The expectation of E

By Corollary 1 and Lemmas 3 and 4, for the asymmetric cases, we have

$$\begin{aligned} \lim _{r\rightarrow \infty }{\mathbb {E}}_{m,n}^{F_{r},G_{r}}\left[ {{\,\textrm{E}\,}}\right]&=\lim _{r\rightarrow \infty }{\mathbb {E}}_{m,n}^{F_{r},G_{r}}\left[ {\mathbb {E}}_{m,n}^{F_{r},G_{r}}\left[ {{\,\textrm{E}\,}}| {{\,\textrm{PR}\,}} \right] \right] \\&=\lim _{r\rightarrow \infty }{\mathbb {E}}_{m,n}^{F_{r},G_{r}}\left[ 2^{BC}-1|{{\,\textrm{PR}\,}}\right] \\&=\lim _{r\rightarrow \infty }{\mathbb {E}}_{m,n}^{F}\left[ {{\,\textrm{PR}\,}}\right] \\&=\sum _{k=1}^{n}\frac{m!}{\left( m-k\right) !m^{k}}\frac{n!}{\left( n-k\right) !n^{k}}. \end{aligned}$$

And for the symmetric cases, we have

$$\begin{aligned} \lim _{r\rightarrow 0}{\mathbb {E}}_{n}^{F_{r}}\left[ {{\,\textrm{E}\,}}\right]&=\lim _{r\rightarrow 0}{\mathbb {E}}_{n}^{F_{r}}\left[ {\mathbb {E}}_{n}^{F_{r}}\left[ {{\,\textrm{E}\,}}|{{\,\textrm{PR}\,}}\right] \right] \\&=\frac{1}{4}{\mathbb {E}}_{n}\left[ {{\,\textrm{PR}\,}}^{2}\right] +\mathbb {{\mathbb {E}}}_{n}\left[ {{\,\textrm{PR}\,}} \right] +\frac{1}{8}{\mathbb {E}}_{n}\left[ \left( -1\right) ^{ {{\,\textrm{PR}\,}} }-1\right] \\ {}&=\sum _{k=1}^{n}\frac{n!\left( k^{2}/4+k+\left( -1\right) ^{k}/8-1/8\right) }{\left( n-k\right) !n^{k+1}}. \end{aligned}$$

1.1.3 Asymptotic analysis

We use the Laplace method in the asymptotic analysis. Details of the theorems and techniques used in this subsection can be found in Sections 4.5, 4.6, and 4.7 of Sedgewick and Flajolet (1996). To make this part self-contained, we list these results as lemmas here:

The analysis is closely related to Ramanujan $Q-$distribution:

$$\begin{aligned} \frac{n!}{\left( n-k\right) !n^{k}} \end{aligned}$$

and Ramanujan Q-function:

$$\begin{aligned} \sum _{k=n}^{n}&\frac{n!}{\left( n-k\right) !n^{k}}. \end{aligned}$$

The following lemma, which is Theorem 4.4 of Section 4.6 in Sedgewick and Flajolet (1996), gives the approximation for the Ramanujan Q-distribution.

Lemma 5

As $n\rightarrow \infty$, the following (relative) approximation holds for $k=o\left( n^{2/3}\right)$:

$$\begin{aligned} \frac{n!}{\left( n-k\right) !n^{k}}=&\exp \left( -\frac{k^{2}}{2n}\right) \left( 1+O\left( \frac{k}{n}\right) +O\left( \frac{k^{3}}{n^{2}}\right) \right) . \end{aligned}$$

In addition, the following (absolute) approximation holds for all k:

$$\begin{aligned} \frac{n!}{\left( n-k\right) !n^{k}}=&\exp \left( -\frac{k^{2}}{2n}\right) +O\left( \frac{1}{\sqrt{n}}\right) . \end{aligned}$$

In Page 181 of Sedgewick and Flajolet (1996), a simpler form of the Euler–Maclaurin summation formula is provided. We summarize the result in the following lemma:

Lemma 6

Let $f\left( x\right)$ be a function defined on an interval $\left[ a,b\right]$ where a and b are integers. Suppose that the derivatives $f^{\prime \prime }\left( x\right)$ exist and are continuous. Then

$$\begin{aligned} \sum _{k=a}^{k=b}f\left( k\right)&=\int _{a}^{b}f\left( x\right) dx+\frac{f\left( a\right) +f\left( b\right) }{2}+\frac{1}{12}\left( f^{\prime }\left( b\right) -f^{\prime }\left( a\right) \right) +R, \end{aligned}$$

where R is a remainder term satisfying

$$\begin{aligned} \left| R\right|&<\frac{1}{\pi ^{2}}\int _{a}^{b}\left| f^{\prime \prime }\left( x\right) \right| dx. \end{aligned}$$

Now, we are ready to derive the asymptotic analysis of Proposition 1. We focus on the asymmetric cases first. Recall that

$$\begin{aligned} \lim _{r\rightarrow \infty }{\mathbb {E}}_{m,n}^{F_{r},G_{r}}\left[ {{\,\textrm{E}\,}}\right]&=\sum _{k=1}^{n}\frac{m!}{\left( m-k\right) !m^{k}}\frac{n!}{\left( n-k\right) !n^{k}}. \end{aligned}$$

For the asymptotic analysis, following the proof of Lemma 5 in Sedgewick and Flajolet (1996), here we use a more accurate form of the relative approximation of the Ramanujan $Q-$distribution for $k=o\left( n^{2/3}\right)$:

$$\begin{aligned} \frac{n!}{\left( n-k\right) !n^{k}}=&\exp \left( \sum _{j=1}^{k-1}\ln \left( 1-\frac{j}{n}\right) \right) \nonumber \\ =&\exp \left( \sum _{j=1}^{k-1}\left( -\frac{j}{n}-\frac{j^{2}}{2n^{2}}-\frac{j^{3}}{3n^{3}}-\cdots \right) \right) \nonumber \\ =&\exp \left( -\frac{k\left( k-1\right) }{2n}-\frac{k\left( k-1\right) \left( 2k-1\right) }{12n^{2}}+O\left( \frac{k^{4}}{n^{3}}\right) \right) \nonumber \\ =&\exp \left( -\frac{k^{2}}{2n}\right) \exp \left( \frac{k}{2n} -\frac{k^{3}}{6n^{2}}+o\left( \frac{k^{3}}{n^{2}}\right) \right) \nonumber \\ =&\exp \left( -\frac{k^{2}}{2n}\right) \left( 1+\frac{k}{2n}-\frac{k^{3}}{6n^{2}} +o\left( \frac{k^{3}}{n^{2}}\right) \right) . \end{aligned}$$

(1)

Note that $\frac{m!}{\left( m-k\right) !m^{k}}\frac{n!}{\left( n-k\right) !n^{k}}$ is decreasing with k. Let $k_{0}$ be the nearest integer to $\sqrt{n}\log n$, we have $k_{0}=o\left( n^{2/3}\right)$. Note that $m\ge n$. Then by Eq. (1),

$$\begin{aligned}&\frac{m!}{\left( m-k_{0}\right) !m^{k_{0}}}\frac{n!}{\left( n-k_{0}\right) !n^{k_{0}}}\nonumber \\&\quad = \exp \left( -\frac{k_{0}^{2}}{2m}-\frac{k_{0}^{2}}{2n}\right) \left( 1+\frac{k_{0}}{2n}-\frac{k_{0}^{3}}{6n^{2}}+o\left( \frac{k_{0}^{3}}{n^{2}}\right) \right) \left( 1+\frac{k_{0}}{2m}-\frac{k_{0}^{3}}{6m^{2}}+o\left( \frac{k_{0}^{3}}{m^{2}}\right) \right) \nonumber \\&\quad = \exp \left( -\frac{\left( m+n\right) \left( \log n\right) ^{2}}{2m}\right) \left( 1+O\left( \frac{\left( \log n\right) ^{3}}{\sqrt{n}}\right) \right) \nonumber \\&\quad = n^{-\frac{\left( m+n\right) \log n}{2m}}\left( 1+O\left( \frac{\left( \log n\right) ^{3}}{\sqrt{n}}\right) \right) . \end{aligned}$$

(2)

Hence, for all $k\ge k_{0}$, since each term is exponentially small, the contribution to the sum is $o\left( 1\right)$, as well as the error terms. Therefore, we can use the relative approximation to all k from 1 to $k_{0}$, and the error terms are determined by

$$\begin{aligned} \exp \left( -\frac{\left( m+n\right) k^{2}}{2mn}\right) \left( \frac{k}{2n}-\frac{k^{3}}{6n^{2}}\right) . \end{aligned}$$

By applying the Euler–Maclaurin summation to the functions $x\exp \left( -\left( m+n\right) x^{2}/2m\right)$ and $x^{3}\exp \left( -\left( m+n\right) x^{2}/2m\right)$ with step $1/\sqrt{n}$, we find that both $\sum _{k=1}^{k_{0}}\left( k/n\right) \exp \left( -\left( m+n\right) k^{2}/\left( 2mn\right) \right)$ and $\sum _{k=1}^{k_{0}}\left( k^{3}/n^{2}\right) \exp \left( -\left( m+n\right) k^{2}/\left( 2mn\right) \right)$ are $O\left( 1\right)$. Hence,

$$\begin{aligned} {\mathbb {E}}_{m,n}^{F_{r},G_{r}}\left[ {{\,\textrm{E}\,}}\right]&=\sum _{k=1}^{n}\frac{m!}{\left( m-k\right) !m^{k}}\frac{n!}{\left( n-k\right) !n^{k}}+o\left( 1\right) \\&=\sum _{k=1}^{n}\exp \left( -\frac{\left( m+n\right) k^{2}}{2mn}\right) +O\left( 1\right) . \end{aligned}$$

Then we apply the Euler–Maclaurin summation to the remaining sum, i.e., the sum of the values of the function $\exp \left( -\left( m+n\right) x^{2}/\left( 2m\right) \right)$ at regularly spaced points with step $1/\sqrt{n}$. The approximation is

$$\begin{aligned} \sum _{k=1}^{n}\exp \left( -\frac{\left( m+n\right) k^{2}}{2m}\right) =&\sqrt{n}\int _{0}^{\sqrt{n}}\exp \left( \frac{-\left( m+n\right) k^{2}}{2m}\right) dk+O\left( 1\right) \nonumber \\ =&\sqrt{n}\int _{0}^{\infty }\exp \left( \frac{-\left( m+n\right) k^{2}}{2m}\right) dk+O\left( 1\right) \nonumber \\ =&\sqrt{\frac{\pi mn}{2\left( m+n\right) }}+O\left( 1\right) . \end{aligned}$$

(3)

The second equality holds as the integrand for $k>\sqrt{n}$ is exponentially small, which means that the integral for $k>\sqrt{n}$ is O(1).

For random $n\times n$ symmetric games, recall that

$$\begin{aligned} \lim _{r\rightarrow 0}{\mathbb {E}}_{n}^{F_{r}}\left[ {{\,\textrm{E}\,}}\right]&=\lim _{r\rightarrow 0}{\mathbb {E}}_{n}^{F_{r}}\left[ {\mathbb {E}}_{n}^{F_{r}}\left[ {{\,\textrm{E}\,}}|{{\,\textrm{PR}\,}}\right] \right] \\&=\frac{1}{4}{\mathbb {E}}_{n}\left[ {{\,\textrm{PR}\,}}^{2}\right] +\mathbb {{\mathbb {E}}}_{n}\left[ {{\,\textrm{PR}\,}} \right] +\frac{1}{8}{\mathbb {E}}_{n}\left[ \left( -1\right) ^{ {{\,\textrm{PR}\,}} }-1\right] . \end{aligned}$$

Note that ${\mathbb {E}}_{n}\left[ \left( -1\right) ^{ {{\,\textrm{PR}\,}} }-1\right]$ is $O\left( 1\right)$. And ${\mathbb {E}}_{n}\left[ {{\,\textrm{PR}\,}} \right] =\sum _{k=1}^{n}n!/\left( \left( n-k\right) !n^{k}\right)$ is the Ramanujan Q-function. Hence, by Theorem 4.8 in Sedgewick and Flajolet (1996),

$$\begin{aligned} {\mathbb {E}}_{n}\left[ PR \right] =&\sqrt{\frac{\pi n}{2}}+O\left( 1\right) . \end{aligned}$$

(4)

Now we have

$$\begin{aligned} {\mathbb {E}}_{n}\left[ {{\,\textrm{PR}\,}} ^{2}\right]&=\sum _{k=1}^{n}k^{2}k!\left( {\begin{array}{c}n\\ k\end{array}}\right) kn^{-k-1}=\sum _{k=1}^{n}\frac{n!}{\left( n-k\right) !n^{k}}\frac{k^{3}}{n}. \end{aligned}$$

Similarly, for the asymptotic analysis in asymmetric cases, we have

$$\begin{aligned} \sum _{k=1}^{n}\frac{n!}{\left( n-k\right) !n^{k}}\frac{k^{3}}{n}&=\sum _{k=1}^{n}\frac{k^{3}}{n}\exp \left( -\frac{k^{2}}{2n}\right) +O\left( 1\right) . \end{aligned}$$

We apply the Euler–Maclaurin summation to the remaining sum, i.e., sum of values of the function $x^{3}\exp \left( -x^{2}/2\right)$ at regularly spaced points with step $1/\sqrt{n}$ to obtain:

$$\begin{aligned} \sum _{k=1}^{n}\frac{k^{3}}{\sqrt{n^{3}}}\exp \left( -\frac{k^{2}}{2n}\right)&=2\sqrt{n}+O\left( 1\right) . \end{aligned}$$

Hence, $\sum _{k=1}^{n}\frac{k^{3}}{n}\exp \left( -\frac{k^{2}}{2n}\right) =2n+O\left( \sqrt{n}\right)$. And we have

$$\begin{aligned} {\mathbb {E}}_{n}^{F_{r}}\left[ {{\,\textrm{E}\,}}\right]&=\frac{1}{4}{\mathbb {E}}_{n}\left[ {{\,\textrm{PR}\,}}^{2}\right] +\mathbb {{\mathbb {E}}}_{n}\left[ {{\,\textrm{PR}\,}}\right] +\frac{1}{8}{\mathbb {E}}_{n}\left[ \left( -1\right) ^{{{\,\textrm{PR}\,}}}-1\right] \\&=\frac{n}{2}+O\left( \sqrt{n}\right) . \end{aligned}$$

1.2 A.2 Proof of Proposition 2

For the sake of completeness, we provide a self-contained proof of Proposition 2. In addition, we would like to thank an anonymous referee for suggesting a simpler proof of the uniqueness of the Nash equilibrium.

For player p’s mixed strategy $\sigma _p$ and $\sigma _p^\prime$, given a subset T of the opponent’s strategy space, we write $\sigma _p >_T\sigma _p^\prime$ if $\sigma _p$ weakly dominates $\sigma _p^\prime$ when opposing profiles are restricted to T. Then the following lemma, which is used in the proof, is Corollary 2 in Weinstein (2020):

Lemma 7

A two-player game has a Nash equilibrium with supports $\left( T_{1},T_{2}\right)$, if and only if there is no pair $\sigma _1\in \Delta \left\{ 1,\ldots ,m\right\} ,\sigma _1^\prime \in \Delta (T_1)$ such that $\sigma _1>_{T_{2}}\sigma _1 ^\prime$ and no pair $\sigma _2\in \Delta \left\{ 1,\ldots ,n\right\} ,\sigma _2^\prime \in \Delta (T_2)$ such that $\sigma _2>_{T_{1}}\sigma _2 ^\prime$. When such pairs do exist, they can be chosen with disjoint support, that is, $\sigma _{1i}\sigma _{1i}^\prime =0$ for all $i\in \left\{ 1,\ldots ,m\right\}$.

To show that the Nash equilibrium supported by $\left( T_{1},T_{2}\right)$ with the point best reply property is unique if the game satisfies the outstanding best reply condition, we present the following Lemma.

Lemma 8

In a generic $m\times n$ game $\left( U,V\right)$ that satisfies the outstanding best reply condition, each of player p’s pure best replies to $\sigma _{-p}$ must be a best reply to some strategy in the support of $\sigma _{-p}$, i.e., the pure best reply to $\sigma _{-p}$ belongs to $B_{p}\left( {{\,\textrm{Supp}\,}}\left( \sigma _{-p}\right) \right)$.

Proof

For player 1 and the mixed strategy $\sigma _2$, let $S_2={{\,\textrm{Supp}\,}}(\sigma _2)$. Consider the mixed strategy $\sigma _1$ that has equal weight on all strategies in $B_1(S_2)$. Then for any pure strategy $i\notin B_1(S_2)$, as the game satisfies the outstanding best reply condition,

$$\begin{aligned} \sum _{j\in S_2}\sigma _{2j}(u_{ij}-{\underline{u}})<\sum _{j\in S_2}\sigma _{2j}\frac{{\bar{u}}_j-{\underline{u}}}{m-1}\le \sum _{j\in S_2}\sigma _{2j}\frac{{\bar{u}}_j-{\underline{u}}}{\left| B_1(S_2)\right| }\le \sum _{j\in S_2} \sum _{i^\prime \in B_1(S_2)} \sigma _{2j} \sigma _{1i^\prime } (u_{i^\prime j}-{\underline{u}}) \end{aligned}$$

i.e., the payoff for i against $\sigma _{2}$ is strictly less than the payoff for i against $\sigma _1$. Hence, the pure best reply to $\sigma _2$ belongs to $B_1({{\,\textrm{Supp}\,}}(\sigma _{2}))$.

A similar argument applies to player 2. And we conclude the proof. $\square$

Proof of Proposition 2

Suppose $\left( T_{1},T_{2}\right)$ has the point best-reply property. We first show that there is some Nash equilibrium supported by $\left( T_{1},T_{2}\right)$. If $T_1=\left\{ 1,\ldots ,m\right\}$ or $T_2=\{1,\ldots ,n\}$, then the condition following ‘if and only if’ in Lemma 7 is satisfied. Therefore, the lemma yields a Nash equilibrium. Now suppose $T_1\subsetneq \{1,\ldots ,m\}$ and $T_2\subsetneq \{1,\ldots ,n\}$. Then $\left| T_1\right| \le m-1$ and $\left| T_2\right| \le n-1$. For any $i\in T_1$, it is a best reply to some $j\in T_2$. As the game satisfies the definition of the outstanding best reply condition, we have

$$\begin{aligned} {u}_{ij}-{\underline{u}}&>\left( m-1\right) \left( u_{i^\prime j}-{\underline{u}}\right) \,\text {for all}\,\, i^\prime \ne i. \end{aligned}$$

(5)

Suppose $\sigma _1>_{T_2}\sigma _1^\prime$ with $\sigma _1\in \Delta \left\{ 1,\ldots ,m\right\} ,\sigma _1^\prime \in \Delta (T_1)$ having disjoint support, as in Lemma 7. Suppose $\sigma _{1i}^\prime \ge 1/(m-1)$. Disjoint support implies $\sigma _{1i}=0$. Then the payoff of $\sigma _{1i}^\prime$ against j is strictly higher than the payoff of $\sigma _{1i}$ against j as

$$\begin{aligned} \sigma _{1i}^\prime ({u}_{ij}-{\underline{u}})+\sum _{i^\prime \ne i}\sigma _{1i^\prime }^\prime ({u}_{i^\prime j}-{\underline{u}})&\ge \frac{1}{m-1}({u}_{ij}-{\underline{u}})>\sum _{i^\prime \in T}\sigma _{1i}^\prime ({u}_{i^\prime j}-{\underline{u}}), \end{aligned}$$

(6)

which is a contradiction. Hence, $\sigma _{1i}^\prime < 1/(m-1)$ for all $i\in T_1$, which contradicts $\sum _{i\in T_1}\sigma _{1i}^\prime =1$. The analogous conditions hold for the reversed role. Therefore, we have proven the existence of the Nash equilibrium with supports $\left( T_{1},T_{2}\right)$.

Now we show that the equilibrium supported by $\left( T_{1},T_{2}\right)$ is unique. Suppose $\left( T_{1},T_{2}\right)$ supports two different Nash equilibria $\left( \sigma _{1},\sigma _{2}\right)$ and $\left( \sigma _{1}^{\prime },\sigma _{2}^{\prime }\right)$. Without loss of generality, assume $\sigma _{1}\ne \sigma _{1}^{\prime }$. Then all elements of $T_2$ are best replies to both $\sigma _1$ and $\sigma _1^\prime$, and to any point on the line containing these points, including a $\sigma _1^{\prime \prime }$ where this line leaves the simplex. However, some element of $T_2$ is not a pure best response to any element of the support of $\sigma _1^{\prime \prime }$, contrary to Lemma 8.

Lastly, note that by Lemma 8, if a game satisfies the outstanding best reply condition, then the support of any Nash equilibrium must have the point best-reply property. Hence, the second part of the proposition holds. We conclude the proof.

1.3 Random symmetric games with more than two players

In this appendix, we discuss the lower bounds for the expected number of Nash equilibria in random symmetric games with more than two players. Consider a w-player random symmetric game with each player having n strategies. To calculate lower bounds based on Proposition 3, we need to know the expected number of pure strategy Nash equilibria, denoted as ${\mathbb {E}}_{n}^{w}\left[ {{\,\textrm{PE}\,}}\right]$, and the probability of the game having no Nash equilibrium, denoted as ${\mathbb {P}}_{n}^{w}\left( {{\,\textrm{PE}\,}}=0\right)$.

We characterize strategy profiles by the partition of the w players. For example, when $w=3$, there are five partitions: $\left\{ \left\{ 1,2,3\right\} \right\}$, $\left\{ \left\{ 1\right\} ,\left\{ 2,3\right\} \right\}$, $\left\{ \left\{ 1,2\right\} ,\left\{ 3\right\} \right\}$, $\left\{ \left\{ 1,3\right\} ,\left\{ 2\right\} \right\}$, and $\left\{ \left\{ 1\right\} ,\left\{ 2\right\} ,\left\{ 3\right\} \right\}$. Each partition characterizes a group of strategy profiles where players in the same subset play the same strategy, and players in different subsets play different strategies, e.g., $\left\{ \left\{ 1,2\right\} ,\left\{ 3\right\} \right\}$ is the group of strategies where player 1 and 2 choose the same strategy and player 3 chooses a different strategy.

Proposition 5

${\mathbb {E}}_{n}^{w}\left[ {{\,\textrm{PE}\,}}\right] \rightarrow B_{w}$ as $n\rightarrow \infty$, where $B_{w}$ is the Bell number.¹⁴

Proof

Suppose A is a partition of the w players. For any strategy profile s characterized as A, the probability that s is a pure strategy Nash equilibrium is $1/n^{\left| A\right| }$. There are $n!/\left( n-\left| A\right| \right) !$ strategy profiles that induce partition A. Therefore, the expected number of pure strategy Nash equilibria that induce partition A is $n!/\left( n^{\left| A\right| }\left( n-\left| A\right| \right) !\right) \rightarrow 1$ as $n\rightarrow \infty$. Hence, the expected number of pure strategy Nash equilibria equals the number of partitions of the w players, $B_{w}$. $\square$

We now provide a method to calculate ${\mathbb {P}}_{n}^{w}\left( {{\,\textrm{PE}\,}}=0\right)$ as $n\rightarrow \infty$, which is an application of the Chen–Stein method as indicated by Arratia et al. (1989).¹⁵ Note that in a symmetric game, if an action profile is a pure strategy Nash equilibrium, then all permutations of the action profile are also pure strategy Nash equilibria. Hence, we cannot directly apply the Chen–Stein method to the set of all action profiles. Instead, we replicate their application with a slight change to the index set. Below, we use $w=3$ as an example. For $w>3$, the logic is the same, but the computation is more tedious.

When $w=3$, we divide action profiles into three sets. Set $A_{1}$ consists of element $\left\{ \left( x,x,x\right) \right\}$: all three players employ the same action. The expected number of elements in $A_{1}$ that are pure strategy Nash equilibria is one. Set $A_{2}$ consists of element $\left\{ \left( x,x,y\right) ,\left( x,y,x\right) ,\left( y,x,x\right) \right\}$: only two players employ the same action. The expected number of elements in $A_{2}$ that are pure strategy Nash equilibria is $n\left( n-1\right) /n^{2}\rightarrow 1$. Set $A_{3}$ consists of element $\left\{ \left( x,y,z\right) ,\left( x,z,y\right) ,\left( y,x,z\right) ,\left( y,z,x\right) ,\left( z,x,y\right) ,\left( z,y,x\right) \right\}$: all three players employ different actions. The expected number of elements in $A_3$ that are pure strategy Nash equilibria is $n\left( n-1\right) \left( n-1\right) /\left( 6n^{3}\right) \rightarrow 1/6$. We then apply the Chen–Stein method to the union of all elements in $A_1$, $A_2$, and $A_3$. Therefore, the number of elements that are pure strategy Nash equilibria asymptotically follows a Poisson distribution with mean $1+1+1/6=13/6$. And we have ${\mathbb {P}}_{n}^{3}\left( {{\,\textrm{PE}\,}}=0\right) \rightarrow e^{-13/6}$.

Hence, by Proposition 3, we obtain a lower bound for random symmetric games with three players $2B_{3}-1+2e^{-13/6}\approx 9.23$.