Structure coefficients and strategy selection in multiplayer games

Abstract

Evolutionary processes based on two-player games such as the Prisoner’s Dilemma or Snowdrift Game are abundant in evolutionary game theory. These processes, including those based on games with more than two strategies, have been studied extensively under the assumption that selection is weak. However, games involving more than two players have not received the same level of attention. To address this issue, and to relate two-player games to multiplayer games, we introduce a notion of reducibility for multiplayer games that captures what it means to break down a multiplayer game into a sequence of interactions with fewer players. We discuss the role of reducibility in structured populations, and we give examples of games that are irreducible in any population structure. Since the known conditions for strategy selection, otherwise known as $\sigma$-rules, have been established only for two-player games with multiple strategies and for multiplayer games with two strategies, we extend these rules to multiplayer games with many strategies to account for irreducible games that cannot be reduced to those simpler types of games. In particular, we show that the number of structure coefficients required for a symmetric game with $d$-player interactions and $n$ strategies grows in $d$ like $d^{n-1}$. Our results also cover a type of ecologically asymmetric game based on payoff values that are derived not only from the strategies of the players, but also from their spatial positions within the population.

Appendices

Appendix A: Reducibility in well-mixed populations

Proof of Proposition 1

If $u$ is reducible, then we can find

$$\begin{aligned} \left\{ \Big \{ v^{\left\{ i_{1},\ldots ,i_{m}\right\} } : S^{m}\rightarrow {\mathbb {R}}^{m} \Big \}_{\left\{ i_{1},\ldots ,i_{m}\right\} \subseteq \left\{ 1,\ldots ,d\right\} } \right\} _{m=2}^{d-1} \end{aligned}$$

(47)

such that, for each $i=1,\ldots ,d$ and $\left( s_{1},\ldots ,s_{d}\right) \in S^{d}$,

$$\begin{aligned} u_{i}\left( s_{1} , \ldots , s_{d} \right)&= \sum _{m=2}^{d-1}\sum _{\left\{ i_{1},\ldots ,i_{m-1}\right\} \subseteq \left\{ 1,\ldots ,d\right\} -\left\{ i\right\} } v_{i}^{\left\{ i,i_{1},\ldots ,i_{m-1}\right\} }\left( s_{i} , s_{i_{1}} , \ldots ,s_{i_{m-1}} \right) . \end{aligned}$$

(48)

Suppose that $u$ is also symmetric. We will show that the right-hand side of Eq. 48 can be “symmetrized” in a way that preserves the left-hand side of the equation. If ${\mathfrak {S}}_{d}$ denotes the symmetric group on $N$ letters, then

$$\begin{aligned}&u_{i} \left( s_{1} , \ldots , s_{d} \right) \nonumber \\&\quad = \frac{1}{\left( d-1\right) !}\sum _{\begin{array}{c} \pi \in {\mathfrak {S}}_{d} \\ \pi \left( i\right) =i \end{array}} u_{i}\left( s_{\pi \left( 1\right) } , \ldots , s_{\pi \left( d\right) } \right) \nonumber \\&\quad = \frac{1}{\left( d-1\right) !}\sum _{\begin{array}{c} \pi \in {\mathfrak {S}}_{d} \\ \pi \left( i\right) =i \end{array}}\left( \sum _{m=2}^{d-1}\sum _{\left\{ i_{1},\ldots ,i_{m-1}\right\} \subseteq \left\{ 1,\ldots ,d\right\} -\left\{ i\right\} } v_{i}^{\left\{ i,i_{1},\ldots ,i_{m-1}\right\} }\left( s_{i} , s_{\pi \left( i_{1}\right) } , \ldots ,s_{\pi \left( i_{m-1}\right) } \right) \right) \nonumber \\&\quad = \sum _{m=2}^{d-1}\sum _{\left\{ i_{1},\ldots ,i_{m-1}\right\} \subseteq \left\{ 1,\ldots ,d\right\} -\left\{ i\right\} } \left( \frac{1}{\left( d-1\right) !}\sum _{\begin{array}{c} \pi \in {\mathfrak {S}}_{d} \\ \pi \left( i\right) =i \end{array}}v_{i}^{\left\{ i,i_{1},\ldots ,i_{m-1}\right\} }\left( s_{i} , s_{\pi \left( i_{1}\right) } , \ldots ,s_{\pi \left( i_{m-1}\right) } \right) \right) . \end{aligned}$$

(49)

Since

$$\begin{aligned}&\sum _{\begin{array}{c} \pi \in {\mathfrak {S}}_{d} \\ \pi \left( i\right) =i \end{array}} v_{i}^{\left\{ i,i_{1},\ldots ,i_{m-1}\right\} }\left( s_{i} , s_{\pi \left( i_{1}\right) } , \ldots ,s_{\pi \left( i_{m-1}\right) } \right) \nonumber \\&\quad = \left( d-m\right) !\sum _{\left\{ j_{1},\ldots ,j_{m-1}\right\} \subseteq \left\{ 1,\ldots ,d\right\} -\left\{ i\right\} } \sum _{\tau \in {\mathfrak {S}}_{m-1}} v_{i}^{\left\{ i,i_{1},\ldots ,i_{m-1}\right\} }\left( s_{i} , s_{j_{\tau \left( 1\right) }} , \ldots ,s_{j_{\tau \left( m-1\right) }} \right) ,\nonumber \\ \end{aligned}$$

(50)

it follows that

$$\begin{aligned} u_{i}\left( s_{1} , \ldots , s_{d} \right)&= \sum _{m=2}^{d-1}\sum _{\left\{ i_{1},\ldots ,i_{m-1}\right\} \subseteq \left\{ 1,\ldots ,d\right\} -\left\{ i\right\} }w_{i}^{m}\left( s_{i};s_{i_{1}},\ldots ,s_{i_{m-1}}\right) , \end{aligned}$$

(51)

where

$$\begin{aligned}&w_{i}^{m} \left( s_{i};s_{i_{1}},\ldots ,s_{i_{m-1}}\right) \nonumber \\&\quad := \frac{\left( d-m\right) !}{\left( d-1\right) !}\sum _{\left\{ j_{1},\ldots ,j_{m-1}\right\} \subseteq \left\{ 1,\ldots ,d\right\} -\left\{ i\right\} }\sum _{\tau \in {\mathfrak {S}}_{m-1}} v_{i}^{\left\{ i,j_{1},\ldots ,j_{m-1}\right\} }\left( s_{i} , s_{i_{\tau \left( 1\right) }} , \ldots ,s_{i_{\tau \left( m-1\right) }} \right) .\nonumber \\ \end{aligned}$$

(52)

(The semicolon in $w_{i}^{m}\left( s_{i};s_{i_{1}},\ldots ,s_{i_{m-1}}\right)$ is used to distinguish the strategy of the focal player from the strategies of the opponents.) If $\pi$ is a transposition of $i$ and $j$, then the symmetry of $u$ implies that

$$\begin{aligned} u_{i}\left( s_{1} , \ldots , s_{i} , \ldots ,s_{j} , \ldots ,s_{d} \right)&= u_{j}\left( s_{1} , \ldots , s_{j} , \ldots ,s_{i} , \ldots ,s_{d} \right) . \end{aligned}$$

(53)

Therefore, with $v^{m}:=\frac{1}{d}\sum _{i=1}^{d}w_{i}^{m}$ for each $m$, we see that

$$\begin{aligned} u_{i}\left( s_{1} , \ldots , s_{d} \right)&= \sum _{m=2}^{d-1}\sum _{\left\{ i_{1},\ldots ,i_{m-1}\right\} \subseteq \left\{ 1,\ldots ,d\right\} -\left\{ i\right\} }v^{m}\left( s_{i};s_{i_{1}},\ldots ,s_{i_{m-1}}\right) . \end{aligned}$$

(54)

The payoff functions $v^{m}$ are clearly symmetric, so we have the desired result.

We now compare the notion of “degree” of a game (Ohtsuki 2014) to reducibility. For $m\in {\mathbb {R}}$ and $k\in {\mathbb {Z}}_{\geqslant 0}$, consider the (generalized) binomial coefficient

$$\begin{aligned} \left( {\begin{array}{c}m\\ k\end{array}}\right)&:= \frac{m\left( m-1\right) \cdots \left( m-k+1\right) }{k!} . \end{aligned}$$

(55)

(We use this definition of the binomial coefficient so that we can make sense of $\left( {\begin{array}{c}m\\ k\end{array}}\right)$ if $m<k$.) A symmetric $d$-player game with two strategies is $k$-reducible if and only if there exist real numbers $\alpha _{i}^{\ell }$ and $\beta _{i}^{\ell }$ for $\ell =1,\ldots ,k-1$ and $i=0,\ldots ,\ell$ such that

$$\begin{aligned} a_{j}&= \sum _{\ell =1}^{k-1}\sum _{i=0}^{\ell }\left( {\begin{array}{c}j\\ i\end{array}}\right) \left( {\begin{array}{c}d-1-j\\ \ell -i\end{array}}\right) \alpha _{i}^{\ell } ; \end{aligned}$$

(56a)

$$\begin{aligned} b_{j}&= \sum _{\ell =1}^{k-1}\sum _{i=0}^{\ell }\left( {\begin{array}{c}j\\ i\end{array}}\right) \left( {\begin{array}{c}d-1-j\\ \ell -i\end{array}}\right) \beta _{i}^{\ell } . \end{aligned}$$

(56b)

$\ell$ denotes the number of opponents in a smaller game, and $\alpha _{i}^{\ell }$ (resp. $\beta _{i}^{\ell }$) is the payoff to an $a$-player (resp. $b$-player) against $i$ players using $a$ in an $\left( \ell +1\right)$-player game. Ohtsuki (2014) notices that both $a_{j}$ and $b_{j}$ can be written uniquely as polynomials in $j$ of degree at most $d-1$, and he defines the degree of the game to be the maximum of the degrees of these two polynomials. In order to establish a formal relationship between the reducibility of a game and its degree, we need the following lemma:

Lemma 1

If $q\left( j\right)$ is a polynomial in $j$ of degree at most $k\geqslant 1$, then there exist coefficients $\gamma _{i}^{\ell }$ with $\ell =1,\ldots ,k$ and $i=0,\ldots ,\ell$ such that for each $j=0,\ldots ,d-1$,

$$\begin{aligned} q\left( j\right)&= \sum _{\ell =1}^{k}\sum _{i=0}^{\ell }\left( {\begin{array}{c}j\\ i\end{array}}\right) \left( {\begin{array}{c}d-1-j\\ \ell -i\end{array}}\right) \gamma _{i}^{\ell } . \end{aligned}$$

(57)

Proof

If $k=1$, then let $q\left( j\right) =c_{0}+c_{1}j$. For any collection $\gamma _{i}^{\ell }$, we have

$$\begin{aligned} \sum _{\ell =1}^{1}\sum _{i=0}^{\ell }\left( {\begin{array}{c}j\\ i\end{array}}\right) \left( {\begin{array}{c}d-1-j\\ \ell -i\end{array}}\right) \gamma _{i}^{\ell }&= \left( d-1\right) \gamma _{0}^{1} + \left( \gamma _{1}^{1}-\gamma _{0}^{1}\right) j , \end{aligned}$$

(58)

so we can set $\gamma _{0}^{1}=c_{0}/\left( d-1\right)$ and $\gamma _{1}^{1}=c_{0}/\left( d-1\right) + c_{1}$ to get the result. Suppose now that the lemma holds for polynomials of degree $k-1$ for some $k\geqslant 2$, and let

$$\begin{aligned} q\left( j\right)&= c_{0} + c_{1}j + \cdots + c_{k}j^{k} \end{aligned}$$

(59)

with $c_{k}\ne 0$. If $\gamma _{i}^{k}=0$ for $i=0,\ldots ,k-1$ and $\gamma _{k}^{k}=k!c_{k}$, then

$$\begin{aligned} q\left( j\right) - \sum _{i=0}^{k}\left( {\begin{array}{c}j\\ i\end{array}}\right) \left( {\begin{array}{c}d-1-j\\ k-i\end{array}}\right) \gamma _{i}^{k} \end{aligned}$$

(60)

is a polynomial in $j$ of degree at most $k-1$, and thus there exist coefficients $\gamma _{i}^{\ell }$ with

$$\begin{aligned} q\left( j\right) - \sum _{i=0}^{k}\left( {\begin{array}{c}j\\ i\end{array}}\right) \left( {\begin{array}{c}d-1-j\\ k-i\end{array}}\right) \gamma _{i}^{k}&= \sum _{\ell =1}^{k-1}\sum _{i=0}^{\ell }\left( {\begin{array}{c}j\\ i\end{array}}\right) \left( {\begin{array}{c}d-1-j\\ \ell -i\end{array}}\right) \gamma _{i}^{\ell } \end{aligned}$$

(61)

by the inductive hypothesis. The lemma for $k$ follows, which completes the proof. $\square$

We have the following equivalence for two-strategy games:

Proposition 2 If $n=2$, then a game is $k$-reducible if and only if its degree is $k-1$.

Proof

If the game is $k$-reducible, then at least one of the polynomials

$$\begin{aligned} a_{j}&= \sum _{\ell =1}^{k-1}\sum _{i=0}^{\ell }\left( {\begin{array}{c}j\\ i\end{array}}\right) \left( {\begin{array}{c}d-1-j\\ \ell -i\end{array}}\right) \alpha _{i}^{\ell } ; \end{aligned}$$

(62a)

$$\begin{aligned} b_{j}&= \sum _{\ell =1}^{k-1}\sum _{i=0}^{\ell }\left( {\begin{array}{c}j\\ i\end{array}}\right) \left( {\begin{array}{c}d-1-j\\ \ell -i\end{array}}\right) \beta _{i}^{\ell } \end{aligned}$$

(62b)

must have degree $k-1$. Indeed, if they were both of degree at most $k-2$, then, by Lemma 1, one could find coefficients $\gamma _{i}^{\ell }$ and $\delta _{i}^{\ell }$ such that

$$\begin{aligned} a_{j}&= \sum _{\ell =1}^{k-2}\sum _{i=0}^{\ell }\left( {\begin{array}{c}j\\ i\end{array}}\right) \left( {\begin{array}{c}d-1-j\\ \ell -i\end{array}}\right) \gamma _{i}^{\ell } ; \end{aligned}$$

(63a)

$$\begin{aligned} b_{j}&= \sum _{\ell =1}^{k-2}\sum _{i=0}^{\ell }\left( {\begin{array}{c}j\\ i\end{array}}\right) \left( {\begin{array}{c}d-1-j\\ \ell -i\end{array}}\right) \delta _{i}^{\ell } , \end{aligned}$$

(63b)

which would mean that the game is not $k$-reducible. Therefore, the degree of the game must be $k-1$. Conversely, if the degree of the game is $k-1$, then, again by Lemma 1, one can find coefficients $\alpha _{i}^{\ell }$ and $\beta _{i}^{\ell }$ satisfying (62a) and (62b). Since the polynomials in (63a) and (63b) are of degree at most $k-2$ in $j$, and at least one of $a_{j}$ and $b_{j}$ is of degree $k-1$, it follows that the game is not $k'$-reducible for any $k<k'$. In particular, the game is $k$-reducible, which completes the proof. $\square$

Note that we require $n=2$ in Proposition 2 since “degree” is defined only for games with two strategies. (However, $k$-reducibility makes sense for any game.)

Proposition 4

If $p\in \left[ -\infty ,+\infty \right]$ and $S$ is a subset of $\left[ 0,\infty \right)$ that contains at least two elements, then the Hölder public goods game with payoff functions

$$\begin{aligned} u_{i}^{p}&: S^{d} \longrightarrow {\mathbb {R}} \nonumber \\&: \left( x_{1},\ldots ,x_{d}\right) \longmapsto r\left( \frac{x_{1}^{p}+\cdots +x_{d}^{p}}{d}\right) ^{\frac{1}{p}} - x_{i} , \end{aligned}$$

(64)

is irreducible if and only if $\frac{1}{p}\not \in \left\{ 1,2,\ldots ,d-2\right\}$.

Proof

Without a loss of generality, we may assume that $S=\left\{ a,b\right\}$ for some $a,b\in \left[ 0,\infty \right)$ with $b>a$. (If the game is irreducible when there are two strategies, then it is certainly irreducible when there are many strategies.) Since

$$\begin{aligned} a_{j}&= r\left( \frac{\left( j+1\right) a^{p}+\left( d-1-j\right) b^{p}}{d}\right) ^{\frac{1}{p}} - a , \end{aligned}$$

(65)

we see that for $p\ne 0,\pm \infty$,

$$\begin{aligned} \frac{d^{d-1}a_{j}}{dj^{d-1}}&= r\left( \frac{\left( j+1\right) a^{p}+\left( d-1-j\right) b^{p}}{d}\right) ^{\frac{1}{p}-d+1}\left( \frac{a^{p}-b^{p}}{d}\right) ^{d-1}\prod _{i=0}^{d-2}\left( \frac{1}{p}-i\right) , \end{aligned}$$

(66)

which vanishes if and only if $\frac{1}{p}\in \left\{ 1,2,\ldots ,d-2\right\}$. Thus, for $p\ne 0,\pm \infty$, the degree of the Hölder public goods game is $d-1$ if and only if $\frac{1}{p}\not \in \left\{ 1,2,\ldots ,d-2\right\}$.

If $p=0$ and $a\ne 0$, then

$$\begin{aligned} \frac{d^{d-1}b_{j}}{dj^{d-1}}&= rb\left( \frac{a}{b}\right) ^{\frac{j}{d}}\ln \left( \left( \frac{a}{b}\right) ^{\frac{1}{d}}\right) ^{d-1} \ne 0 . \end{aligned}$$

(67)

If $p=0$ and $a=0$, then

$$\begin{aligned} b_{j}&= rb\left( \frac{\left( 1-j\right) \left( 2-j\right) \cdots \left( \left( d-1\right) -j\right) }{\left( d-1\right) !}\right) - b . \end{aligned}$$

(68)

Thus, for $p=0$, the degree of the game is $d-1$. If $p=-\infty$, then

$$\begin{aligned} b_{j}&= r\left( b-a\right) \left( \frac{\left( 1-j\right) \left( 2-j\right) \cdots \left( \left( d-1\right) -j\right) }{\left( d-1\right) !}\right) + ra - b , \end{aligned}$$

(69)

and again the degree is $d-1$. Similarly, if $p=+\infty$, then

$$\begin{aligned} a_{j}&= r\left( a-b\right) \left( \frac{j\left( j-1\right) \cdots \left( j-\left( d-2\right) \right) }{\left( d-1\right) !}\right) + rb - a. \end{aligned}$$

(70)

Since we have shown that the degree of the Hölder public goods game is $d-1$ if and only if $\frac{1}{p}\not \in \left\{ 1,2,\ldots ,d-2\right\}$, the proof is complete by Proposition 2. $\square$

Remark 3

The irreducibility of the games in Examples 2 and 4 follow from the same type of argument used in the proof of Proposition 4.

Appendix B: Selection conditions

We generalize the proof of Theorem 1 of Tarnita et al. (2011) to account for more complicated games:

Asymmetric games Consider an update rule that is symmetric with respect to the strategies and has smooth transition probabilities at $\beta =0$. Suppose that

$$\begin{aligned} \bigcup _{j\in J} \left\{ u_{i}^{j} : S^{d\left( j\right) } \longrightarrow {\mathbb {R}} \right\} _{i=1}^{d\left( j\right) } \end{aligned}$$

(71)

is the collection of all distinct, irreducible payoff functions needed to determine payoff values to the players. The indexing set $J$ is finite if the population is finite, and for our purposes we do not need any other information about $J$. In a game with $n$ strategies, this collection of payoff functions is determined by an element ${\mathbf {u}}\in {\mathbb {R}}^{\sum _{j\in J}\sum _{i=1}^{d\left( j\right) }n^{d\left( j\right) }}$. The assumptions on the process imply that the average abundance of strategy $r\in \left\{ 1,\ldots ,n\right\}$ may be written as a function

$$\begin{aligned} F_{r}&: {\mathbb {R}}^{\sum _{j\in J}\sum _{i=1}^{d\left( j\right) }n^{d\left( j\right) }} \longrightarrow {\mathbb {R}} . \end{aligned}$$

(72)

The coordinates of ${\mathbb {R}}^{\sum _{j\in J}\sum _{i=1}^{d\left( j\right) }n^{d\left( j\right) }}$ will be denoted by ${\mathbf {a}}_{i}^{j}\left( s_{i};s_{i_{1}},\ldots ,s_{i_{d\left( j\right) -1}}\right)$, where $j\in J$, $i\in \left\{ 1,\ldots ,d\left( j\right) \right\}$, and $s_{i},s_{i_{1}},\ldots ,s_{i_{d\left( j\right) -1}}\in \left\{ 1,\ldots ,n\right\}$. (The semicolon is used to separate the strategy of the focal player, $i$, from the strategies of the opponents.) By the chain rule, the selection condition for strategy $r$ has the form

$$\begin{aligned} 0&< \frac{d}{d\beta }\Bigg \vert _{\beta =0}F_{r}\left( \beta {\mathbf {u}}\right) \nonumber \\&= \sum _{j\in J}\sum _{i=1}^{d\left( j\right) }\sum _{s_{i};s_{i_{1}},\ldots ,s_{i_{d\left( j\right) -1}}=1}^{n}\frac{\partial F_{r}}{\partial {\mathbf {a}}_{i}^{j}\left( s_{i};s_{i_{1}},\ldots ,s_{i_{d\left( j\right) -1}}\right) }\Bigg \vert _{{\mathbf {a}}={\mathbf {0}}}u_{i}^{j}\left( s_{i};s_{i_{1}},\ldots ,s_{i_{d\left( j\right) -1}}\right) . \end{aligned}$$

(73)

Let $\displaystyle \alpha _{r}^{ij}\left( s_{i};s_{i_{1}},\ldots ,s_{i_{d\left( j\right) -1}}\right) :=\frac{\partial F_{r}}{\partial {\mathbf {a}}_{i}^{j}\left( s_{i};s_{i_{1}},\ldots ,s_{i_{d\left( j\right) -1}}\right) }\Bigg \vert _{{\mathbf {a}}={\mathbf {0}}}$. Since the update rule is symmetric with respect to the strategies, it follows that

$$\begin{aligned} \alpha _{r}^{ij}\left( s_{i};s_{i_{1}},\ldots ,s_{i_{d\left( j\right) -1}}\right)&= \alpha _{\pi \left( r\right) }^{ij}\left( \pi \left( s_{i}\right) ;\pi \left( s_{i_{1}}\right) ,\ldots ,\pi \left( s_{i_{d\left( j\right) -1}}\right) \right) \end{aligned}$$

(74)

for each $\pi \in {\mathfrak {S}}_{n}$. We now need the following lemma:

Lemma 2

The group action of ${\mathfrak {S}}_{n}$ on $\left[ n\right] ^{\oplus m}$ defined by

$$\begin{aligned} \pi \cdot \left( i_{1},\ldots ,i_{m}\right)&= \left( \pi \left( i_{1}\right) ,\ldots ,\pi \left( i_{m}\right) \right) . \end{aligned}$$

(75)

partitions $\left[ n\right] ^{\oplus m}$ into

$$\begin{aligned} \left| \left[ n\right] ^{\oplus m}{/}{\mathfrak {S}}_{n} \right|&= \sum _{k=0}^{n} S\left( m ,k\right) \end{aligned}$$

(76)

equivalence classes, where $S\left( - ,-\right)$ is the Stirling number of the second kind.

Proof

Let ${\mathbf {P}}\left\{ 1,\ldots ,m\right\}$ be the set of partitions of $\left\{ 1,\ldots ,m\right\}$ and consider the map

$$\begin{aligned} \varPhi&: \left[ n\right] ^{\oplus m} \longrightarrow {\mathbf {P}}\left\{ 1,\ldots ,m\right\} \nonumber \\&: \left( i_{1},\ldots ,i_{m} \right) \longmapsto \varPhi \left( i_{1},\ldots ,i_{m}\right) , \end{aligned}$$

(77)

where, for $\varDelta _{j}\in \varPhi \left( i_{1},\ldots ,i_{m}\right)$, we have $s,t\in \varDelta _{j}$ if and only if $i_{s}=i_{t}$. This map satisfies

$$\begin{aligned} \varPhi \left( \pi \cdot \left( i_{1},\ldots ,i_{m}\right) \right)&= \varPhi \left( i_{1},\ldots ,i_{m}\right) \end{aligned}$$

(78)

for any $\pi \in {\mathfrak {S}}_{n}$, and the map $\varPhi ':\left[ n\right] ^{\oplus m}/{\mathfrak {S}}_{n}\rightarrow {\mathbf {P}}\left\{ 1,\ldots ,m\right\}$ is injective. Thus,

$$\begin{aligned} \left| \left[ n\right] ^{\oplus m}{/}{\mathfrak {S}}_{n} \right|&= \left| \text {Im}\left( \varPhi \right) \right| = \sum _{k=0}^{n} S\left( m ,k\right) , \end{aligned}$$

(79)

which completes the proof. $\square$

By the lemma, the number of equivalence classes of the relation induced by the group action

$$\begin{aligned} \varphi&: {\mathfrak {S}}_{n} \times \left[ n\right] ^{1+d\left( j\right) } \longrightarrow \left[ n\right] ^{1+d\left( j\right) } \nonumber \\&: \left( \pi , \left( r, s_{i}, s_{i_{1}}, \ldots ,s_{i_{d\left( j\right) -1}} \right) \right) \longmapsto \left( \pi \left( r\right) , \pi \left( s_{i}\right) , \pi \left( s_{i_{1}}\right) ,\ldots ,\pi \left( s_{i_{d\left( j\right) -1}}\right) \right) \end{aligned}$$

(80)

is $\sum _{k=0}^{n}S\left( 1+d\left( j\right) ,k\right)$.

Let ${\mathbf {P}}_{n}\left\{ -1,0,1,\ldots ,d\left( j\right) -1\right\}$ be the set of partitions of $\left\{ -1,0,1,\ldots ,d\left( j\right) -1\right\}$ with at most $n$ parts. $-1$ is the index of the strategy in question (in this case, $r$), 0 is the index of the strategy of the focal player, and $1,\ldots ,d\left( j\right) -1$ are the indices of the strategies of the opponents. For $\varDelta \in {\mathbf {P}}_{n}\left\{ -1,0,1,\ldots ,d\left( j\right) -1\right\}$, we let $\overline{u}_{i}^{j}\left( \varDelta ,r\right)$ denote the quantity obtained by averaging $u_{i}^{j}$ over the strategies, once for each equivalence class induced by $\varDelta$ that does not contain $-1$. Stated in this way, the definition is perhaps difficult to digest, so we give a simple example to explain the notation: if $d\left( j\right)$ is even and

$$\begin{aligned} \varDelta&= \left\{ \left\{ -1,0,1\right\} , \left\{ 2,3\right\} , \left\{ 4,5\right\} , \ldots , \left\{ d\left( j\right) -2, d\left( j\right) -1\right\} \right\} , \end{aligned}$$

(81)

then

$$\begin{aligned} \overline{u}_{i}^{j}\left( \varDelta ,r\right)&= n^{-d\left( j\right) /2+1}\!\!\sum _{s_{1},\ldots ,s_{d\left( j\right) /2-1}=1}^{n}u_{i}^{j}\left( r;r,s_{1},s_{1},s_{2},s_{2},\ldots ,s_{d\left( j\right) /2-1},s_{d\left( j\right) /2-1}\right) . \end{aligned}$$

(82)

Using this new notation, we can write

$$\begin{aligned}&\frac{d}{d\beta }\Bigg \vert _{\beta =0}F_{r}\left( \beta {\mathbf {u}}\right) \nonumber \\&\quad = \sum _{j\in J}\sum _{i=1}^{d\left( j\right) }\sum _{s_{i};s_{i_{1}},\ldots ,s_{i_{d\left( j\right) -1}}=1}^{n} \alpha _{r}^{ij}\left( s_{i};s_{i_{1}},\ldots ,s_{i_{d\left( j\right) -1}}\right) u_{i}^{j}\left( s_{i};s_{i_{1}},\ldots ,s_{i_{d\left( j\right) -1}}\right) \nonumber \\&\quad = \sum _{j\in J}\sum _{i=1}^{d\left( j\right) }\sum _{\varDelta \in {\mathbf {P}}_{n}\left\{ -1,0,1,\ldots ,d\left( j\right) -1\right\} } \lambda _{r}^{ij}\left( \varDelta \right) \overline{u}_{i}^{j}\left( \varDelta ,r\right) , \end{aligned}$$

(83)

where each $\lambda _{r}^{ij}\left( \varDelta \right)$ is a linear function of the coefficients $\alpha _{r}^{ij}\left( s_{i};s_{i_{1}},\ldots ,s_{i_{d\left( j\right) -1}}\right)$ (the precise linear expression is unimportant). As a consequence of Eq. (74), we see that $\lambda _{r}^{ij}\left( \varDelta \right) =\lambda _{r'}^{ij}\left( \varDelta \right)$ for each $r,r'\in \left\{ 1,\ldots ,n\right\}$, so we may relabel these coefficients using the notation $\lambda ^{ij}\left( \varDelta \right)$. Since $\sum _{r=1}^{n}F_{r}\left( \beta {\mathbf {u}}\right) =1$, it follows that

$$\begin{aligned} 0&= \sum _{r=1}^{n}\frac{d}{d\beta }\Bigg \vert _{\beta =0}F_{r}\left( \beta {\mathbf {u}}\right) = \sum _{j\in J}\sum _{i=1}^{d\left( j\right) }\sum _{\varDelta \in {\mathbf {P}}_{n}\left\{ -1,0,1,\ldots ,d\left( j\right) -1\right\} } \lambda ^{ij}\left( \varDelta \right) \sum _{r=1}^{n}\overline{u}_{i}^{j}\left( \varDelta ,r\right) . \end{aligned}$$

(84)

Let $R_{j}:=\left\{ \varDelta \in {\mathbf {P}}_{n}\left\{ -1,0,1,\ldots ,d\left( j\right) -1\right\} \ :\ -1\sim 0 \right\}$ and write

$$\begin{aligned}&\sum _{\varDelta \in {\mathbf {P}}_{n}\left\{ -1,0,1,\ldots ,d\left( j\right) -1\right\} }\lambda ^{ij}\left( \varDelta \right) \sum _{r=1}^{n}\overline{u}_{i}^{j}\left( \varDelta ,r\right) \nonumber \\&\quad = \sum _{\varDelta \in R_{j}} \lambda ^{ij}\left( \varDelta \right) \sum _{r=1}^{n}\overline{u}_{i}^{j}\left( \varDelta ,r\right) + \sum _{\varDelta \in R_{j}^{c}} \lambda ^{ij}\left( \varDelta \right) \sum _{r=1}^{n}\overline{u}_{i}^{j}\left( \varDelta ,r\right) . \end{aligned}$$

(85)

For $\varDelta \in R_{j}^{c}$, let $\varDelta _{-1}$ and $\varDelta _{0}$ be the sets containing $-1$ and $0$, respectively. Since $\varDelta \in R_{j}^{c}$, we know that $\varDelta _{-1}\cap \varDelta _{0}=\varnothing$. Now, let $\eta \left( \varDelta \right)$ be the partition whose sets are equal to those in $\varDelta$ with the exception of $\varDelta _{-1}$ and $\varDelta _{0}$, which are replaced by $\left\{ -1\right\} \cup \varDelta _{0}$ and $\varDelta _{-1}-\left\{ -1\right\}$, respectively. For example, if $d\left( j\right) =5$ and

$$\begin{aligned} \varDelta&= \Big \{ \left\{ -1,2,3\right\} , \left\{ 0,1\right\} , \left\{ 4\right\} \Big \}, \end{aligned}$$

(86)

then

$$\begin{aligned} \eta \left( \varDelta \right)&= \Big \{ \left\{ -1,0,1\right\} , \left\{ 2,3\right\} , \left\{ 4\right\} \Big \} . \end{aligned}$$

(87)

This assignment defines a surjective map $\eta :R_{j}^{c}\rightarrow R_{j}$.

For fixed $j$ and $i$, consider the equivalence relation on ${\mathbf {P}}_{n}\left\{ -1,0,1,\ldots ,d\left( j\right) -1\right\}$ defined by

$$\begin{aligned} \varDelta \sim \varDelta '&\iff \sum _{r=1}^{n}\overline{u}_{i}^{j}\left( \varDelta ,r\right) = \sum _{r=1}^{n}\overline{u}_{i}^{j}\left( \varDelta ' ,r\right) . \end{aligned}$$

(88)

The map $\eta :R_{j}^{c}\rightarrow R_{j}$ satisfies $\varDelta \sim \eta \left( \varDelta \right)$ for $\varDelta \in R_{j}^{c}$. Therefore, it follows that

$$\begin{aligned} \frac{d}{d\beta }\Bigg \vert _{\beta =0}F_{r}\left( \beta {\mathbf {u}}\right)&= \sum _{j\in J}\sum _{i=1}^{d\left( j\right) }\sum _{\varDelta \in R_{j}^{c}} \lambda ^{ij}\left( \varDelta \right) \Big ( \overline{u}_{i}^{j}\left( \varDelta ,r\right) - \overline{u}_{i}^{j}\left( \eta \left( \varDelta \right) , r\right) \Big ) , \end{aligned}$$

(89)

so the selection condition for strategy $r$ is

$$\begin{aligned} \sum _{j\in J}\sum _{i=1}^{d\left( j\right) }\sum _{\varDelta \in R_{j}^{c}} \lambda ^{ij}\left( \varDelta \right) \Big ( \overline{u}_{i}^{j}\left( \varDelta ,r\right) - \overline{u}_{i}^{j}\left( \eta \left( \varDelta \right) , r\right) \Big )&>0 , \end{aligned}$$

(90)

which, for each $j$ and $i$, involves $\sum _{k=0}^{n}\Big ( S\left( 1+d\left( j\right) , k\right) - S\left( d\left( j\right) , k\right) \Big )$ structure coefficients. To be consistent with the existing literature on selection conditions, we let $\sigma ^{ij}:=-\lambda ^{ij}$ and write the selection condition as

$$\begin{aligned} \sum _{j\in J}\sum _{i=1}^{d\left( j\right) }\sum _{\varDelta \in R_{j}^{c}} \sigma ^{ij}\left( \varDelta \right) \Big ( \overline{u}_{i}^{j}\left( \eta \left( \varDelta \right) ,r\right) - \overline{u}_{i}^{j}\left( \varDelta ,r\right) \Big )&> 0 . \end{aligned}$$

(91)

As long as $n\geqslant 1+\max _{j\in J}d\left( j\right)$, then the number of structure coefficients in selection condition (91) is independent of $n$. The same argument used in the proof of Theorem 1 of Tarnita et al. (2011) shows that each $\sigma ^{ij}\left( \varDelta \right)$ may be chosen to be independent of $n$ for all games with at least $1+\max _{j\in J}d\left( j\right)$ strategies. One could calculate the structure coefficients for a game with exactly $1+\max _{j\in J}d\left( j\right)$ strategies and still obtain the selection condition for games with fewer strategies in exactly the same way that Tarnita et al. (2011) deduce the result for $n=2$ strategies from the result for $n\geqslant 3$ strategies.

Symmetric games If we assume that the payoff functions of display (71) are all symmetric, then for each $j\in J$ the function $u_{i}^{j}$ is independent of $i$: the index $j$ is used to denote the particular group of players involved in the interaction (of size $d\left( j\right)$), and $u_{i}^{j}$ denotes the payoff function to the $i$th player in this group. If the game specified by $j$ is symmetric, then we need know only the payoff values to one of these players. Therefore, we assume in this setting that the collection of all payoff functions needed to determine payoff values is

$$\begin{aligned} \bigcup _{j\in J} \Big \{ u^{j} : S^{d\left( j\right) } \longrightarrow {\mathbb {R}} \Big \} . \end{aligned}$$

(92)

Condition (91) then takes the form

$$\begin{aligned} \sum _{j\in J}\sum _{\varDelta \in R_{j}^{c}} \sigma ^{j}\left( \varDelta \right) \Big ( \overline{u}^{j}\left( \eta \left( \varDelta \right) ,r\right) - \overline{u}^{j}\left( \varDelta ,r\right) \Big )&> 0 . \end{aligned}$$

(93)

For fixed $j$ and $\varDelta \in R_{j}^{c}$, we again let $\varDelta _{-1}$ and $\varDelta _{0}$ be the sets in $\varDelta$ that contain $-1$ and $0$, respectively. The collection $\varDelta -\left\{ \varDelta _{-1},\varDelta _{0}\right\}$ defines a partition of the number $1+d\left( j\right) -\left| \varDelta _{-1}\right| -\left| \varDelta _{0}\right|$ whose parts are the sizes of the sets in the collection $\varDelta -\left\{ \varDelta _{-1},\varDelta _{0}\right\}$. For example, if $d\left( j\right) =12$ and

$$\begin{aligned} \varDelta&= \Big \{ \left\{ -1,2,3\right\} \!, \left\{ 0,1\right\} \!, \left\{ 4,5\right\} \!, \left\{ 6,7,8\right\} \!, \left\{ 9,10,11\right\} \Big \}\!, \end{aligned}$$

(94)

then $\varDelta$ defines the partition $2+3+3=8$ of the number $1+12-3-2=8$. An equivalence relation may then be defined on $R_{j}^{c}$ by letting $\varDelta \sim \varDelta '$ if and only if the following three conditions hold:

• (a) $\left| \varDelta _{-1}\right| =\left| \varDelta _{-1}'\right|$;

• (b) $\left| \varDelta _{0}\right| =\left| \varDelta _{0}'\right|$;

• (c) the partitions of $1+d\left( j\right) -\left| \varDelta _{-1}\right| -\left| \varDelta _{0}\right|$ defined by $\varDelta$ and $\varDelta '$ are the same.

The symmetry of $u^{j}$ implies that if $\varDelta ,\varDelta '\in R_{j}^{c}$ and $\varDelta \sim \varDelta '$, then

$$\begin{aligned} \overline{u}^{j}\left( \varDelta , r\right)&= \overline{u}^{j}\left( \varDelta ', r\right) ; \end{aligned}$$

(95a)

$$\begin{aligned} \overline{u}^{j}\left( \eta \left( \varDelta \right) , r\right)&= \overline{u}^{j}\left( \eta \left( \varDelta '\right) , r\right) . \end{aligned}$$

(95b)

Therefore, condition (93) becomes

$$\begin{aligned} \sum _{j\in J}\sum _{\varDelta \in R_{j}^{c}/\sim } \sigma ^{j}\left( \varDelta \right) \Big ( \overline{u}^{j}\left( \eta \left( \varDelta \right) ,r\right) - \overline{u}^{j}\left( \varDelta , r\right) \Big )&> 0. \end{aligned}$$

(96)

For each $j$, the number of structure coefficients contributed to this condition by payoff function $u^{j}$ is

$$\begin{aligned} \left| R_{j}^{c} / \sim \right|&= \sum _{c+k=d\left( j\right) -1}1 + \sum _{0\leqslant c+k<d\left( j\right) -1}q\Big (d\left( j\right) -1-c-k,n-2\Big ) \nonumber \\&= d\left( j\right) + \sum _{m=1}^{d\left( j\right) -1}\Big (d\left( j\right) -m\Big ) q\left( m,n-2\right) , \end{aligned}$$

(97)

where $q\left( m,k\right)$ denotes the number of partitions of $m$ with at most $k$ parts.

Appendix C: Explicit calculations

Let ${\mathbf {M}}$ be the transition matrix for an evolutionary process with mutations. The existence of nontrivial strategy mutations ensures that this chain is irreducible, so there is a unique stationary distribution, $\mu$, by the Perron–Frobenius theorem. Let ${\mathbf {M}}':={\mathbf {M}}-{\mathbf {I}}$, and let ${\mathbf {M}}'\left( i,\nu \right)$ be the matrix obtained by replacing the $i$th column of ${\mathbf {M}}'$ by $\nu$. Press and Dyson (2012) show that this stationary distribution satisfies

$$\begin{aligned} \mu \cdot \nu&= \frac{\det {\mathbf {M}}'\left( i,\nu \right) }{\det {\mathbf {M}}'\left( i,{\mathbf {1}}\right) } \end{aligned}$$

(98)

for any vector, $\nu$. (${\mathbf {1}}$ is the vector of ones.) Therefore, if $\psi _{r}$ is the vector indexed by ${\mathcal {S}}$ with $\psi _{r}\left( s\right)$ being the density of strategy $r$ in state $s$, then the selection function (72) may be written

$$\begin{aligned} F_{r}&= \mu \cdot \psi _{r} = \frac{\det {\mathbf {M}}'\left( i,\psi _{r}\right) }{\det {\mathbf {M}}'\left( i,{\mathbf {1}}\right) }. \end{aligned}$$

(99)

By the quotient rule and Jacobi’s formula for the derivative of a determinant,

$$\begin{aligned} \frac{dF_{r}}{d\beta }\Bigg \vert _{\beta =0}&= \frac{\det {\mathbf {M}}'\left( i,\psi _{r}\right) }{\det {\mathbf {M}}'\left( i,{\mathbf {1}}\right) }\Bigg \vert _{\beta =0} \nonumber \\&\quad \times \text {tr}\left( {\mathbf {M}}'\left( i,\psi _{r}\right) \vert _{\beta =0}^{-1}\frac{d}{d\beta }\Big \vert _{\beta =0}{\mathbf {M}}'\left( i,\psi _{r}\right) - {\mathbf {M}}'\left( i,{\mathbf {1}}\right) \vert _{\beta =0}^{-1}\frac{d}{d\beta }\Big \vert _{\beta =0}{\mathbf {M}}'\left( i,{\mathbf {1}}\right) \right) \nonumber \\&= \frac{1}{n}\text {tr}\left( {\mathbf {M}}'\left( i,\psi _{r}\right) \vert _{\beta =0}^{-1}\frac{d}{d\beta }\Big \vert _{\beta =0}{\mathbf {M}}'\left( i,\psi _{r}\right) - {\mathbf {M}}'\left( i,{\mathbf {1}}\right) \vert _{\beta =0}^{-1}\frac{d}{d\beta }\Big \vert _{\beta =0}{\mathbf {M}}'\left( i,{\mathbf {1}}\right) \right) \nonumber \\&= \frac{1}{n}\text {tr}\left( \left( {\mathbf {M}}'\left( i,\psi _{r}\right) \vert _{\beta =0}^{-1} - {\mathbf {M}}'\left( i,{\mathbf {1}}\right) \vert _{\beta =0}^{-1}\right) \frac{d}{d\beta }\Big \vert _{\beta =0}{\mathbf {M}}'\left( i,{\mathbf {0}}\right) \right) \end{aligned}$$

(100)

since all strategies have equilibrium density $1/n$ when $\beta =0$.

In general, the dimension of ${\mathbf {M}}$ is quite large, so this method is not feasible for large structured populations. However, in well-mixed populations, one can greatly reduce the size of the state space of the Markov chain by keeping track of only the number of each strategy present in the population. If there are $n$ strategies and $N=d$ players, then a state of the population may effectively be described by an $n$-tuple $\left( k_{1},\ldots ,k_{n}\right)$, where $k_{r}$ is the number of players using strategy $r$ for $r=1,\ldots ,n$. Clearly $k_{1}+\cdots +k_{n}=d$, so the total size of the state space is $\left( {\begin{array}{c}d+n-1\\ n-1\end{array}}\right)$.

Using (100), we may explicitly calculate the selection conditions for well-mixed populations as long as $d$ and $n$ are small. These selection conditions could be calculated directly from (99), but (100) is more efficient on computer algebra systems. Each data point in Fig. 5 in the main text was generated using a $d$-player game in a population of size $N=d$, ie. every player in the population participates in every interaction. The growth clearly supports the prediction of Proposition 3.