In this section, we introduce our classification. Section 4.1 presents a theorem stating that any \(2 \times 2\) game can be decomposed into a common interest game and a zero-sum game, and show how this can be used to describe the tension between the common interest and the self interest of the players. Section 4.2 then considers the classification of symmetric \(2 \times 2\) games this gives rise to.
Decomposition of symmetric \(2 \times 2\) games
Any game can be decomposed into a common interest game and a zero-sum game (Kalai & Ehud, 2013). This is shown in Proposition 4.1 using the following natural convention for addition of bimatrices:
$$\begin{aligned} (P_1,P_2) + (P'_1,P'_2) = (P_1 + P'_1,P_2 + P'_2). \end{aligned}$$
Proposition 4.1
(Decomposition) A 2-player game with payoff bimatrix P can always be decomposed into the sum of a zero-sum game with payoff bimatrix Z and a game of common interest with payoff bimatrix C:
$$\begin{aligned} P = C + Z. \end{aligned}$$
Proof
Let P be the payoff bimatrix of an arbitrary 2-player game. P can be decomposed as follows:
$$\begin{aligned} P&= (P_1,P_2)\\ &= \frac{1}{2}\left( (P_1 + P_2), ~(P_1 + P_2)\right) + \frac{1}{2}\left( (P_1 - P_2), ~(P_2 - P_1)\right) \\ &= C + Z. \end{aligned}$$
Here \(C = \frac{1}{2}((P_1 + P_2), ~(P_1 + P_2))\) is a common interest game (Definition 2.4), and \(Z = \frac{1}{2}((P_1 - P_2), ~(P_2 - P_1))\) is a zero-sum game (Definition 2.3). \(\square \)
Theorem 4.1 shows how a general 2-player game bimatrix can be decomposed into one common interest game and a zero-sum game. We will interchangeably refer to the zero-sum game as the conflict part of the game, since the players’ incentives are completely opposed in these games. In the special case of \(2 \times 2\) symmetric games, the decomposition of the payoff matrix has the following form:
$$\begin{aligned} \begin{bmatrix} (a,a) &{} (b,c) \\ (c,b) &{} (d,d) \end{bmatrix} = \begin{bmatrix} (a,a) &{} (\frac{b+c}{2}, \frac{b+c}{2}) \\ (\frac{b+c}{2}, \frac{b+c}{2}) &{} (d,d) \end{bmatrix} + \begin{bmatrix} (0,0) &{} (\frac{b-c}{2}, - \frac{b-c}{2}) \\ (- \frac{b-c}{2}, \frac{b-c}{2}) &{} (0,0) \end{bmatrix}. \end{aligned}$$
(2)
The decomposition separates the common interest part from the conflict part of the game. We define the variables \(c_1\), \(c_2\) and \(z_1\) for the incentives in the common interest and the zero-sum parts (see Fig. 4). In the definition, \(u_i^C\) and \(u_i^Z\) are the payoffs for player i in the common interest part and the zero-sum part, respectively.
$$\begin{aligned} c_1= u_1^C(10) - u_1^C(00) = \frac{b+c}{2} - a \end{aligned}$$
(3)
$$\begin{aligned} c_2= u_2^C(11) - u_2^C(10) = d - \frac{ b + c }{ 2 } \end{aligned}$$
(4)
$$\begin{aligned} z_1= u_1^Z(10) - u_1^Z(00) = - \frac{b-c}{2}-0 = -\frac{ b - c }{ 2 }. \end{aligned}$$
(5)
Remark 4.2
In symmetric \(2 \times 2\) games we only need \(c_1, c_2\) and \(z_1\) to describe the incentives in the common interest and the zero-sum parts. For general \(2 \times 2\) games one could also define \(c_3 = u_2^C(01) - u_2^C(00), c_4 = u_1^C(11) - u_1^C(01)\) and define \(z_2, z_3\) and \(z_4\) analogously. However in symmetric \(2 \times 2\) games \(c_1 = c_3, c_2 = c_4\) and \(z_k = z_l~\forall ~k,l \in \{1,2,3,4\}\) as shown in Fig. 4.
The variables \(c_1, c_2\) and \(z_1\) allow us to compare the strength of the incentives in the common interest part with the strengths of the incentives in the conflict part, and to check whether the incentives are aligned or not. For example, if \(c_1>0\), then player 1 prefers outcome \(\{1,0\}\) in the common interest matrix over outcome \(\{0,0\}\). The sign of the variables determine what outcomes the players prefer. In Fig. 4 we represent these preferences as arrows, where a positive sign of the variables mean that the arrows point in the direction shown in the figure, and in the opposite direction if the sign is negative. The vertical arrows represent the preferences of player 1 and the horizontal arrows represent the preferences of player 2. The absolute value of the variables \(c_1\), \(c_2\) and \(z_1\) represent the strengths of the players’ preferences.
In the five standard games (Prisoner’s Dilemma, Chicken, Stag Hunt, Leader and Hero), the conflict part of the decomposition counteracts the common interest part of the decomposition. Expressed in our variables, this means that the conflict variable \(z_1\) and at least one of the common interest variables \(c_1\) or \(c_2\) have opposite signs and that the value of \(z_1\) is greater than this variable. Consider for example the decomposition of Prisoner’s Dilemma in Example 4.3.
Example 4.3
In Prisoner’s Dilemma, the players have to choose between cooperation and defection. The temptation of the selfish choice is stronger than the incentive to cooperate and as result \(\{1,1\}\) (both players defect) is the unique NE even though both players could get a higher payoff from cooperating. The game below is an example of a Prisoner’s Dilemma game. In Fig. 5 below, the decomposition of the Prisoner’s Dilemma bimatrix is presented. As suggested by the arrows, the conflict part of the decomposition draws the players toward the selfish outcome and it is strong enough to counteract both of the common interest variables that draw the players toward the cooperation outcome.
In the common interest game both players prefer outcome \(\{0,0\}\) (cooperation) and hence the arrows point in this direction. In the conflict game player 1 prefers the \(\{1,0\}\) outcome and player 2 prefers the \(\{0,1\}\) outcome. As a result, their combined incentives draw them toward the \(\{1,1\}\) outcome, as the arrows suggest. Because the payoff differences in the conflict game is larger than those in the common interest game, the incentives in the conflict game counteracts the incentives in the common interest game. This can be told from the conflict and common interest variables. We have \(z_1 = 1.5\) and \(c_1 = c_2 = -0.5\). Because the \(z_1\) variable and the \(c_1\) and \(c_2\) variables have different signs the conflict counteracts the common interest and since \(|z_1| > \max (|c_1|,|c_2|)\), the conflict is strong enough to overpower the common interest. Therefore the selfish outcome \(\{1,1\}\) is the NE of the game.
Strategic equivalence Games with different payoffs and tension vectors may be strategically equivalent. Indeed, except for the zero-conflict game with \(z_1=0\), all games are equivalent to a game with \(z_1=1\).
Proposition 4.4
Every game with non-zero conflict is strategically equivalent with a game with
\(z_1=1.\)
Proof
According to Definition 2.5, if the payoff matrix of a game can be constructed by permuting the rows and columns of another game’s payoff matrix, then the games are strategically equivalent. This means that the two matrices in Eq. 6 are strategically equivalent:
$$\begin{aligned} A = \begin{bmatrix} a &{} b \\ c &{} d \end{bmatrix} \sim \begin{bmatrix} d &{} c \\ b &{} a \end{bmatrix} = A'. \end{aligned}$$
(6)
These matrices have the following decomposition.
$$\begin{aligned}A = \begin{bmatrix} a &{} b \\ c &{} d \end{bmatrix} = \begin{bmatrix} a &{} \frac{b+c}{2} \\ \frac{b+c}{2} &{} d \end{bmatrix} + \begin{bmatrix} 0 &{} \frac{b-c}{2} \\ -\frac{b-c}{2} &{} 0 \end{bmatrix} \end{aligned}$$
(7)
$$\begin{aligned}A' = \begin{bmatrix} d &{} c \\ b &{} a \end{bmatrix} = \begin{bmatrix} d &{} \frac{b+c}{2} \\ \frac{b+c}{2} &{} a \end{bmatrix} + \begin{bmatrix} 0 &{} -\frac{b-c}{2} \\ \frac{b-c}{2} &{} 0 \end{bmatrix}. \end{aligned}$$
(8)
That is, \(c_1 = -c'_2\), \(c_2 = -c'_1\) and \(z_1 = - z'_1\) and from this we can tell that every point in \({{\mathbb{R}}}^3\) with \(z_1 < 0\) represents a game that is strategically equivalent with a game such that \(z_1 > 0\). Hence we only need to consider games with \(z_1 \ge 0\).
Scalar invariance defined in Definition 2.5 states that to games with payoff matrices A and \(A'\) respectively are strategically equivalent if there exists \(\alpha > 0\) such that \(A' = \alpha A\). Multiplying the game matrix with a constant \(\alpha \) is equivalent with multiplying all of the payoffs with \(\alpha \). Since \( z_1 = -\frac{ b - c }{ 2 }\), the relation \(A' = \alpha A\) implies that \( z_1' = -\frac{ \alpha b - \alpha c }{ 2 } = \alpha z_1\). This means that every game with \(z_1 > 0\) has a strategically equivalent game with \(z_1 = 1\). Combined with the observation above, this completes the proof. \(\square \)
Classification
The common interest and conflict variables \(c_1\), \(c_2\), \(z_1\) locate games in \({\mathbb{R}}^3\) via the vector \([c_1, c_2, z_1]^{\mathrm{t}}\). It is easily shown that every game with negative \(z_1\) is strategically equivalent with a game with positive \(z_1\) (see proof of Proposition 4.4). A partition of \({\mathbb{R}}^2\times {\mathbb{R}}_+\) thereby groups games into classes. The arguably simplest partitioning of \({\mathbb{R}}^2\times {\mathbb{R}}_+\) is given by the planes spanned by any two coordinate axes:
$$\begin{aligned} c_1= 0 \end{aligned}$$
(9)
$$\begin{aligned} c_2= 0. \end{aligned}$$
(10)
This turns out to be a far too coarse classification, with significantly different games such as Prisoner’s Dilemma and Stag Hunt sharing regions. The second simplest set of planes is arguably the diagonal planes:
$$\begin{aligned} c_1 = c_2\quad c_1 = -c_2 \end{aligned}$$
(11)
$$\begin{aligned} c_1 = z_1 \quad c_1 = -z_1 \end{aligned}$$
(12)
$$\begin{aligned} c_2 = z_1\quad c_2 = -z_1. \end{aligned}$$
(13)
Splitting \({\mathbb{R}}^2\times {\mathbb{R}}_+\) with the planes defined in Eqs. 9–13 gives a accurate game classification.
Proposition 4.5
The planes \(c_1 = 0, c_2 = 0, z_1 = \pm c_1, z_1 = \pm c_2\) and \(c_1 = \pm c_2\) divide \({\mathbb{R}}^2 \times {\mathbb{R}}_+\) into 24 different regions.
Proof
The first two planes defines 2 regions each and since they are pairwise orthogonal this results in \(2^2 = 4\) regions in total. The lengths \(|c_1|,|c_2|\) and \(|z_1|\) can be ordered in \(3! = 6\) ways independently of the signs of \(c_1\) and \(c_2\). Hence the planes divide \({{\mathbb{R}}}^3\) into \(2^2 \times 3! = 24\) regions in total. \(\square \)
By employing Proposition 4.4 we can visualise the classification through its intersection with \(z_1 = 1\). The visualisation is shown in Fig. 6.
As we will establish in the next subsection, all of the standard games have their own regions. For example, the four Prisoner’s Dilemma regions can be found in the high-conflict part of the map. The Common Interest regions consist of games where the conflict variable is weaker than both of the common interest variables, or where the conflict variable and the common interest variables have the same sign. In the Partial Conflict regions the conflict variable and one of the common interest variables have opposite signs and the conflict variable is strong enough to dominate the common interest variable.
Interpreting the boundaries
What explains the success of this classification principle is that the planes generated by Eqs. 9–13 capture points where either the alignment between common interest and conflict turns to disalignment, or the relative strength between two components \(c_1\), \(c_2\), \(z_1\) shifts. In the following subsections we analyse interpretations and implications of the planes more closely, and link our 24 regions to the some of the standard games. We also determine where Nash equilibria (NE), Definition 2.6, and Altruistic Equilibria (AE), Definition 2.7, are located in different regions.
Bijective transformation
As a first step for our more careful analysis, we will add an auxiliary variable x in addition to \(c_1\), \(c_2\) and \(z_1\), to make the transformation from the payoff parameters a, b, c, d bijective. In Fig. 4 we define the variables \(c_1, c_2 \) and \(z_1\) according to the transformation
$$\begin{aligned} \begin{bmatrix} c_1 \\ c_2 \\ z_1 \end{bmatrix} = \frac{1}{2} \begin{bmatrix} -2 &{} 1 &{} 1 &{} 0 \\ 0 &{} -1 &{} -1 &{} 2 \\ 0 &{} -1 &{} 1 &{} 0 \end{bmatrix} \begin{bmatrix} a \\ b \\ c \\ d \end{bmatrix}. \end{aligned}$$
(14)
The transformation in Eq. 14 is not invertible. To make it invertible, we add the auxiliary variable \(x \triangleq \frac{a+b+c+d}{2}\). The resulting transformation
$$\begin{aligned} \begin{bmatrix} x \\ c_1 \\ c_2 \\ z_1 \end{bmatrix} = A \begin{bmatrix} a \\ b \\ c \\ d \end{bmatrix} = \frac{1}{2} \begin{bmatrix} 1 &{} 1 &{} 1 &{} 1 \\ -2 &{} 1 &{} 1 &{} 0 \\ 0 &{} -1 &{} -1 &{} 2 \\ 0 &{} -1 &{} 1 &{} 0 \end{bmatrix} \begin{bmatrix} a \\ b \\ c \\ d \end{bmatrix} \end{aligned}$$
(15)
is invertible, with determinant \(|A| = -1\).
Note that changing the auxiliary variable x is equivalent to adding a constant to the payoffs. It preserves strategic equivalence. This in combination with the fact that the transformation in Eq. 15 is invertible and Proposition 4.4 implies that every point in the map in Fig. 6 represents exactly one strategic equivalence class of games. It also means that every strategic equivalence class of games is represented by exactly one point in the map. This is desirable since there is no interesting difference between strategically equivalent games. Hence it is not meaningful to represent the same strategically equivalence class with more than one point in a classification map.
By inverting the transformation matrix A in Eq. 15, we can express the payoffs a, b, c, d, in terms of \(c_1, c_2, z_1\) and x:
$$\begin{aligned} \begin{bmatrix} a \\ b \\ c \\ d \end{bmatrix} = A^{-1} \begin{bmatrix} x \\ c_1 \\ c_2 \\ z_1 \end{bmatrix} = \frac{1}{4} \begin{bmatrix} 2 &{} -3 &{} -1 &{} 0\\ 2 &{} 1 &{} -1 &{} -4\\ 2 &{} 1 &{} -1 &{} 4\\ 2 &{} 1 &{} 3 &{} 0 \end{bmatrix} \begin{bmatrix} x \\ c_1 \\ c_2 \\ z_1 \end{bmatrix}. \end{aligned}$$
(16)
Being able to convert between our variables \(c_1,c_2,z_1\) and the payoffs a, b, c, d will be useful in establishing Propositions 4.7 and 4.8 below. The next three subsections will investigate each of the planes generated by Eqs. 9–13 in turn.
The \(c_i=\pm z_1\) conditions, and NE and AE regions
We next state and prove two propositions expressing conditions for NE and AE in terms of our cooperation and conflict variables \(c_1, c_2\) and \(z_1\). The concept of NE is standard in game theory. The concept of AE is not as commonly known as NE, but it does nevertheless play an important part in many games, for example Deadlock, Chicken and in the symmetric version of Battle of the Sexes.
To be able to express the NE and AE conditions in terms of \(c_1\), \(c_2\) and \(z_1\) we need to introduce Lemma 4.6.
Lemma 4.6
The following list states the NE conditions expressed in terms of a, b, c and d.
-
1.
\(\{0,0\}\) is NE iff \(a \ge c.\)
-
2.
\(\{0,1\}\) is NE iff \(b \ge d\) and \(c \ge a.\)
-
3.
\(\{1,0\}\) is NE iff \(c \ge a\) and \(b \ge d.\)
-
4.
\(\{1,1\}\) is NE iff \(d \ge b.\)
Proof
We provide the proof for the first case. The proof for the other three cases are analogous. According to Definition 2.6, \(\{0,0\}\) is NE if and only if \(u_1(0,0) \geqslant u_1 (1,0)\) and \(u_2(0,0) \geqslant u_2(0,1)\). But \(u_1(0,0) \geqslant u_1(1,0) \Leftrightarrow a \ge c\) and \(u_2(0,0) \geqslant u_2(0,1) \Leftrightarrow a \ge c\). That is \(\{0,0\}\) is NE iff \(a \ge c\). \(\square \)
Proposition 4.7
In any symmetric \(2\times 2,\) two-player game G with cooperation and conflict variables \(c_1, c_2\) and \(z_1,\) the following statements are true:
-
(i)
\(\{0,0\}\) is NE if and only if \(z_1 + c_1 \le 0,\)
-
(ii)
\(\{0,1\}\) and \(\{1,0\}\) are NE if and only if \(z_1 + c_1 \ge 0\) and \(z_1 + c_2 \le 0,\)
-
(iii)
\(\{1,1\}\) is NE if and only if \(z_1 + c_2 \ge 0.\)
Proof
Let \({\vec{y}} = (a, b, c, d)^{\mathrm{T}}\) be a vector with the payoffs for G, and let
$$\begin{aligned} {\vec{v}} = (x, c_1, c_2, z_1)^{\mathrm{T}} = A {\vec{y}}, \end{aligned}$$
where A is the matrix in Eq. 15. Equation 16 gives us that the payoff vector \({\vec{y}}\) can be written \( {\vec{y}} = A^{-1} {\vec{v}}\), and hence
$$\begin{aligned} a&= \frac{1}{4}(2x-3c_1-c_2)\quad b= \frac{1}{4}(2x+c_1-c_2-4z_1)\\ c&=\frac{1}{4}(2x+c_1-c_2+4z_1)\quad d= \frac{1}{4}(2x+c_1+3c_2). \end{aligned}$$
(17)
Combining the list and Eq. 17 with Lemma 4.6, we can express these conditions in terms of the variables \(c_1\), \(c_2\) and \(z_1\).
-
(i)
\(\{0,0\}\) is NE if and only if
$$\begin{aligned} a \ge c \iff \frac{1}{4}(2x-3c_1-c_2) \ge \frac{1}{4}(2x+c_1-c_2+4z_1) \iff z_1 + c_1 \le 0. \end{aligned}$$
-
(ii)
\(\{0,1\}\) and \(\{1,0\}\) are NE if and only if \(b \ge d\) and \(c \ge a\).
$$\begin{aligned} b \ge d \iff \frac{1}{4}(2x+c_1-c_2-4z_1) \ge \frac{1}{4}(2x+c_1+3c_2) \iff z_1 + c_2 \le 0. \end{aligned}$$
Similarly, \(c \ge a \iff z_1 + c_1 \ge 0\).
-
(iii)
\(\{1,1\}\) is NE if and only if
$$\begin{aligned} d \ge b \iff z_1 +c_2 \ge 0. \end{aligned}$$
\(\square \)
Proposition 4.7 implies that the planes \(z_1 +c_1 =0\) and \(z_1 + c_2 =0\) divide \({{\mathbb{R}}}^3\) into four regions with different types of NE. Both of the planes that separates games with different types of NE are used in our model to partition the game space. This in every region in our model, all of the games have the same type of NE. Recall that in Fig. 6, \(z_1=1\), which means that the lines \(c_1 = -1\) and \(c_2 = -1\) divide the map into the four NE regions (see Fig. 7).
Proposition 4.8 provides the analogous analysis for AE. The lines \(c_1 = 1\) and \(c_2 = 1\) divide the games according to their AE types in the same way that \(c_1 = -1\) and \(c_2 = -1\) divide the NE regions.
Proposition 4.8
In any symmetric \(2\times 2,\) two-player game G with cooperation and conflict variables \(c_1, c_2\) and \(z_1,\) the following statements are true:
-
(i)
\(\{0,0\}\) is AE iff \(z_1 \ge c_1,\)
-
(ii)
\(\{0,1\}\) and \(\{1,0\}\) are AE iff \(z_1 \le c_1\) and \(z_1 \ge c_2,\)
-
(iii)
\(\{1,1\}\) is AE iff \(z_1 \le c_2.\)
Proof
Let \({\vec{y}} = (a, b, c, d)^{\mathrm{T}}\) be a vector with the payoffs for G, and let
$$\begin{aligned} {\vec{v}} = (x, c_1, c_2, z_1)^{\mathrm{T}} = A {\vec{y}}, \end{aligned}$$
where A is the matrix in Eq. 15. We can express the payoffs a , b, c and d associated with \({\vec{v}}\) as in Eq. 17. In a \(2 \times 2\) symmetric game AE is equivalent to NE in the transposed payoff matrix. This means that
-
1.
\(\{0,0\}\) is AE iff \(a \ge b\),
-
2.
\(\{0,1\}\) and \(\{1,0\}\) are AE iff \(b \ge a\) and \(c \ge d\),
-
3.
\(\{1,1\}\) is AE iff \(d \ge c\).
Using Eq. 17, these conditions can be expressed in the variables \(c_1\), \(c_2\) and \(z_1\) as follows.
-
(i)
\(\{0,0\}\) is AE if and only if
$$\begin{aligned} a \ge b \iff \frac{\alpha }{4}(2x-3c_1-c_2) \ge \frac{\alpha }{4}(2x+c_1-c_2-4z_1) \iff z_1 \ge c_1. \end{aligned}$$
-
(ii)
\(\{0,1\}\) and \(\{1,0\}\) is AE if and only if \(b \ge a\) and \(c \ge d\).
$$\begin{aligned} b \ge a \iff z_1 \le c_1 \\ c \ge d \iff \frac{\alpha }{4}(2x+c_1-c_2+4z_1) \ge \frac{\alpha }{4}(2x+c_1+3c_2) \iff z_1 \ge c_2. \end{aligned}$$
-
(iii)
\(\{1,1\}\) is AE if and only if
$$\begin{aligned} d \ge c \iff z_1 \le c_2. \end{aligned}$$
\(\square \)
Proposition 4.8 states that all four types of AE are described by the planes \(z_1 - c_1 = 0\) and \(z_1 - c_2=0\). Just as the NE planes, they divide \({{\mathbb{R}}}^3\) into four different regions with different types of AE and the planes are used in our model to describe the strength relationship between \(|z_1|\), \(|c_1|\) and \(|c_2|\). In Fig. 7 the four NE regions and the four AE regions in the map from Fig. 6 are shown.
Propositions 4.7 and 4.8 show that our method of classification, which is intended to divide games into groups depending on their type of decomposition, captures all different types of NE and AE. Since NE and AEFootnote 4 have been argued to be important aspects of games, the results support our hypothesis that tension between the common interest and self-interest is what makes a game interesting, and that games are interesting in different ways because they have different common interest–self-interest tensions.
The \(c_1=\pm c_2\) conditions
Now we know the meaning of the planes \(z_1 = \pm c_1,c_2\). The next step is to analyze the meaning of the planes \(c_1 = \pm c_2\). The plane \(c_1 = -c_2\) determines the difference between the two diagonal outcomes, as
$$\begin{aligned} c_1> -c_2 \iff -2a + b + c> b + c -2d \iff a > d. \end{aligned}$$
Thus, the players prefer outcome \(\{0,0\}\) over \(\{1,1\}\) when \(c_1 > -c_2\), and the \(\{1,1\}\) outcome over the \(\{0,0\}\) outcome otherwise. An interpretation of this condition is that when \(c_1 + c_2 > 0\), the common interests are more aligned than disaligned with the zero-sum incentive.
This difference is exactly the difference between Prisoner’s Dilemma and Deadlock. In Prisoner’s Dilemma \(c_1 + c_2 < 0\), meaning that the summed common interests are disaligned with the zero-sum incentive. In Deadlock on the other hand \(c_1 + c_2 > 0\) and hence the summed common interests are aligned with the zero-sum incentive. Indeed, in Fig. 6 the \(c_1 = -c_2\) line is the border between Prisoner’s Dilemma and Deadlock.
The plane \(c_1 = c_2\) is not a border between any standard games. It does however have some interesting properties. Note that
$$\begin{aligned} c_1> c_2 \iff -2a + b + c> -b -c +2d \iff b + c > a + d. \end{aligned}$$
This means that when \(c_1 > c_2\), the mean of the payoffs in the anti-diagonal positions are higher than the mean of the diagonal positions. Therefore, in some of the regions \(c_1 > c_2\), the players can cooperate by alternating between the two anti-diagonal positions in iterated play. However the \(c_1 + c_2 = 0\) condition is arguably not as important as the \(c_1 - c_2 = 0\) condition. Remember that the latter decides whether or not the common interest incentives are more aligned than disaligned with the zero sum incentives. The \(c_1 - c_2 = 0\) condition has no direct connection to the zero sum incentives and therefore one might expect this condition to have a smaller impact on the games than the \(c_1 + c_2 =0\) condition.
The \(c_i=0\) conditions
The \(c_i = 0\) conditions decide the common interest incentives and in general the direction of the \(c_i\) arrows are important. However, since the \(c_i\) incentives are small near the boarder \(c_i = 0\) they will be dominated by the zero sum incentives. Therefore it seems likely that the change of direction of the \(c_i\) arrows will not make a clear distinction between interestingly different games. Indeed, in the map in Fig. 6 these condition does not separate any standard games. Instead one might expect a gradual change in the games as \(c_i\) changes signs.
Comparison with other classifications
In this section we compare our proposed classification with those reviewed in Sect. 3 that have a similar approach as us. We prove that our classification is at least as fine-grained as the classifications by Harris and Huertas-Rosero. Moreover, we capture the conditions which Harris add ad hoc to divide certain regions further. Since our approach to classifying games differs quite a bit from the classifications by Rapoport et al. and Robinson and Goforth, we do not compare our classification with theirs. The classification by Borm is also hard to compare with, since he classifies mixed extension \(2\times 2\) games. We leave these comparisons as open questions in Sect. 8.
Huertas-Rosero (2003)
Huertas-Rosero (2003) classifies non-zero sum symmetric \(2 \times 2\) games based on NE and AE locations, and on whether or not the payoffs in the NE outcomes are larger than the payoffs in the AE outcomes. As stated in Propositions 4.7 and 4.8, the NE and AE locations are determined by our classes. Further, in each of our 24 classes one can tell whether the payoffs in the NE outcomes are larger than the payoffs in the AE outcomes. A proof for this can be found in Böörs and Wängberg (2017), Chapter 3.2. This means that all of Huertas-Rosero’s classes except one can be found in our classification. However, this remaining class is empty (Böörs and Wängberg 2017, chapter 2.2.4). Thus, our classification contains all 11 non-empty classes of Huertas-Rosero. Figure 8 shows Huertas-Rosero’s 11 non-empty classes in our classification map.
All the aspects of the classification provided by Huertas-Rosero are found in our classification as well, despite his rather different starting point based on NE and AE locations. This supports our hypothesis that the interesting differences between games are explained by the differences in their decomposition.
Harris (1969)
Harris uses a geometrical approach to classifying games (see Sect. 3.2). In that respect his method is similar to our classification method. We can therefore easily compare the parameters that are used in Harris1969’ classification to ours, and prove that they are in fact equivalent. Harris defines the two parameters
$$\begin{aligned} r_3 = \frac{d-b}{c-b}\quad r_4 = \frac{c-a}{c-b}. \end{aligned}$$
He partitions the space of symmetric \(2 \times 2\) games by dividing the \(r_3r_4-\)plane with the lines defined by
$$\begin{aligned} r_3= 1\quad r_4= 0\quad r_3= 0\quad r_3 + r_4= 1\quad r_4= 1. \end{aligned}$$
(18)
Supposing that \(c \ne b\), we can express these lines in terms of our variables using the definition of \(r_3\) and \(r_4\) and Eq. 17. If \(c = b\), the game lies on the border between two of our classes:
$$\begin{aligned} r_3 = 1 \iff c&= d \iff z_1 = c_2\\ r_4 = 0 \iff c&= a \iff z_1 = -c_1\\ r_3 = 0 \iff d&= b \iff z_1 = -c_2\\ r_4 = 1 \iff b&= a \iff z_1 = c_1\\ r_3 + r_4 = 1 \iff d&= a \iff c_2 = -c_1. \end{aligned}$$
(19)
Harris also uses two ad hoc conditions to divide only his Prisoner’s Dilemma region. These extra conditions also correspond to conditions in our classification, as can be easily demonstrated:
$$\begin{aligned} r_4 &= \frac{1}{2} \iff b + c = 2a \iff c_1 = 0\nonumber \\ r_3 &= \frac{1}{2} \iff b + c = 2d\iff c_2 = 0. \end{aligned}$$
(20)
In Fig. 9 Harris’ classes are drawn in our map to show the similarities and the differences between his classification and ours. Harris’ classes are a subset to the set of our classes. This again supports the idea that the interesting elements of a game is defined by its decomposition. Our classification also captures the different variants of Prisoner’s Dilemma and Chicken without the need for ad hoc conditions.