The Iterated Hawk–Dove Game Revisited: The Effect of Ownership Uncertainty on Bourgeois as a Pure Convention

Abstract

Classical evolutionary game theory shows that respect for ownership (“Bourgeois” behavior) can arise as an arbitrary convention to avoid costly disputes, but the same theory also predicts that a paradoxical disrespect for ownership (“anti-Bourgeois” behavior) can evolve under the same conditions. Given the rarity of the latter strategy in the natural world, it is clear that the classical model is lacking in some important biological details. For instance, the classical model assumes that roles of owner and intruder can be recognized unambiguously. However, in the natural world there is often confusion over ownership, mediated for example by the temporary absence of the owner. We show that if intruders sometimes believe themselves to be owners, then the resulting confusion over ownership can broaden the conditions under which Bourgeois behavior is evolutionarily stable in the one-shot Hawk–Dove game. Likewise, introducing mistakes over ownership into a more realistic game with repeated interactions facilitates the evolution of Bourgeois behavior where previously such a result could arise only if owners are intrinsically more likely to win fights than intruders. Collectively, therefore, we find that mistakes over ownership facilitate the evolution of Bourgeois behavior. Nevertheless, relaxing the assumption that ownership is unambiguously recognized does not appear to completely explain the extreme rarity of anti-Bourgeois behavior in nature.

Appendices

Appendix 1

Here we present details of the mathematical model described in Sect. 3. Although we consider only deterministic strategies in the main body of the paper, it will facilitate analysis in this appendix to first define a set of probabilistic strategies of which $\{H,B,X,D\}$ forms a subset. Accordingly, let an animal’s strategy be represented by the vector

$$\begin{aligned} u = (u_1, u_2) = {\left\{ \begin{array}{ll} \hbox {If role perceived as owner, escalate with probability }u_1\\ \quad \hbox {and display with probability } 1-u_1\\ \hbox {If role perceived as intruder, escalate with probability }u_2\\ \quad \hbox {and display with probability } 1-u_2 \end{array}\right. } \end{aligned}$$

(9)

so that the strategies in Table 3 are embedded as

$$\begin{aligned} H = (1,1),\quad B = (1,0),\quad X = (0,1),\quad D = (0,0). \end{aligned}$$

(10)

To determine whether $u$ is an ESS, we must obtain an expression for the reward $f(u,v) = f(u_1, u_2, v_1, v_2)$ to a representative $u$-strategist in a population of $v$-strategists. We will assume in the first instance that this population is unintrusive, but later modify the results to allow for intrusiveness. Let the random variable $X_u(t)$ denote a $u$-strategist’s state at time $t$. We will suppose that animals can change their state only at discrete instants of time, $t = k$, where $k$ is a positive integer. Let $x_j(k)$ be the probability that a $u$-strategist is in state $j$ at time $k$ (i.e., immediately after time $k$), and let $y_j(k)$ be the probability that a $v$-strategist is in state $j$ at time $k$; that is, define

$$\begin{aligned} x_j(k) = \hbox {Prob}(X_u(k) = j),\quad y_j(k) = \hbox {Prob}(X_v(k) = j), \quad j = 1,\ldots ,4. \end{aligned}$$

(11)

Then, because any animal, whether $u$-strategist or $v$-strategist, must be in some state at time $k$, we have

$$\begin{aligned} x_1(k) + x_2(k) + x_3(k) + x_4(k) = 1 = y_1(k) + y_2(k) + y_3(k) + y_4(k) \end{aligned}$$

(12)

for $k = 0, \ldots , K$.

For $1 \le i, j \le 4$, let $\phi _{ij}(k,u,v)$ be the conditional probability that a $u$-strategist in a population of $v$-strategists is in state $j$ at time $k + 1$, given that it is in state $i$ at time $k$, that is, define

$$\begin{aligned} \phi _{ij}(k,u,v) = \hbox {Prob}(X_u(k+1) = j|X_u(k) = i),\quad i, j = 1, \ldots , 4. \end{aligned}$$

(13)

Note that three parameters and two subscripts are necessary to distinguish between, on the one hand, the probability that a $u$-strategist goes from state $i$ to state $j$, which is $\phi _{ij}(k,u,v)$; and, on the other hand, the probability that a $v$-strategist goes from state $i$ to state $j$, which is

$$\begin{aligned} \phi _{ij}(k,v,v) = \hbox {Prob}(X_v(k+1) = j|X_v(k) = i),\quad i, j = 1, \ldots , 4. \end{aligned}$$

We assume that animals and sites are both so numerous that $N \rightarrow \infty$ and $M \rightarrow \infty$, but in such a way that the ratio $\sigma$ defined by (7) is finite. Then every $v$-strategist is effectively playing only against $v$-strategists: the probability that a $v$-strategist will meet a $u$-strategist is negligible. For $j = 1,\ldots ,4$ and $0 \le k \le K - 1$, we now have

$$\begin{aligned} x_j(k+1) = \sum _{i=1}^4\hbox {Prob}(X_u(k\!+\!1) = j|X_u(k) = i)\cdot \hbox {Prob}(X_u(k) = i)\!=\! \sum _{i=1}^4 x_i(k)\phi _{ij}(k,u,v) \end{aligned}$$

(14)

and

$$\begin{aligned} y_j(k+1) = \sum _{i=1}^4\hbox {Prob}(X_v(k+1) \!=\! j|X_v(k) = i)\cdot \hbox {Prob}(X_v(k) \!=\! i) = \sum _{i=1}^4 y_i(k)\phi _{ij}(k,v,v). \end{aligned}$$

(15)

Note that

$$\begin{aligned} \sum _{j=1}^4 \phi _{ij}(k,u,v) = 1 \end{aligned}$$

(16)

for all $1 \le i, j \le 4$ and $0 \le u, v \le 1$, because if a $u$-strategist is in state $i$ at time $k$, then at time $k+1$ it must either remain in state $i$ or enter one of the other states.

An animal’s payoff will be highest if it ends the contest as owner of a site, but the site is worth less to the animal if it is injured. Thus it is reasonable to regard

$$\begin{aligned} F_K(u,v) = {\left\{ \begin{array}{ll} 1&{}\quad \hbox {if } X_u(K) = 1\\ \alpha &{}\quad \hbox {if } X_u(K) = 2\\ 0&{}\quad \hbox {if }X_u(K) \ge 3 \end{array}\right. } \end{aligned}$$

(17)

as the payoff to a $u$-strategist against $v$-strategists if the contest lasts for $K$ periods. $F_K$, so defined, is a random variable, whose expected value

$$\begin{aligned} 1\cdot \hbox {Prob}(X_u(K) = 1) + \alpha \cdot \hbox {Prob}(X_u(K) = 2) + 0 \cdot \hbox {Prob}(X_u(K) \ge 3) = x_1(K) + \alpha x_2(K) \end{aligned}$$

is the reward to strategy $u$ against strategy $v$ if the contest lasts for $K$ periods. Thus, if the length of the contest is random, then the reward to strategy $u$ against strategy $v$ is the expected value of $x_1(K) + \alpha x_2(K)$, calculated over the distribution of $K$, that is,

$$\begin{aligned} f(u,v) = \sum _{k=1}^\infty \{x_1(k) + \alpha x_2(k)\}\hbox {Prob}(K = k) = (1-w)\sum _{k=1}^\infty w^{k-1}\{x_1(k) + \alpha x_2(k)\}\qquad \end{aligned}$$

(18)

because (6) implies that $\hbox {Prob}(K = k) = \hbox {Prob}(K \ge k) - \hbox {Prob}(K \ge k+1) = w^{k-1}-w^k$.

We can deduce the vectors $x(k)$ and $y(k)$ from (14) and (15) for any value of $k$ if we first obtain an explicit expression for the $4 \times 4$ matrix $\Phi (k,u,v)$ defined by (13). All possible transitions from states 1 or 2 are shown in Fig. 6. We will first calculate the transition probabilities for the unintrusive case, and subsequently indicate which terms require modification for the intrusive case.

Fig. 6

Flow chart of all possible transitions from owner states 1 (uninjured) or 2 (injured), where $q = q(y_3(k))$ and $p_v$ are defined by (19) and (20). All arc probabilities are conditional (on reaching the node from which the arc emanates) probabilities, and are equal to $1$ for unlabeled arcs. The diagram is drawn for the intrusive case, using thick solid lines and larger arrowheads to indicate arcs of the network that must be deleted for the unintrusive case (with the unused weight of $\frac{1}{2}$ on a deleted arc transferred to the other arc emanating from the same parent node to increase its weight from $\frac{1}{2}$ to $1$), and using thick dashed lines to surround nodes that are irrelevant to the unintrusive case

Accordingly, let us first suppose that the $u$-strategist is in state $1$ (uninjured owner) at time $k$. Then it can move into state $4$ (injured non-owner) only if it is intruded upon, is attacked, attacks back, and then loses; whereas it can move into state $3$ (uninjured non-owner) only if it is intruded upon, is attacked, and promptly surrenders; and it can move into state $2$ (injured owner) only if it is intruded upon, is attacked, attacks back, and then wins, but sustains an injury. At time $k$, the expected number of uninjured non-owners is $Ny_3(k)$, because all of them are $v$-strategists; and each finds a site during the interval $k < t < k + 1$ with probability $\epsilon$. Because they are all equally likely to find any site, the probability that one of them finds the particular site now occupied by the $u$-strategist is $\epsilon /M$. The probability that one of them does not find the u-strategist is therefore $1-\epsilon /M$, the probability that none of them finds the $u$-strategist is that number raised to the power of $Ny_3(k)$, and so the probability that at least one of them intrudes is

$$\begin{aligned} 1 - \biggl (1-\frac{\epsilon }{M}\biggr )^{Ny_3(k)} = 1 - \biggl (1-\frac{\epsilon }{M}\biggr )^{M \sigma y_3(k)}. \end{aligned}$$

Thus, in the limit as $M \rightarrow \infty$, the probability that at least one uninjured non-owner locates the $u$-strategist is $q(y_3(k))$, where we define

$$\begin{aligned} q(y_3) = 1 - e^{-\epsilon \sigma y_3}. \end{aligned}$$

(19)

Having assumed that at most one animal intrudes upon a site per unit of time, we interpret $q(y_3(k))$ as the probability that an uninjured non-owner intrudes during the interval $k < t < k + 1$. (In the event that more than one animal located the site, we could suppose that the $u$-strategist interacted with the first, and that later arrivals ignored them both.) The smaller the value of $\epsilon$, the more reasonable this interpretation; and we note in passing that for sufficiently small $\epsilon$ we can approximate (19) by $q(y_3) = \epsilon \sigma y_3$. Then, because another animal attacks as intruder with probability

$$\begin{aligned} p_v = \theta v_1 + (1-\theta )v_2, \end{aligned}$$

(20)

and because the $u$-strategist attacks as owner with probability $u_1$ and loses with probability $1-\mu$ (conditional upon attacking), the probability that a non-owner intrudes and attacks, and that the $u$-strategist attacks and loses, is $q(y_3(k))$ times $p_v$ times $u_1$ times $1-\mu$; all conditional, of course, on the $u$-strategist being in state $1$, and assuming that the intruder’s attack probability is independent of the $u$-strategist’s surrender probability. We have thus established that

$$\begin{aligned} \phi _{14}(k,u,v) = (1-\mu )u_1p_v\,q(y_3(k)). \end{aligned}$$

(21)

Similarly, because the $u$-strategist surrenders as owner with probability $1 - u_1$, we have

$$\begin{aligned} \phi _{13}(k,u,v) = (1-u_1)p_v\,q(y_3(k)), \end{aligned}$$

(22)

and because the $u$-strategist wins with probability $\mu$ but (conditional upon winning) sustains an injury with probability $\lambda$, we have

$$\begin{aligned} \phi _{12}(k,u,v) = \mu \lambda u_1 p_v\,q(y_3(k)). \end{aligned}$$

(23)

The probability that the $u$-strategist remains in state $1$ is now readily deduced from (16) with $i = 1$:

$$\begin{aligned} \phi _{11}(k,u,v) = 1-\{1-\mu (1-\lambda )u_1\}p_v\,q(y_3(k)). \end{aligned}$$

(24)

Alternatively, (24) can be obtained by summing the products of conditional probabilities around all three circuits from State 1 back to itself in Fig. 6—one through NO INTRUSION, one through UNINJURED, and one through NOT ATTACKED (with the ESCALATE and NO ESCALATE nodes deleted for the time being, because for now we deal only with the unintrusive case).

Second, let us suppose that the $u$-strategist is in state 2 (injured owner). Then the $u$-strategist is injured and does not recover, whence

$$\begin{aligned} \phi _{21}(k,u,v) = 0 = \phi _{23}(k,u,v). \end{aligned}$$

(25)

The $u$-strategist either remains an injured owner or becomes an injured non-owner. Given that it is already injured, it surrenders its site if it is intruded upon, which happens with probability $q(y_3(k))$; and attacked, which happens with probability $\{\theta v_1 + (1-\theta )v_2\}$. Thus, on using (16) with i$= 2$, we have

$$\begin{aligned} \phi _{22}(k,u,v) = 1- p_v\,q(y_3(k)), \quad \phi _{24}(k,u,v) = p_v\,q(y_3(k)). \end{aligned}$$

(26)

All possible transitions from states 3 or 4 are shown in Fig. 7. Let us first suppose that the $u$-strategist is in state 3 (uninjured non-owner). Then it will descend into state 4 (injured non-owner) only if it intrudes upon an uninjured owner, attacks, is attacked back, and then loses. At time $k$, the expected number of uninjured owners is $Ny_1(k)$, because every one of them is a $v$-strategist; whence, if the $u$-strategist finds a site, then the probability that it is occupied by an uninjured owner is $Ny_1(k)/M = \sigma y_1(k)$—provided, of course, that this number does not exceed 1.

Fig. 7

Flow chart of all possible transitions from non-owner states 3 (uninjured) or 4 (injured), where $s_1 = s_1(y(k))$, $s_2 = s_2(y(k))$ and $p_u$ are defined by (27) and (28). All arc probabilities are conditional probabilities, and are equal to $1$ for unlabeled arcs. The diagram is drawn for the intrusive case, using thick solid lines and larger arrowheads to indicate arcs of the network that must be deleted for the unintrusive case (with the unused weight of $\frac{1}{2}$ on a deleted arc transferred to the other arc emanating from the same parent node to increase its weight from $\frac{1}{2}$ to $1$), and using thick dashed lines to surround nodes that are irrelevant to the unintrusive case

We must here digress to observe that the expected number of owners should not exceed the number of sites available; i.e., $N(y_1+y_2) \le M$, or $\sigma (y_1+y_2) \le 1$. Because our model underestimates the true probability of intrusion, and hence overestimates the probability of ownership, however, it is possible for $N(y_1+y_2)$—as calculated from (15)—to exceed $M$, but only if $\sigma$ is significantly greater than 1. Although such cases are not considered in this paper, for completeness we correct for the anomaly by simply assuming that if (15) predicts $N(y_1+y_2) > M$, then all $M$ sites are occupied with $My_1$ owners uninjured and $My_2$ injured, so that the proportions are consistent with the probabilities. Thus the probability that a site is occupied at time $k$ by an uninjured owner is $\sigma y_1(k)$ if $\sigma y_1(k) + \sigma y_2(k) \le 1$, but $y_1(k)/\{y_1(k)+y_2(k)\}$ if $\sigma y_1(k) + \sigma y_2(k) > 1$; that is, the probability is $s_1(y(k))$, where we define

$$\begin{aligned} s_m(y) = \min \Bigl (\sigma y_m, \frac{y_m}{y_1+y_2}\Bigr ) \end{aligned}$$

(27)

for $m = 1, 2$. Similarly, the probability that a site is occupied at time $k$ by an injured owner is $s_2(y(k))$.

Let us now return to calculating the probability that a $u$-strategist in state 3 will descend into state 4 during the interval $k < t < k + 1$. The $u$-strategist finds a site with probability $\epsilon$, and it is occupied by an uninjured owner with probability $s_1(y(k))$; in which case, the $u$-strategist attacks as intruder with probability

$$\begin{aligned} p_u = \theta u_1 + (1-\theta )u_2, \end{aligned}$$

(28)

is attacked back by the owner with probability $v_1$ and subsequently loses with probability $\mu$. Thus, multiplying all the conditional probabilities together, we have

$$\begin{aligned} \phi _{34}(k,u,v) = \mu \epsilon p_u v_1s_1(y(k)). \end{aligned}$$

(29)

Similarly, because (conditional upon an intrusion and engagement) the $u$-strategist wins and is injured with probability $(1-\mu )\lambda$, we have

$$\begin{aligned} \phi _{32}(k,u,v) = (1-\mu )\lambda \epsilon p_u v_1s_1(y(k)). \end{aligned}$$

(30)

The probability that the $u$-strategist finds a site whose occupier is either injured or uninjured, and does not attack, is $\epsilon$ times $s_1(y(k)) + s_2(y(k))$ times $1-p_u$. The probability that the $u$-strategist does not find a site is $1-\epsilon$. In either case, it remains in state 3. Thus

$$\begin{aligned} \phi _{33}(k,u,v) = 1-\epsilon + \epsilon \{1-p_u\}\{s_1(y(k)) + s_2(y(k))\} \end{aligned}$$

(31)

and (16) with $i = 3$ implies

$$\begin{aligned} \phi _{31}(k,u,v) = \epsilon \bigl \{ 1-(1-p_u)\{s_1(y(k)) + s_2(y(k))\} - (\mu + \lambda (1-\mu ))p_u v_1s_1(y(k)) \bigr \}.\qquad \end{aligned}$$

(32)

Finally, because an injured animal is unable to search, even for unoccupied sites, we have

$$\begin{aligned} \phi _{41}(k,u,v) = \phi _{42}(k,u,v) = \phi _{43}(k,u,v) = 0, \quad \phi _{44}(k,u,v) = 1. \end{aligned}$$

(33)

The matrix $\phi$ has now been defined for an unintrusive population.

Six elements of this matrix, namely, $\phi _{11}$, $\phi _{13}$, $\phi _{22}$ , $\phi _{24}$, $\phi _{31}$, and $\phi _{33}$, require modification if the population is intrusive. In (22), conditional on being intruded upon and not fighting, a $u$-strategist’s probability of descending from state 1 to state 3 in an unintrusive population is simply $p_v$, the probability that the intruder attacks. In an intrusive population, however, if the intruder does not attack, then owner and intruder are equally likely to obtain the site. Thus, conditional on being intruded upon and not fighting, a $u$-strategist’s probability of descending from state 1 to state 3 is, not $p_v$, but $p_v + \frac{1}{2}(1-p_v) = \frac{1}{2}(1+p_v)$. Thus (22) is replaced by

$$\begin{aligned} \phi _{13}(k,u,v) = \tfrac{1}{2}(1-u_1)\{1+p_v\}\,q(y_3(k)), \end{aligned}$$

(34)

and from (16) with $i = 1$, (24) is replaced by

$$\begin{aligned} \phi _{11}(k,u,v) = 1-\tfrac{1}{2}\bigl \{1-u_1 + \{1+(1-2\mu (1-\lambda ))u_1\}p_v \bigr \}q(y_3(k)). \end{aligned}$$

(35)

Similarly, because of our assumption that an injured owner in an intrusive population is equally likely to lose or keep its site when $B$ or $D$ intrudes, (26) is replaced by

$$\begin{aligned} \phi _{22}(k,u,v) = 1- \tfrac{1}{2}\{1+p_v\}q(y_3(k)),\qquad \phi _{24}(k,u,v) = \tfrac{1}{2}\{1+p_v\}q(y_3(k)). \end{aligned}$$

(36)

In state $3$, the focal $u$-strategist finds a site and then does not fight for it with probability $\epsilon \cdot (1-p_u)$. Conditional thereupon, it remains in state $3$ with probability $\frac{1}{2}$ if the site is occupied by an injured owner or with probability $v_1\cdot 1 + (1-v_1) \cdot \frac{1}{2} = \frac{1}{2}(1+v_1)$ if it is occupied by an uninjured owner, that is, with net conditional probability $\frac{1}{2}\cdot s_2(y(k)) + \frac{1}{2}(1+v_1)\cdot s_1(y(k)) = \tfrac{1}{2}\{(1+v_1)s_1(y(k)) + s_2(y(k))\}$. Thus (31) is replaced by

$$\begin{aligned} \phi _{33}(k,u,v) = 1-\epsilon + \tfrac{1}{2}\epsilon \{1-p_u\}\{(1+v_1)s_1(y(k)) + s_2(y(k))\}, \end{aligned}$$

(37)

while (32) becomes

$$\begin{aligned} \phi _{31}(k,u,v) = \epsilon \bigl \{ 1-\tfrac{1}{2}\{1-p_u\}\{(1+v_1)s_1(y(k)) + s_2(y(k))\} - \{\mu + \lambda (1-\mu )\}p_u v_1s_1(y(k)) \bigr \} \end{aligned}$$

(38)

by (16).

Substitution of $\phi$ into (14) and (15) leads to a set of eight first-order, nonlinear difference equations. Throughout Sect. 3 we assume that all individuals are initially in state 3, and so

$$\begin{aligned} x(0) = (0,0,1,0) = y(0). \end{aligned}$$

(39)

In Sect. 4, however, for $\sigma > 1$ we also discuss the alternative initial conditions

$$\begin{aligned} x(0) = (1/\sigma ,0,1-1/\sigma ,0) = y(0) \end{aligned}$$

(40)

of maximum initial occupancy. Given (39) or (40), the vectors $x(k)$ and $y(k)$ are now readily calculated from $k$ successive recursions of (14)–(15) for any strategy combination $(u,v)$, and for any value of $k$; and so $f(u,v)$ in (18) can be approximated to an arbitrarily high degree of accuracy by truncating the infinite series after a sufficiently large number of terms. A series of such calculations yields—again to arbitrarily high accuracy—the payoff matrix

$$\begin{aligned} A = \begin{bmatrix} f(1, 1, 1, 1)&\quad f(1, 1, 1, 0)&\quad f(1, 1, 0, 1)&\quad f(1, 1, 0, 0)\\ f(1, 0, 1, 1)&\quad f(1, 0, 1, 0)&\quad f(1, 0, 0, 1)&\quad f(1, 0, 0, 0)\\ f(0, 1, 1, 1)&\quad f(0, 1, 1, 0)&\quad f(0, 1, 0, 1)&\quad f(0, 1, 0, 0)\\ f(0, 0, 1, 1)&\quad f(0, 0, 1, 0)&\quad f(0, 0, 0, 1)&\quad f(0, 0, 0, 0) \end{bmatrix} \end{aligned}$$

(41)

(in which $f(u_1, u_2, v_1, v_2)$ is used as an alternative notation for $f(u,v)$). We note in passing that $a_{14}$ is the only payoff for which we have found an analytic expression: $a_{14} = \epsilon + (1-\epsilon )w\epsilon + (1-\epsilon )^2w^2\epsilon + \ldots = \epsilon /\{1-w(1-\epsilon )\}$ because the payoff to a mutant Hawk in a population of Doves is just the probability of finding a site. Strictly, however, it is unnecessary to compute $a_{14}$ or any other term in the last column of $A$, in view of the result in Appendix 2.

Appendix 2

Here we establish that Dove is never an ESS of the iterated game described in Appendix 1. Because there are no fights and hence no injuries in a population of Doves, neither state $2$ nor state $4$ is ever entered. The second row and column and the fourth row and column of the transition matrix $\Phi (k,u,v)$ are therefore irrelevant; and the four remaining relevant entries follow from (22), (24) and (31), (32) for the unintrusive case, or from (34)–(35) and (37)–(38) for the intrusive one. Noting that $y_2 = 0$ implies $s_2 = 0$ by (27), for a Hawk in a Dove population we set $u_1 = u_2 = 1$ (implying $p_u = 1)$ and $v_1 = v_2 = 0$ (implying $p_v = 0)$ to obtain $\phi _{11}(k,u,v) = 1$, $\phi _{13}(k,u,v) = 0$, $\phi _{31}(k,u,v) = \epsilon$, and $\phi _{33}(k,u,v) = 1-\epsilon$ (for either case); whereas for a Dove in a Dove population we set $u_1 = u_2 = 0$ (implying $p_u = 0)$ and $v_1 = v_2 = 0$ to obtain $\phi _{11}(k,u,v) = 1$, $\phi _{13}(k,u,v) = 0$, $\phi _{31}(k,u,v) = \epsilon \{1-s_1(y(k))\}$, and $\phi _{33}(k,u,v) = 1-\epsilon \{1-s_1(y(k))\}$ for the unintrusive case or $\phi _{11}(k,u,v) = 1 - \frac{1}{2}q(y_3(k))$, $\phi _{13}(k,u,v) = \frac{1}{2}q(y_3(k))$, $\phi _{31}(k,u,v) = \epsilon \{1-\frac{1}{2}s_1(y(k))\}$ and $\phi _{33}(k,u,v) = 1-\epsilon \{1-\frac{1}{2}s_1(y(k))\}$ for the intrusive one. Either way, the probability of transition from state 3 to state 1 is never larger for a Dove in a Dove population than for a Hawk in a Dove population and becomes strictly lower as soon as any sites are occupied, so that $x_1(k)$ in (18) is never larger for a Dove and is strictly lower at sufficiently large $k$ than for a Hawk (with $x_2(k) = 0$ for either strategy). Hence $a_{14} > a_{44}$, on summing over the distribution of $K$. That is, Hawk can always invade.