Journal of Game Theory
p-ISSN: 2325-0046 e-ISSN: 2325-0054
2016; 5(1): 9-15
doi:10.5923/j.jgt.20160501.02

Essam EL-Seidy
Department of Mathematics, Faculty of Science, Ain Shams University, Cairo, Egypt
Correspondence to: Essam EL-Seidy, Department of Mathematics, Faculty of Science, Ain Shams University, Cairo, Egypt.
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Copyright © 2016 Scientific & Academic Publishing. All Rights Reserved.
This work is licensed under the Creative Commons Attribution International License (CC BY).
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The emergence of cooperative behavior in human and animal societies is one of the fundamental problems in biology and social sciences. In this article, we study the evolution of cooperative behavior in the hawk dove game. There are some mechanisms like kin selection, group selection, direct and indirect reciprocity can evolve the cooperation when it works alone. Here we combine two mechanisms together in one population. The transformed matrices for each combination are determined. Some properties of cooperation like risk-dominant (RD) and advantageous (AD) are studied. The property of evolutionary stable (ESS) for strategies used in this article is discussed.
Keywords: Hawk-dove game, Evolutionary game dynamics, of evolutionary stable strategies (ESS), Kin selection, Direct and Indirect reciprocity, Group selection
Cite this paper: Essam EL-Seidy, On the Behavior of Strategies in Hawk-Dove Game, Journal of Game Theory, Vol. 5 No. 1, 2016, pp. 9-15. doi: 10.5923/j.jgt.20160501.02.
. If a Hawk meets a Dove the Dove immediately withdraws, so its payoff is zero, while the payoff of the Hawk is B. If two Doves meet, the one who first gets hold of the resource keeps it while the other does not fight for it. Thus, the average payoff for a Dove meeting a Dove is
. The strategic form of the game is given by the payoff matrix:![]() | (1) |
to be an
. EitherI.
that is, the payoff for playing
against (another playing)
is greater than that for playing any other strategy
against
for all
OrII.
and
, that is, the payoff of playing
against itself is equal to that of playing
against
but the payoff of playing
against
is less than that of playing
against
for all
Note that either (I) or (II) will do and that the previous is a more grounded condition than the recent. Clearly, if (I) gets, the Y invader commonly loses against X, and along these lines it can't even start to increase with any achievement. If (II) gets, the Y invader does as well against X as X itself, but it loses to X against other Y invaders, and therefore it cannot multiply. In short,
players cannot successfully invade a population of X players. It is conceivable to present a strategy that is stronger than an
, namely, an unbeatable strategy. Strategy
is unbeatable if, given whatever other strategy
:
An unbeatable strategy is the most powerful strategy, because it strictly dominates any other strategy; however it is additionally uncommon, and subsequently in exceptionally constrained use.
and
, given by the payoff matrix:![]() | (2) |
obtains payoff a when playing another
player, but payoff b when playing a
player. Likewise, strategy
obtains payoff c when playing an
player and payoff d when playing a
player.1. If
and
, then
dominates
. In this case, it is always better to use strategy
. The expected payoff of
players is greater than that of
players for any composition of a well-mixed population.
is an unbeatable strategy in the sense of. If instead
and
, then
dominates
and we have exactly the reverse situation .2. If
and
, then both strategies are best replies to themselves, which leads to a ‘coordination game’. In a population where most players use
, it is best to use
. In a population where most players use
, it is best to use
. A coordination game leads to bi-stability: both strategies are stable against invasion by the other strategy.3. If
and
, then both strategies are best replies to each other, 4. If
then
is risk-dominant (RD). If both strategies are ESS, then the risk dominant strategy has the bigger basin of attraction.5. If
then
is advantageous (AD).6. If
then
is a strict Nash equilibrium. Likewise, if
then
is a strict Nash equilibrium. A strategy which is a strict Nash equilibrium is always an evolutionarily stable strategy (ESS) ([9], [23]).
, a population of doves can be invaded by hawks. Because of
and
, H is an ESS if
. But what if
? Neither H nor D is an ESS. But we could ask: What would happen to a population of individuals which are able to play mixed strategies? Maybe there exists a mixed strategy which is evolutionary stable.Consider a population consisting of a species, which is able to play a mixed strategy, i.e. sometimes Hawk and sometimes Dove with probabilities
and
respectively. For a mixed ESS S to exist the following must hold:
Suppose that there exists an ESS in which H and D, which are played with positive probability, have different payoffs. Then it is worthwhile for the player to increase the weight given to the strategy with the highest payoff since this will increase expected utility. But this means that the original mixed strategy was not a best response and hence not part of an ESS, which is a contradiction. Therefore, it must be that in an ESS all strategies with positive probability yield the same payoff. Thus:
Thus a mixed strategy with a probability
of playing Hawk and a probability
of playing Dove is evolutionary stable, i.e. that it cannot be invaded by players playing one of the pure strategies Hawk or Dove.To find the Nash equilibrium point of this game, Let
be the probability of playing hawk if you are player 1 and let
be the probability of playing hawk if you are player 2. The payoffs to the two players are:
Which simplifies to
Thus
So the optimal
is given by
Similarly,
So the optimal
is given by
This gives the diagram depicted in Figure 1. The best response functions intersect in three places, each of which is a Nash equilibrium. However, the only symmetric Nash equilibrium, in which the players cannot condition their moves on whether they are player 1 or player 2, is the mixed-strategy Nash equilibrium
.![]() | Figure 1. Nash equilibria in the Hawk-Dove Game |
where r is the genetic relatedness of two interacting agents, b is the fitness benefit to the beneficiary, and c is the fitness cost to the altruist. This rule suggests that agents should show more altruism and less aggression toward closer kin ([21]). A simple way to study games between relatives was proposed by Maynard Smith for the Hawk- Dove game . In this section, we will study the Hawk-Dove game in which there is e relationship between the players. Consider a population where the average relatedness between players is given by r, which is a number between 0 and 1. There are two possible methods to study the games between relatives. The "inclusive fitness " method adds to the payoff of a player r times the payoff to his co-player. The personal fitness method, proposed by Grafen ([28]) modifies the fitness of the player by allowing for the fact that a player is more likely than other players of the population to meet co-player adopting the same strategy as himself. We regard the inclusive fitness method to study the Hawk-Dove game ([16] , [10] , [26]). If we assume that there is a relationship between the players, then by using the inclusive fitness method, the payoff matrix of the Hawk-Dove game is given by:![]() | (3) |
Implies to
.● Hawk’player will be risk-dominant (RD) whenever 
Implies to
● Hawk’playerwill be advantageous (AD) if:
i.e. if:
Implies to 
the game is over. Consequently, the average number of interactions between two individuals is
.In order to determine a fundamental condition for the evolution of cooperation in the repeated Hawk-Dove game, we can study the interaction between the dove who play with a strategy ‘always-defect’ (AllD) (i.e. Always cooperate and give away ) and the hawk who play with a strategy Tit -For- Tat (TFT). TFT starts with cooperation and then does whatever the opponent has done in the past move. On the off chance that two hawks (i.e. defectors) meet, they defect constantly. If two doves meet, they cooperate and give away all the time, so:
If a hawk meets a dove, the TFT’player fighting in the first round and give away a short time later, while the AllD’player gives away in every round, so:
While, the AllD’player will get
And the two TFT’players will fight in all the interactions, so:
Thus, the payoff matrix is given by:![]() | (4) |
Implies to ![]() | (4.1) |
● Fighting will be advantageous (AD) if:
![]() | (5) |
● The fighting will be risk-dominant (RD) if 
● TFT’player has advantageous (AD) if:
i.e. whenever
![]() | (6) |
, while the groups which give away have a constant payoff
. Hence, in a sense between groups the game can take the form as follows:![]() | (7) |
![]() | (8) |
and hawks will invade doves if
. If
and
respectively. Hawks are RD if
and will be AD if
.![]() | (9) |
● The fighting between relatives in the same group will be (RD) if the inequality:
Implies to
● Also the fighting between relatives will be (AD) if:
i.e. when
Therefore, group and kin selection together can evolve strong fighting than either of them working alone, especially when average relatedness is low, groups are large and the number of groups is small.
, the population of cooperators will be RD if:
. Where r is the average relatedness between individuals , which is a number between 0 and 1, and w is the probability of next round.When the group selection works with kin selection, then our fundamental conditions, that we derived , showed that fighting can be maintained in the population, even when the average relatedness is low, groups are large and even if the benefits of fighting are low, if
, fighting between players can emerge, otherwise the cooperation (give way) behavior will be maintain in this situation. Hawks strategy will be risk-dominant (RD) whenever
, and will be advantageous (AD) if
.For cooperation (give way) to prove stable, the future must have a sufficiently large shadow. An indefinite number of interactions, therefore, is a condition under which cooperation (give way) can emerge.