Editing Evolutionary game theory (section)

===The evolutionarily stable strategy===
[[File:AssessorGraph.jpg|thumb|300px| The payoff matrix for the hawk dove game, with the addition of the assessor strategy. This "studies its opponent", behaving as a hawk when matched with an opponent it judges "weaker", like a dove when the opponent seems bigger and stronger. Assessor is an ESS, since it can invade both hawk and dove populations, and can withstand invasion by either hawk or dove mutants.]]
{{main|Evolutionarily Stable Strategy}}

The [[evolutionarily stable strategy]] (ESS) is akin to the Nash equilibrium in classical game theory, but with mathematically extended criteria. Nash equilibrium is a game equilibrium where it is not rational for any player to deviate from their present strategy, provided that the others adhere to their strategies. An ESS is a state of game dynamics where, in a very large population of competitors, another mutant strategy cannot successfully enter the population to disturb the existing dynamic (which itself depends on the population mix). Therefore, a successful strategy (with an ESS) must be both effective against competitors when it is rare – to enter the previous competing population, and successful when later in high proportion in the population – to defend itself. This in turn means that the strategy must be successful when it contends with others exactly like itself.<ref>Taylor, P. D. (1979). ''Evolutionarily Stable Strategies with Two Types of Players'' J. Appl. Prob. 16, 76–83.</ref><ref>Taylor, P. D., and Jonker, L. B. (1978). ''Evolutionarily Stable Strategies and Game Dynamics'' Math. Biosci. 40, 145–156.</ref><ref>Osborn, Martin, Introduction to Game Theory, 2004, Oxford Press, p. 393-403 {{ISBN|0-19-512895-8}}</ref>

An ESS is not:

* An optimal strategy: that would maximize fitness, and many ESS states are far below the maximum fitness achievable in a fitness landscape. (See hawk dove graph above as an example of this.)
* A singular solution: often several ESS conditions can exist in a competitive situation. A particular contest might stabilize into any one of these possibilities, but later a major perturbation in conditions can move the solution into one of the alternative ESS states.
* Always present: it is possible for there to be no ESS. An evolutionary game with no ESS is "rock-scissors-paper", as found in species such as the side-blotched lizard (''[[Uta stansburiana]]'').
* An unbeatable strategy: the ESS is only an uninvadeable strategy.

[[File:Agelenopsis actuosa fem sp.jpg|thumb|Female funnel web spiders (Agelenopsis aperta) contest with one another for the possession of their desert spider webs using the assessor strategy.<ref>{{cite journal | last1=Riechert | first1=S.|author1-link= Susan Riechert |last2= Hammerstein | first2=P. |year=1995 |title= Putting Game Theory to the Test |doi= 10.1126/science.7886443 | pmid=7886443 |journal= Science |volume= 267 | issue=5204 |pages= 1591–1593 |bibcode= 1995Sci...267.1591P | s2cid=5133742 }}</ref>]]

The ESS state can be solved for by exploring either the dynamics of population change to determine an ESS, or by solving equations for the stable stationary point conditions which define an ESS.<ref>{{cite journal | last1=Chen | first1=Z | last2=Tan | first2=JY | last3=Wen | first3=Y | last4=Niu | first4=S | last5=Wong | first5=S-M | year=2012 | title=A Game-Theoretic Model of Interactions between Hibiscus Latent Singapore Virus and Tobacco Mosaic Virus | journal=PLOS ONE | volume=7 | issue=5| page=e37007 | doi=10.1371/journal.pone.0037007 |bibcode=2012PLoSO...737007C | pmid=22623970 | pmc=3356392| doi-access=free }}</ref>  For example, in the hawk dove game we can look for whether there is a static population mix condition where the fitness of doves will be exactly the same as fitness of hawks (therefore both having equivalent growth rates – a static point). 
  
Let the chance of meeting a hawk=p so therefore the chance of meeting a dove is (1-p)

Let Whawk equal the payoff for hawk...

Whawk=payoff in the chance of meeting a dove + payoff in the chance of meeting a hawk

Taking the payoff matrix results and plugging them into the above equation:

{{math|<var>Whawk</var>{{=}} <var>V·(1-p)+(V/2-C/2)·p</var>}}

Similarly for a dove:

{{math|<var>Wdove</var>{{=}} <var>V/2·(1-p)+0·(p)</var>}}

so....

{{math|<var>Wdove</var>{{=}} <var> V/2·(1-p) </var>}}

Equating the two fitnesses, hawk and dove

{{math|<var>V·(1-p)+(V/2-C/2)·p</var>{{=}} <var> V/2·(1-p) </var>}}

... and solving for p

{{math|<var>p</var>{{=}} <var>V/C</var>}}

so for this "static point" where the ''population percent'' is an ESS solves to be ESS<sub>(percent Hawk)</sub>=''V/C''

Similarly, using inequalities, it can be shown that an additional hawk or dove mutant entering this ESS state eventually results in less fitness for their kind – both a true Nash and an ESS equilibrium.  This example shows that when the risks of contest injury or death (the cost C) is significantly greater than the potential reward (the benefit value V), the stable population will be mixed between aggressors and doves, and the proportion of doves will exceed that of the aggressors.  This explains behaviours observed in nature.