Game Theory was fist applied to evolution by John Maynard-Smith and George Price in 1973. It differs from traditional game theory is that it focusses on dynamics of strategy change more than the properties of strategy equilibria, although equilibria still exist within EGT but are know as Evolutionary Stable Strategies as opposed to Nash Equilibria.
Imagine a situation in which 2 members of a species come into conflict over a resource. Within this conflict each animal has the optional to ‘fight’, ‘display’ or ‘run away’. There are 2 strategies within this species, either the Dove strategy or the Hawk strategy. In the Dave strategy, upon meeting someone also adopting the Dove strategy both “Doves” display and share the resource or upon meeting a “Hawk” the Dove runs away. Adopting the Hawk strategy entails always fighting. So upon meeting a Dove the Hawk will fight and the Dove will run away and the Hawk will take all of the resource, and upon meeting another Hawk, both will fight and one will win out. On average across many interactions with other Hawks, the payoff gained ends up being (v/2)-c where v=value of resource and c=cost.
|Dove||v/2, v/2||v, 0|
|Hawk||0, v||(v-c)/2, (v-c)/2|
The question to ask of this game is, given values v and c, which strategy will evolutionarily win out?
In answering this question I will take the following assumptions:
1. That the strategies are genetically fixed
2. That the organisms in the species reproduce asexually
3. That payoffs are equal to incremental fitness (so successful strategies reproduce faster)
4. The population is random mixing and infinite
So. If the proportion of Hawks in the population is x and 0≤x≤1 then:
fitness of hawk: w(H) = x((v-c)/2) + (1-x)v
fitness of dove: w(D) = x(0) + (1-x)v/2
Evolutionary Stable Strategy (ESS)
An ESS is an instance of a Nash Equilibrium within an evolutionary game. These are present when one strategy is stable against invasion by other strategies OR if there is a stable mix of these strategies.
The ESS in the dove-hawk game does not lie with a 100% Dove population because if one Hawk enters the population then it will kick ass.
If the population is 100% Hawks, on the other hand, this will be an ESS but only if (v-c)/2 > 0 (i.e. if v>c)
But in an instance where c>v the ESS will lie where w(D) = w(H).
Lets see an example of this:
w(H) = x((4-6)/2)+(1-x)4= 4-5x
w(D) = x(0)+(1-x)2=2-2x
So the ESS in this instance is where 2/3 of the population are Hawks. Simple!
The missing Nash Equilibrium from last time
This, incidentally, is how we’d find the 3rd missing Nash Equilibrium from the Driving Game in the last post. In the Driving Game if the proportion of drivers staying on the left is x and 0≤x≤1 then the payoff of staying left is x(100)+(1-x)0 and the payoff of staying right is x(0)+(1-x)100.
The Nash Equilibrium is when x(100)+(1-x)0 = x(0)+(1-x)100
So with the driving game there as Nash Equilibria where x=1, x=0 and x=1/2
I hope all that made sense! If it didn’t feel free to comment to ask questions.
Here’s a video from Yale University on Evolutionary Game Theory!