> From: Chan Dorothy <firstname.lastname@example.org>
> I am reading the commentary for Mealey's article and I don't quite get
> the "game of Chicken" by Maynard Smith(which appears in Andre M.
> Colman's comm.). The model talks about a mixed-strategy equlibrium
> point which the population will evolve into. What is this point
> got to do with the probability of choosing to cooperate or to
> defect? Is this model different from the Prisoner's Dillema because
> it takes into account the no. of people who chosse either to
> cooperate or to defect tend to fluctuate? How does this model
> determine what is the payoff for choosing to cooperate (or to
The table was garbled and the figure was missing. I've fixed the table
and put a verbal description of the figure into the commentary. (Of
course the published version in the resource room is correct; the error
was only in the email version.)
In both Classical Prisoner's Dilemma and the "Chicken" variant, the
game can be one-time or iterated (repeated many times); it can also be
one against one, or collective, testing what happens if one person
defects and everyone else cooperates, and so on. For the classical
Prisoner's Dilemma, the payoffs are, in decreasing order T(emptation)
(I defect, everyone else cooperates) R(eward) (I cooperate and everyone
else does too) P(unishment) (I defect, and everyone else does too)
S(ucker's punishment) (I cooperate and everyone else defects).
In the "Chicken" variant, the order, instead of T R P S is T R S P.
That means I do worse if everyone defects than if I cooperate and
everyone else defects.
In the classical TRPS version of the game, defecting is always the
stable strategy for everyone. In the "chicken" TRSP version, depending on
how many defectors there are, up to a certain proportion, cooperating
is better; beyond that point, defecting is better. So the stable
strategy converges on a mixture of both types (and exactly what
proportion of each there will be depends on the actual cost/reward
numbers you give to T, R, etc.).
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