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I'll start by explaining 'Nash equilibria' for 2-person games. These are situations where neither player can profit by changing what they're doing. Then I'll introduce 'mixed strategies', where the players can choose among several strategies with different probabilities. Then I'll introduce evolutionary game theory, where we think of each strategy as a species, and its probability as the fraction of organisms that belong to that species.
Back in Part 9, I told you about the 'replicator equation', which says how these fractions change with time thanks to natural selection. Now we'll see how this leads to the idea of an 'evolutionarily stable strategy'. And finally, we'll see that when evolution takes us toward such a stable strategy, the amount of information the organisms have 'left to learn' keeps decreasing!
We can describe a certain kind of two-person game using a payoff matrix, which is an
Note that in this kind of game, there's no significant difference between the 'first player' and the 'second player': either player wins an amount
We say a strategy
for all
For example, suppose our matrix is
Then we've got the Prisoner's Dilemma exactly as described last time! Here strategy 1 is cooperate and strategy 2 is defect. If a player cooperates and so does his opponent, he wins
meaning he gets one month in jail. We include a minus sign because 'winning a month in jail' is not a good thing. If the player cooperates but his opponent defects, he gets a whole year in jail:
If he defects but his opponent cooperates, he doesn't go to jail at all:
And if they both defect, they both get three months in jail:
You can see that defecting is a Nash equilibrium, since
So, oddly, if our prisoners know game theory and believe Nash equilibria are best, they'll both be worse off than if they cooperate and don't betray each other.
So far we've been assuming that with 100% certainty, each player chooses one strategy
Now let's throw some probability theory into the stew! Let's allow the players to pick different pure strategies with different probabilities. So, we define a mixed strategy to be a probability distribution on the set of pure strategies. In other words, it's a list of
that sum to one:
Say I choose the mixed strategy
so the expected value of my winnings is
or using vector notation
where the dot is the usual dot product on
We can easily adapt the concept of Nash equilibrium to mixed strategies. A mixed strategy
This means that if both you and I are playing the mixed strategy
If this were a course on game theory, I would now do some examples. But it's not, so I'll just send you to page 6 of Sandholm's paper: he looks at some famous games like 'hawks and doves' and 'rock paper scissors'.
We're finally ready to discuss evolutionarily stable strategies. To do this, let's reinterpret the 'pure strategies'
Similarly, we'll reinterpret the 'mixed strategy'
We'll reinterpret the payoff matrix
where the fitness function
But in evolutionary game theory it's common to start by looking at a simple special case where
where
is the fraction of replicators who belong to the
What does this mean? The idea is that we have a well-mixed population of game players—or replicators. Each one has its own pure strategy—or species. Each one randomly roams around and 'plays games' with each other replicator it meets. It gets to reproduce at a rate proportional to its expected winnings.
This is unrealistic in all sorts of ways, but it's mathematically cute, and it's been studied a lot, so it's good to know about. Today I'll explain evolutionarily stable strategies only in this special case. Later I'll go back to the general case.
Suppose that we select a sample of replicators from the overall population. What is the mean fitness of the replicators in this sample? For this, we need to know the probability that a replicator from this sample belongs to the
This is just a weighted average of the fitnesses in our earlier formula. But using the magic of vectors, we can write this sum as
We already saw this type of expression in the last section! It's my expected winnings if I play the mixed strategy
John Maynard Smith defined
In simple terms: a small 'invading' population will do worse than the population as a whole.
More precisely:
for all mixed strategies
is the population we get by replacing an
Puzzle: Show that
and also, for all
The first condition says that
Again, I should do some examples... but instead I'll refer you to page 9 of Sandholm's paper, and also these course notes:
Now comes the punchline... but with a slight surprise twist at the end. Last time we let
be a population that evolves with time according to the Lotka–Volterra equation, and we let
obeys
where
for all
We can adapt this to the special case we're looking at now. Remember, right now we're assuming
so
Thus, the relative information will never increase if
or in other words,
Now, this looks very similar to the conditions for an evolutionary stable strategy as stated in the Puzzle above. But it's not the same! That's the surprise twist.
Remember, the Puzzle says that
and also
Note that condition (1), the one we want, is neither condition (2) nor condition (3)! This drove me crazy for almost a day.
I kept thinking I'd made a mistake, like mixing up
But the solution turned out to be this. After Maynard Smith came up with his definition of 'evolutionarily stable state', another guy came up with a different definition:
For him, an evolutionarily stable strategy
and also
Condition (1) is stronger than condition (3), so he renamed Maynard Smith's evolutionarily stable strategies weakly evolutionarily stable strategies. And condition (1) guarantees that the relative information
Except for one thing: why should we switch from Maynard Smith's perfectly sensible concept of evolutionarily stable state to this new stronger one? I don't really know, except that
and
So, it's a small mystery for me to mull over. If you have any good ideas, let me know.
You can read a discussion of this article on Azimuth, and make your own comments or ask questions there!
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