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PROFESSOR: All right, so today what we're going to do
is we're just start with a short review problem
on rugged landscapes, just so that you
get some sense of the kind of thing
that I would expect you to be able to do a week from today.
And then we'll get into the core topic of the class, which
is evolutionary game theory.
And we'll discuss why it is that you
don't need to invoke any notion of rationality, which
is kind of the traditional thing we do when we're
talking about game theory applied to human decision
making.
Then we'll try to understand this difference
to know what a Nash equilibrium is in the context of game
theory versus an evolutionary stable strategy
in this context.
And we'll say something about the evolution of cooperation
and experiments that one can do with microbial populations
in the laboratory.
Are there any questions before I get started?
All right, so just on this question of evolutionary paths,
on Tuesday we discussed the Weinreich paper
where he talked about sort of different models
that you might use to try to make estimates of the path
that evolution might take on that fitness landscape
that he measured.
So he measured this MIC, the minimum inhibitory
concentration, on all 2 to the 5 or 32 different states,
and then tried to say something about the probability
that different paths will be taken.
So I just want to explore this question
about paths in a simpler landscape, where
by construction here, I'm to going to give you some fitness
values just so that we can be clear about why
it is that there might be different paths,
or what determines the probability
that different paths are taken.
So what we want to do is assume that we
are in a population that is experiencing this Moran process
or Moran model, constant population size N equal to,
in this case, we'll say 1,000.
And let's say that the mutation rate is 10 to the minus 6.
So each time that an individual divides, it has a 1
in a million probability of mutating.
And that's a per base pair mutation rate.
And I'll show you what I mean by that.
And in particular, we're going to have genotypes.
Originally when we discussed this,
we were talking about just mutations, maybe A's and B's.
But now, what we're going to have
is just a short genome that's string length 2.
So we might have 0, 0, which has relative fitness 1, 0, 1.
So we're assuming that this is relative fitness
as compared to the 0, 0 state.
We're going to start in the 0, 0 state with 1,000
isogenic individuals, all 0, 0.
And the question is, what's going to happen eventually?
And in particular, what path will be taken on this landscape
here?
In particular, what we want to know
is the probability of taking this path.
You can start thinking about it while I write out
some possibilities that we can vote for,
and I'll give you a minute to think about it.
So don't--
Are there any questions about what I'm trying to ask here?
AUDIENCE: So this is the long time?
So we assume that in the long time,
it will go from 0, 0 to 1, 1?
PROFESSOR: That's right, yes, so if we wait long enough,
the population will get there, and the 1, 1 genotype
will fix in the population.
We can talk a bit later about how long it's
going to take to get there, and so forth.
AUDIENCE: And we're assuming that from 0, 1,
it can't go back to 0, 0?
PROFESSOR: Right.
Yeah, so we'll discuss the situations
when we have to worry about that,
and when we don't, and so forth.
But for now, if you'd like, we can
say that this is even just mu sub b, the beneficial mutation
rate per base pair, assuming that the 0's can only
turn into 1s.
Then after we think about this, we
could figure out if that's important,
or when it's important, and so forth.
Yes.
AUDIENCE: [INAUDIBLE]?
PROFESSOR: No.
All right, so we're starting with all 1,000 individuals
being in the 0, 0 state, because now we're
allowing some mutation rate.
AUDIENCE: [INAUDIBLE].
PROFESSOR: And you also have to think
about this first mutation-- will it fix or not?
AUDIENCE: But it's not [INAUDIBLE] first mutation,
the mutation of one element of that population
will go to 0, 1 [INAUDIBLE]?
PROFESSOR: I'm not sure if I understand your question.
AUDIENCE: So if we start out with all 0, 0, and then one
mutation [INAUDIBLE]?
And if that mutation-- are we assuming
that that mutation is 0, 1 and then figuring out [INAUDIBLE]?
PROFESSOR: Well, OK, you're asking kind
of what I mean by path here.
AUDIENCE: Yeah, I guess.
PROFESSOR: Yeah, all right, so I'll say path
means that that this was the dominant probability
trajectory of the population through there.
We'll also discuss whether it somehow
is very likely is going to kind of have to go through one
or the other of them.
The probability of getting both mutations in one generation
is going to be 10 to the minus 12.
So that's going to be a very rare thing, at least given
these parameters, and so forth.
And then there's another question, which is,
will 0, 1 actually fix in the population
before you later fix this population?
And actually, I think the answers to all these questions
are in principal already on the board.
Because there's a question of do we
have to worry about clonal interference?
Are these things neutral or not?
And really, this is in some ways a very simple problem.
But in another way, you have to keep
track of lots of different things,
and which regime we're in and so forth.
So that's what makes it such a wonderful exam problem.
If you understand what's going on,
you can answer it in a minute.
But if you don't understand what's going on,
it'll take you an hour.
Yes?
No?
Maybe?
Well, I'll give you another 20 seconds.
Hopefully, you've been thinking about it
while we've been talking.
All right, do you need more time?
Why don't we go ahead and vote?
I think it's very likely that we will not
be at the kind of 100% mark, in which case
you'll have a chance to talk about it and think about it
some more.
Ready?
Three, two, one.
OK, all right, so we do have a fair range of answers.
I'd say it might be kind of something like 50-50.
And that's great.
It means that there should be something to talk about.
So turn to a neighbor.
You should be able to find somebody
that disagrees with you.
And if everyone around you agrees, you can maybe--
all right, so there's a group of D's and a group of B's here,
which means that everybody--
AUDIENCE: Let's fight.
PROFESSOR: All right, so everybody
thinks that everybody agrees with them,
but you just need to look a little bit more long distance.
So turn to a pseudo-neighbor.
You should be able to find somebody there.
It's roughly even here, so you should be able to find someone.
[INTERPOSING VOICES]
So I don't see much in the way of vibrant discussion
and argument.
You guys should be passionately defending your choice here.
[INTERPOSING VOICES]
Yeah, that's a higher order point.
I wouldn't worry about that.
[INTERPOSING VOICES]
All right, it looks like people are having a nice discussion.
But I might still go ahead and cut it short,
just so that we can get on to evolutionary game theory.
But I would like to see where people are.
And we'll discuss it as a group, so don't
be too disappointed if you don't get finished there.
But I do want to see kind of where we are.
Ready?
Three, two, one.
OK, so it still is, maybe, split roughly equally between D
and maybe a B-ish and some C's.
All right, does somebody want to volunteer their explanation?
Yes.
AUDIENCE: I'm not sure how good it is,
but I was thinking about what's the probability of going
to 0, 1 instead of 1, 0.
And I just took it as the ratio of the extra benefit of 0, 1
over the benefit of 1, 0.
PROFESSOR: Sure, OK, and just to start out,
which answer are you arguing for?
AUDIENCE: D.
PROFESSOR: D, OK, all right.
So you're saying D. And you're saying, all right,
maybe because of the extra, that 1, 0 is somehow
more fit than 0, 1.
And you've taken some relative rates or ratios
for which reason?
AUDIENCE: Well, I took 0.02 and then 0.1, which is 1/5.
And then I decided that that should
be around what it is, but slightly less, because there's
also a chance that [INAUDIBLE].
PROFESSOR: OK, yes.
I think the arguments there-- there's
a lot of truth to the arguments that you're saying.
Yeah, it's a little-- right and another question
is exactly why might it be 1/6 instead of 1/5 is,
I think, a little bit hazy in this here.
It's OK, but it's close.
Does somebody want to offer an explanation?
So here, that was an argument of roughly maybe why it's D-ish.
Because D is very different from B--
order of magnitude different.
So can somebody offer why their neighbor thought it was B?
Yeah.
AUDIENCE: So I knew that it was B, because I considered the two
paths, both from 0, 0 to 0, 1.
I first checked S, N and it's non-neutral.
So probably [INAUDIBLE] S. So the probability
for that first path would be the S for 0, 1,
so it's 0.02, which is 1/50, multiplied by the probability
that the other [INAUDIBLE] 1, 0 would die out [INAUDIBLE].
PROFESSOR: Right, so there are two related questions.
And I think that this explanation here
is answering a slightly different question.
OK, so let me try to explain what
the two questions are here.
So the question that you're answering
is, if you have kind of 998 individuals that
are 0, 0 individuals, and you have one that's 0, 1,
and you have one individual that is 1, 0.
So this is like these problems that we did a couple weeks ago,
where we said, you imagine in the population you have
a couple different kinds of mutants that are present maybe
in one copy.
And then we were asking, well, what's
the probability that this individual is going to fix?
And what's the probability this individual is going to fix?
And what's the probability that these guys are
going to go extinct, and this one will therefore fix?
And I think that's the calculation that you're
describing, where you say, OK, well,
in order for this individual to fix,
he has to survive stochastic extinction, which happens
with the probability of 2%.
And the 1, 0 individual has to go extinct,
which happens 90% of the time.
And so this is, indeed, answering the question
that if you had one copy of each of these two mutant individuals
in the population, that's the answer to what
is the probability that this 0, 1 mutant
would fix in the population.
Right?
But that's a slightly different question than if we ask,
we're going to start with an entire population at 0, 0,
and now these mutations will be occurring randomly
at some rate.
And then something's going to happen.
Somehow the population is going to climb up
this fitness landscape.
And we're trying to figure out the relative probability
that it's going to take kind of one path or another.
Do you see the difference between these two questions?
So indeed, this is the correct answer to a different question.
And so it's going to end up being D.
And now we want to try to figure out how to get there.
Because I think it is a bit tricky.
And in order in order to figure out to get there,
we have to make sure we keep track of which
parameter regime we're in.
So there are a couple of questions we have to ask.
First of all, we have to remember
that we start out with everybody, all 1,000
individuals in the 0, 0 state.
So there are initially no mutants in the population.
But they're just replicating at some rate.
And every now and then, mutation's going to occur.
Now one thing we have to answer, we have to think about,
is whether these are nearly neutral mutations.
Verbally yes or no?
Ready?
Three, two, one.
AUDIENCE: No.
PROFESSOR: No, right.
And that's because we want to ask
for, if it's nearly neutral, we want to ask,
is the magnitude of S times N much greater or much less
than 1?
In particular, if they're much greater than 1,
as is the case here, then we're in a nice, simple regime.
And it's easy to get paralyzed in this situation,
because there's more than one S. But in both cases,
S times N is much larger than 1.
We can take the smaller S, which is 2 over 100,
and S times N is 20, right?
So S for the 0, 1 state times N is 20,
which is much greater than 1.
What this tells us is that if we do
get this mutant appearing in the population,
then he or she will have a probability S of surviving
stochastic extinction.
So probability of surviving stochastic extinction
if the individual appears is equal to S 0,
1, which is equal to 2%.
Whereas for the 1, 0 state, that's going to be 0.1.
Now, this is assuming that the mutation appears
in the population, that's the probability it will survive
stochastic extinction.
Now, just as a reminder, surviving stochastic extinction
roughly corresponds to this becoming an established idea.
And becoming established was what again?
AUDIENCE: [INAUDIBLE].
PROFESSOR: What's that?
AUDIENCE: S1 is [INAUDIBLE].
PROFESSOR: It's when S1--
AUDIENCE: [INAUDIBLE].
PROFESSOR: Yeah, that's right.
All right, so established-- when we say established,
what we mean is that this corresponds
to saying that this probability that we talked about
before this X sub i is approximately equal to 1.
So this question is, how many individuals do you
have to get to in the population before you're
very likely to fix?
And what we found is that that number established went as 1
over the selection coefficient.
So in this case, you would need to have 50 individuals
before you were kind of more likely to fix than not.
So if you want to be much more likely,
you might need twice that or so.
Do you guys remember that?
This is not important for this question, necessarily,
but it might be important at a later date.
AUDIENCE: And so for nearly neutral mutations,
the whole point is that the number needed
to become established is equal to the population.
PROFESSOR: Yeah, so everything kind of works, right?
OK, so the way that we can think about this
is, now we have this population, 1,000 individuals.
They're dividing at some rate.
Mutations are going to appear.
Now we know if they did appear, the probability they would fix.
This is assuming there's no clonal interference, right?
Because if there's clonal interference,
then surviving stochastic extinction
is not the same thing as fixing.
If they both appear in the population,
and they both survive stochastic extinction,
then this mutant loses to this mutant.
That's the clonal interference.
Do we have to worry about clonal interference in this situation?
So remember, this was comparing the two time scales.
This was comparing the time between
successive establishment events, which went as 1 over mu N S.
And the other one is the time between the time to fix,
which went as 1 over S log of NS, right?
So we can ignore clonal interference
if this is much larger than that.
So no clonal interference corresponds
to mu N log NS much less than 1.
No clonal interference, same as this statement.
Is that right?
Did I do it right?
OK.
So and once again, there are multiple S's, and it's
easy to get kind of upset about this.
But you can just use whichever S would be--
which S would you want to use to be kind of--
AUDIENCE: Small or large [INAUDIBLE]?
PROFESSOR: To be on the safe or conservative side,
we want to take this to be as big as possible.
So we take S actually as big as we can, right?
It's in the log.
So details, right?
But we can see we have 10 to the minus 6, 10 to the 3,
and then this is the log of maybe 100, which
is, like, 4 or 5.
Is it closer to 4 or 5?
I don't know, but it doesn't matter.
We'll say 5.
This is indeed much less than 1.
So indeed, we don't have to worry
about clonal interference.
This is a wonderful simplification.
What it's saying is that the population is dividing.
Every now and then, a mutation occurs in the population.
It could be either the 0, 1 or the 1, 0.
But in either case, the fate of that mutation
is resolved before the next mutation occurs.
So you don't need to worry about them
competing in the population.
Instead, just at some constant rate they're appearing.
And given that they appear, there's some probability
that they're going to fix.
So that leads to effective rates going to each of those two
steps-- going to 0, 1 or 1, 0.
And in particular, this is like a chemical reaction, where we
have some chemical state here.
We have two rates.
There's the k going to 0, 1, the k going to 1, 0.
And what we know is we know the ratio of those rates.
And that's everything we need to know
to calculate the relative probabilities of taking
those states, because the probability of going through
to 0, 1-- we want to go that direction-- 0, 1, this
is going to be given by k 0, 1 divided by k 0, 1 plus k 1, 0.
So this is how we get 1/6 instead of 1/5.
Because this thing is 1/5 of that.
So it's like 1, and then 1, 5.
So this is actually, in principle, not quite answering
the question that I asked, because this
is talking about the relative probability of the first state,
the first mutant to fix.
In principle, it is possible that from there, there's
some rate of coming back.
Or they might not necessarily move forward on up that hill.
Do you guys understand what I'm talking about?
Because it goes from here to there.
Because we really want to know about this next step,
going to the 1, 1 state.
But in this case do we have to worry about going backwards?
No.
And why not?
It's very unlikely.
And in particular, you could think now
that you're here you can talk about the rate of going
to the 1, 1 state as compared to the rate of going to 0, 1.
And those are going to be exponentially different.
Because just as this was a non-neutral beneficial
mutation, that means that going from 0,
1 back is going to be a non-neutral deleterious
mutation.
So the probability of fixing it in the back direction is not 0,
but it's exponentially suppressed.
I think it's very important to understand
all the different pieces of this kind of puzzle,
because it incorporates many different ideas that we've
talked about over the last few weeks.
If there are questions, please ask now.
Yes.
AUDIENCE: What about the [INAUDIBLE]?
[INAUDIBLE] 0, 0 to 1, 0 to 1, 1?
Then it seems like the benefit of 1, 0 versus [INAUDIBLE].
PROFESSOR: All right, so you're wondering about--
so the fitness of the 1, 1 state was 1.2.
So you're pointing out that it's actually easier
to go from the 0, 1 state to the 1, 1 as
compared to the 1, 0 to the 1, 1.
AUDIENCE: Right, which seems like a reason for why
we wouldn't care about [INAUDIBLE].
PROFESSOR: Yeah, OK, right.
So if anything, in some ways, this
actually provides a bias going towards the 0, 1 state,
because it's saying that if we do get to 0, 1,
it's actually easier to move forward as
compared to this other path.
In practice, it doesn't actually matter,
because this acts as a ratchet.
Because all these mutations are non-neutral,
once you fix this state or this one, you can't go back.
So the population will move forward
once it gets to one of those two states.
Now I mean, it would be a very interesting question
to ask if we instead did a different arrangement.
What would the rate of evolution be, and so forth?
Yeah, but what you're saying is certainly true,
that if this took up all of the benefit going here,
then it may not actually be somehow
an optimal path in terms of the rate of evolution or something
like that.
I'll think about that when designing.
Problems.
AUDIENCE: In this system, 0, 0 eventually becomes 1, 1.
PROFESSOR: That's right.
AUDIENCE: So the probability is 1.
PROFESSOR: That's right, so we are guaranteed
that we will eventually evolve to this peak in the fitness
landscape.
And so what we're asking here is which of these two paths
is going to be taken.
AUDIENCE: Yeah, so how to mathematically prove
that the system will go from 0, 0 to 1, 1?
PROFESSOR: I mean, I feel like I kind of proved it,
although I understand that nothing I said was rigorous.
And of course, there are non-zero probabilities
of going backwards.
It's just that they are reduced.
And actually, you can prove, for those of you
who are interested in such things,
that over long time scales, there's
going to be an equilibrium that distribution
over all these states, where the probability of being
in a particular state will-- it goes as the fitness.
It scales as the relative fitness to the Nth power.
So we talk about these fitness landscapes
as energy landscapes.
And indeed, in this regime where you have small mutation rates,
then it's going to be a detailed balance.
And it's actually a thermodynamic system.
So then in that case, you can make a correspondence
between everything that we normally talk about,
where fitness is like energy and population
size is like temperature.
So the relative amplitude of being in this peak as compared
to the other states is going to be, in this case,
the ratios of those things is, indeed, described
by the ratios of the fitnesses.
And it's going to go as kind of like 1.1 to the 1,000th power,
which is big.
Which means that the population has really
cohered at this peak in the finished landscape.
Yeah.
AUDIENCE: So if you want to calculate a problem going
from 0, 1 to 0, 0, then [INAUDIBLE]
that would just be-- I guess I'm not sure.
PROFESSOR: OK, you want to know the rate
that that's going to happen.
AUDIENCE: Yeah.
PROFESSOR: No, that's fine.
Let's do that.
So for example, let's imagine that we
don't have-- so let's imagine that we just have the 0, 0
and the 0, 1 states, just so we don't have to worry
about going up the landscape.
And so what we have is we have r is relative fitness 1 and 1.02.
Now what we want to do is we want to ask, well,
what is the rate of going back and forth?
Well, so the rate of going forward,
well, we sample mutations at a rate mu.
And this is mu only for this one state,
because pretend that we're not going to mutate this other one.
So rate mu N, you have mutations appearing.
And times this s, 0.02, is the probability
that it'll actually fix in the forward direction.
And now what we want to know is the rate of coming back.
Well, the beginning part's the same,
because we have mu N is the rate that you
get this deleterious mutant in the population.
But then we need to multiply it by the probability of fixation.
And the probability of fixation is--
there was this thing X1, which was 1 minus-- now this is r,
but it's r in the other direction, so be careful.
Because the general equation was 1 over.
But now r, instead of being 1.02, is 1/1.02.
So which of these terms is going to be dominant?
This thing gets up to be some really big number
is our problem.
So we should be able to figure this out, though.
Because this new r is 1/1.02.
So we for example, have 1 minus 1.02, 1 minus 1.02
to the 1,000.
All right, so this is a negative number,
but this is a negative number, too.
So we end up with 0.02--
AUDIENCE: 200.
PROFESSOR: Is it 200?
Yeah, you're keeping only the first,
which, since it's much larger than 1, it's bigger than 200.
Right?
I mean do you guys understand what I'm saying?
You can't keep just the first term in a series.
If the terms grow with number.
AUDIENCE: [INAUDIBLE] squared, 3.98 or something.
PROFESSOR: Wait, which one?
AUDIENCE: 1.02 to the 1,000.
PROFESSOR: It's 4?
OK, all right.
OK, so it's 1 minus 4.
So it's 2/300.
OK, so this is teamwork, right?
OK, so there's less than a 1% probability of it fixing.
Is this believable?
AUDIENCE: It's about right.
PROFESSOR: 2, 1,000, 50-- I think
that you did it 1.02 to the 100 rather than 1.02 to the 1,000.
AUDIENCE: OK.
PROFESSOR: No?
Do you not have it in front of you?
AUDIENCE: No, it's 3-- [INAUDIBLE].
AUDIENCE: 4 times 10 to the 8.
AUDIENCE: I never thought that my calculator
would become so controversial.
AUDIENCE: Oh, 4 times 10 to the 8.
PROFESSOR: Yes, sorry, I was just saying this
doesn't-- so this is why I'm saying you always check to make
sure that your calculation makes any sense at all.
So it's not this.
But it's tiny, right?
AUDIENCE: Yes, [INAUDIBLE].
PROFESSOR: Yeah, because this didn't make sense,
because this was of the same order as-- well,
this would be larger than 1 over N, so it's totally nonsensical.
Because 1 over N would be the probability of fixation
of a neutral mutation.
This is a deleterious mutation.
It's not even nearly neutral.
So it has to be much less than 1 over N, right?
So this whole thing is 10 to the minus 10,
or something like that?
OK, 4 times 10 to the minus 8.
Well, OK, whatever.
It's 10 to the minus 9.
It's something small.
So this times the probability of fixation, which is 10 minus 9--
this is how you would calculate the rate of going backwards.
There's some rate that the mutation appears,
and you multiply by the probability that it would fix.
And it's tiny.
OK?
All right, any other questions about how
to think about these sorts of evolutionary dynamics
with presence of mutation, fixation, everything?
Yeah.
AUDIENCE: Can we handle a situation
where [INAUDIBLE] interference is important at this point?
PROFESSOR: Yeah, so this is what you do in your problem set
with simulations.
Yeah.
AUDIENCE: [INAUDIBLE] numerical.
PROFESSOR: You know, I think that it
gets really messy with clonal interference, I'll say.
AUDIENCE: But, like, with basic--
I guess I was thinking about it and you could probably
imagine that [INAUDIBLE] calculate the probability
that 1, 0 doesn't arise first.
PROFESSOR: Right, yeah, OK, this is an important statement.
In the limit, as you get more and more mutations, when
clonal interference is really significant,
then you're pretty much just guaranteed
to take the 1, 0 path.
Because if you have many mutants,
the definition of clonal interference is you
have multiple mutations that have established.
And once you have multiple mutations that
have established, then it's likely that one of them
is going to be this.
And if it's established, it's going to win.
But the other thing is that as you go up in the mutation rate,
you don't even do successive fixations.
So it may be that neither state ever actually fixes,
because it could be that the 1, 0 state
is growing exponentially, but is a minority of the population.
And it gets another mutation that allows it to go to 1, 1.
So as you increase the mutation rate,
you don't have to actually take single steps.
You can kind of move through states.
And there's a whole literature of the rate at which you
cross fitness valleys.
So this is like tunneling in quantum mechanics or so.
And it has a lot of the same behaviors,
in the sense of exponential suppression of probabilities
as a function of the depth and the width of the valley you're
trying to traverse.
And there's some very nice papers,
if you're interested in looking at this stuff.
And one of them is actually in the syllabus that I mentioned.
I'm trying to remember who.
It was Journal of Theoretical Biology,
but I put it on as optional reading for those of you
who are interested.
All right, OK.
So what I want to do now is I want to switch gears,
so we can think about this evolutionary game theory
business.
And I think the most important thing
to stress when thinking about evolutionary game theory
is just that this point that we don't need to assume anything
about rationality.
Because the puzzles that we like to give each other in your dorm
rooms Friday night, you give these logic
puzzles to each other.
Is that-- I don't know.
[LAUGHTER]
OK, let me just say, back when I was in college, that was, like,
all the cool kids were doing it.
But in these puzzles, you assume that there's hyper-rationality.
You assume that if this guy knows
that I did this, and that, and if I
did that, he would do that.
And then you end up, and then you
have the villagers that are jumping off
cliffs on the seventh day.
Have you guys done these puzzles?
No?
OK.
All right.
Well.
[LAUGHTER]
So the point was that people assume
that when we're talking about game theory,
you have to invoke this hyper-rationality even
humans don't engage in.
And I think that it's just very important
to remember that we're talking about evolutionary game
theory in the case of, well, biological evolution.
You don't assume anything about rationality.
Instead, you simply have mutations that sample
different strategies.
And then you have differences in fitness
that just lead to evolution towards the same solutions
of the game.
So it's evolution to the game solutions,
so the Nash equilibrium, for example.
So it's not that we think that the cells are
engaging in any sort of weird puzzle solving.
Instead, they're just mutations.
And the more fit individuals spread in the population.
And somehow, you evolve to the same or similar solutions,
to these Nash equilibria in the context of game theory.
And we'll see how this plays out in a few concrete examples.
Now, there are always different ways of looking at these games.
One thing I want to stress, though, is that all
the selection that we've been talking
about in the last few weeks, that all
is consistent with game theory in the sense
that the idea of the game theory is
that we allow for the possibility
that the fitness of individuals depends
upon the rest of the population.
Whereas in all these calculations we've been doing,
I told you, all right, I just gave you some fitnesses.
So I said, here we have a 0, 0 state that has some fitness.
0, 1 has a higher fitness, and so forth.
But in general, these fitness values
may depend upon what the population composition is.
And in that situation, then you want to use evolutionary game
theory.
In many cases, people just assume
that you can do something like this-- that you can describe it
as some fitness landscape.
But you can't do that if there's this frequency-dependent
selection-- if there's any sort of evolutionary game
interactions going on.
So it's just important.
If the fitnesses depend on composition--
this is the population composition--
then you cannot even define a fitness landscape.
Then no fitness landscape.
For example, you can have situations where the population
evolves to lower fitness.
So you can have a situation where,
if I tell you individual 0, you measure its growth
rate, whatnot, its fitness might be 1.
So this is genome, and this is fitness.
Now, if I go and I measure the fitness
of some other individual, different genome-- so
another strain of bacteria or yeast or whatever--
and you say, oh, well, its fitness is 1.2.
So this strain has higher fitness than this strain.
Now, it would be very natural to assume that this strain will
out-compete this strain.
And indeed, that's been the assumption in everything
we've been talking about.
But it's not necessarily true.
And that's the basic insight of evolutionary game theory,
is that just knowing the fitness of a pure population
is not actually enough information to know that it's
going to be selected for.
Because it's still possible that in a mixed population,
the genome 0 may actually have higher fitness
than the genome 1.
And once you kind of study these things,
it's kind of clear that it can happen.
But then it's easy to then go back to the lab
and forget that it's true.
And so we'll see how this plays out.
AUDIENCE: On this game theory, [INAUDIBLE]?
PROFESSOR: No.
That's the other thing, is that I
like to just draw these things as graphs, because I
think it's much easier to see what's happening.
And it's clear that things can be non-linear.
But the basic insights are all intact.
From my standpoint as kind of an experimentalist--
don't forget about the exam-- I think
that the more formal evolutionary game theory
thing-- these two-player games that you guys just
read about-- I think they're important because they tell you
what are the possible outcomes of measurements or of systems,
even in the most simple situation where everything's
linear.
Now, when things are not linear, of course
you can get even richer dynamics.
But in practice, you basically get the categories of outcomes
that we saw there.
So maybe what I'll do is-- so what we're going to do
is think about competition between two individuals
A and B. And often, we talk about these things
in the context of the two-player games, where we have A, B, C,
D. And because this is really importing the kind of approach,
or the nomenclature, from conventional game theory,
and then immediately applying it to populations where you just
assume that all the individuals have
equal probability of interacting with everybody
in the population.
So it's what you would get if you just
had some two-player game like they study in game theory,
but in a population of 1,000 or whatnot.
You just made a bunch of random pairwise interactions.
You had them play the game.
And then you had them do that again over time.
And then this is the payouts that you read about
in Chapter Four are kind of what would happen
in that sort of situation, where everybody's
interacting with everybody else with equal probability.
Now, remember the way that you read
this is that, depending upon the strategy that these guys are
following, you get different payouts.
And normally what we say is that if this could be,
for example, strategy one and two, strategy one and two.
And this is telling us about the payout
that the A individual gets depending on what he does,
and depending upon what his opponent does.
Now, we're not explicitly saying what the payout to the B
individual is, but we're assuming
that this is a symmetric game, so you could figure it out
by looking at the opposite entry.
So if A follows strategy one, B follows strategy one,
then individual A gets little a fitness,
whereas B also gets little a fitness,
because it's a symmetric game.
So the case it's different is when we're in the diagonals.
And from this framework, you can see
that there are going to be already
a bunch of kind of non-trivial things
that can happen, even in this regime where everything's
linear.
And the probably best well-known of these
is this Prisoner's Dilemma, which
is the standard model of cooperation
in the field of game theory.
So there's a story that goes along with it.
It's this idea that-- I'm sure you guys watch these cop shows,
where you have the cops bring in the two accomplices.
And then they put them in separate rooms.
And they tell them that they have
to confess to committing the crime,
because the guy in the other room is confessing,
and if he doesn't confess, then he's
going to be in trouble, et cetera.
You've seen these cop shows?
And incidentally, in these questions,
when cops are questioning witnesses,
they're actually allowed to lie to the person being questioned,
which feels a little bit weird, actually.
Doesn't it?
I know, I know, this is not relevant.
So the idea of the prisoner's dilemma
is that if you set up these jail sentences in the right way,
then it could be the case that each individual has
the incentive to confess, even though both individuals would
be better off if they cooperated.
And you can come up with some reasonable payout structure
that has that property.
And we'll call this-- so this is for individual one,
say and individual two.
So there are different strategies you can follow.
And do you guys remember from the reading
slash my explanation how to read these charts?
All right, now the question is, just to remind ourselves,
what is the Nash equilibrium of this game?
And I know that you read about it last night.
Well, use your cards.
Is it C or is it D?
Or is there no Nash equilibrium, you can flash something else.
AUDIENCE: Are those negative or positive?
PROFESSOR: These are positive fitness.
I kind of don't like the Prisoner's Dilemma as a story,
because it's not very intuitive, because you
have to actually specify the jail terms,
and you have to remember that jail terms are bad, not good.
So these are good things, OK?
These are years off that you get as a result
of doing one thing or another.
You want to get big numbers.
Ready?
Three, two, one.
So at least we have a majority that are D,
but it's not all of them.
And I think this is basically a reflection--
and D is indeed the Nash equilibrium.
It's to do this strategy D that we're saying here.
All right, now the question is why?
And part of the challenge here is just understanding
how to read these charts.
Now, first of all, the payout that everybody gets if everyone
follows strategy D is what?
Verbally, three, two, one.
AUDIENCE: One.
PROFESSOR: So everybody gets payout one.
Now, if you look at this chart, you
say, well, gee, that is a shame.
Because 1 is just not the biggest number you see here.
And indeed, the important point to note
here is that if both players had followed this strategy
C for cooperate, D for defect, then both individuals would be
getting fitness 3, or payout 3.
So the idea here is that both individuals would
do better if they both played strategy C.
But the problem is that that's not evolutionarily stable.
Or in the context of game theory,
that is cheatable in some ways.
And so the reason that this is a Nash equilibrium
is that you ask-- so a Nash equilibrium, what it means,
if you recall, is that if everyone's
playing that strategy, then nobody has the incentive
to change strategy.
So no incentive to change strategy.
So now you just imagine, let's say,
that you're playing against somebody else,
or in the context of biology, it's
a population of individuals following the D strategy.
The question is whether you as an individual
would have the incentive to switch to the other strategy?
And the answer is no, because what you have control over
is this rows.
The column is specified by the rest of the population.
So if you're in this state, what you have a choice of
is to switch to the cooperate strategy, which
would be to go up here.
So you have a choice to move up to this 0 payout,
but that's not to your advantage.
Now, it's true that your opponent would get payout 5.
So you opponent would actually do wonderfully.
But you would do poorly.
So you'd be selected against, if you imagine
this being in the context of biology--
that you have a genotype that are playing
D. If you're a mutant that starts following this strategy
C, you have lower fitness, so you're selected against.
So that's saying that the strategy D is noninvadable.
We can also think about what happens if we're
a population of cooperators.
Now everybody has high fitness-- fitness 3.
Question is, what happens if there's
a mutation that leads to one individual following the D
strategy?
Is he selected four or not?
AUDIENCE: Yes.
PROFESSOR: Yes, so the point here
is that you always will have higher fitness,
regardless of what your opponent does in the context of a game
theory situation, or regardless of the distribution
of cooperation and defection in the population.
It's always better to be a defector.
So the problem here is, it's always better to play D.
Now, I really like drawing the graphs of these things,
because I think it's just much more clear.
And you can either draw the fitness
of the two types minus each other, or just the raw fitness.
Yes.
AUDIENCE: So what if instead of 5, you have 7?
Because then the population as a whole's fitness
decreases when you [INAUDIBLE].
So how does that--
PROFESSOR: So you're saying if this 5 were a 7 instead?
AUDIENCE: Yeah.
PROFESSOR: Right, then what you're saying
is that-- so it doesn't change.
The Nash equilibrium is still defect.
The subtle thing here is that, in general,
in terms of game theory, we like it when the mean of these two
is smaller than this one.
That's why you're asking, right?
Exactly, because that's right.
Exactly, right.
So yeah, so that's a slightly more complicated situation,
because in that situation, then, if you had two rational agents,
say, playing this game, then you could alternate cooperation
and defection.
And that would actually be the ultimate form of cooperation
in such a game, because you could actually get a higher
payout by alternating.
Right, so we've chosen the numbers
as they are so that this is subtlety is not an issue.
Does everybody understand the issue there?
So in the context of evolutionary game theory, what
we can do is we can plot as a function of the fraction
of the population that's cooperator between 0 and 1,
say.
And we can plot the payout for the cooperator
and for the defector.
For example, I'm going to draw a solid line for the cooperator,
dashed line for the defector.
Now the question is, what should be the y-axes on either ends
and so forth?
Do you understand?
So what should these things look like?
I'd like to encourage you to-- I'll
give you 30 seconds to try to draw
what this should look like.
So this is the payout or the expected payout.
So we're assuming that you're going to interact randomly
with the other members of the population as a function
of the fraction cooperator.
So then 1 minus that will be the fraction defector.
Do you understand what I'm trying to ask you to do?
AUDIENCE: So the scale on the right-hand side supposed
to be for the defectors?
[INAUDIBLE]?
PROFESSOR: This is just a legend, or key, or something.
So I want you to draw something over here that's
a solid line and a dashed line.
AUDIENCE: All right, so it's just one scale.
And you don't--
PROFESSOR: It's one scale.
AUDIENCE: [INAUDIBLE].
PROFESSOR: Oh, yeah, sorry.
I'm just telling you what's going to be a solid line
and what's going to be a dashed line.
And I'll give you a hint, that up here is number 5.
This is going to be the expected payout for a lone individual
given the rest of the population is following
some fraction of cooperator.
Do you guys understand what I'm asking you to do?
Because I'm a little bit concerned that there
are very few plots in front.
AUDIENCE: What is fc?
PROFESSOR: So this is the fraction of the population
that's cooperator.
Well, I was giving you a chance to think about it.
But from looking around, I think that maybe you're
not quite sure what I'm trying to ask you to do.
So I'm trying to plot the expected payout
for an individual that is either cooperating or defecting,
based on the fact that the rest of the population
has some composition between all cooperate or all defect.
So it's the evolution game theory extension
of this simple model.
So first, we can ask, well, if the entire population is
cooperating, we want to know the fitness for a cooperator
or defector.
Well, this is really just saying that we're
all the way over on here, and we just choose between the two.
And the defector is the 5 one.
So this is going to be a dashed line that's
going to start from here.
And then this is 2 and 1/2, 3.
So this is cooperator starts here, and defector starts here.
Do you understand what I'm?
Now, OK, let's see.
So now this is the one where if everybody else is defecting,
well, now, the cooperator line goes to what?
Verbally, three, two, one.
AUDIENCE: 0.
PROFESSOR: 0.
The defector line goes to 1.
OK, that line, I started going the wrong direction,
but that's supposed to be a line.
So this is an example of what this
looks like for the Prisoner's Dilemma.
And what you see is the defector fitness
is always above the cooperator fitness.
So for any population composition,
defectors have higher fitness than cooperators.
So evolution brings you to the pure defecting state,
where you have fitness 1.
And if you want, you could calculate
what the mean fitness of the population is, for example.
And the mean fitness starts out over here,
and ends up over here.
So the mean fitness decreases over time.
Now, you can imagine that in the simple, two-player models,
all these are lines.
But you can imagine that the only thing that's important
are how these lines cross each other.
So for example, there are only a few different things
that can happen.
You can have one strategy that dominates,
which is what occurred here.
And surprisingly, that does not mean
that that strategy is higher fitness,
in the sense that you may evolve to a state of low fitness.
That's what's weird.
You can have coexistence, or you can have bi-stability.
So I'll give you another example of this.
So now we're just going to have two strategies.
The strategies-- we'll just call them A and B.
And the question is, what is the Nash equilibrium?
Is it A, B, or C should be neither, D is both.
Do you understand?
I'm going to ask, because if this is the game
and this is the interaction, is the Nash equilibrium A,
Nash equilibrium B?
If you vote C, it means neither, D means both.
Do you understand the question?
I'll give you 30 seconds to think about it.
All right, are we ready to vote?
Ready, three, two, one.
All right, so we have a fair distribution.
I may not have us vote, but yeah, in this case,
they're actually both Nash equilibria.
So let's see this.
If both individuals, or an entire population,
say, is playing A, they're getting fitness 5.
Question is, as a lone individual,
you can choose to switch over and get fitness 3.
Do you want to do that?
No.
So that means that A is going to be a Nash equilibrium.
Incidentally, the difference between
the so-called regular Nash equilibrium and the strict Nash
equilibrium is that Nash equilibrium
means that no individual has the incentive to change strategy.
A strict Nash equilibrium means that any change in strategy
leads to an actual decrease in fitness.
So it's a question of whether you can make neutral changes
in strategy or not.
Do you understand?
So A is a Nash equilibrium.
What about B?
Well, in that case, everybody's getting to fitness 1.
Now, as a lone individual, what can you do?
All you can do is switch.
As an individual, you can only choose rows.
So you go up to 0.
That's a decrease of fitness.
So that means that strategy B is also a Nash equilibrium.
So there are two Nash equilibria in this game.
And what does that mean about which regime
you're in here, if you convert this into an evolutionary game
theory scenario?
Ready, three, two, one.
OK, so a majority is saying yes.
This is indeed a situation in which you have bi-stability.
So what does that mean in terms of these lines if we draw them?
So this is payout as a function of the fraction that
is playing the A strategy.
Should the lines cross?
Yes or no?
Ready, three, two, one.
AUDIENCE: Yes.
PROFESSOR: Yes, and indeed, in principle,
the math that we do in all these situations
is kind of super simple.
Yet it's easy to get confused about what's going on
in all these situations.
So the idea here is that if the population is A, that
means that the A here is at 5.
But then it goes down to 0.
Whereas over here, B here is 3.
And then it goes to 1.
Because these two lines cross, does that
mean that you have bi-stability?
Ready, yes or no, three, two, one,
AUDIENCE: No.
PROFESSOR: No, and why not?
AUDIENCE: [INAUDIBLE].
PROFESSOR: That's right, because you can also
do the other thing, and then that leads to coexistence.
Now, in some ways coexistence is the most subtle
of the situations.
And that's for an interesting reason.
AUDIENCE: Sorry, sir, you said you can also
do the other thing.
What is the other thing here?
PROFESSOR: I'm saying that these things can
cross in the other orientation.
Let me put a matrix out there, and then--
so this is something that, for example, is what's
known as a Hawk/Dove game.
Or it has many other names.
And we can maybe figure out what would be the Hawk strategy,
and what's the Dove strategy.
Now, we want to ask the same question-- is A a Nash
equilibrium?
Is B a Nash equilibrium?
Is it neither?
Or is it both?
And maybe I shouldn't have covered this up,
so you're not influenced, in case you actually
did do the reading.
Then I don't want to you to be influenced by this.
So think about it for 30 seconds.
Do you need more time?
Let's go see where we are.
Ready, three, two, one.
All right, so most of the group is agreeing in this case,
neither are the Nash equilibrium.
So neither are a Nash equilibrium.
Does that mean that this game has no Nash equilibrium?
Yes or no, verbally-- ready, three, two, one.
AUDIENCE: No.
PROFESSOR: No, it does not mean that.
This game has a Nash equilibrium.
And indeed, all games like this have Nash equilibria.
And this is what Nash won the Nobel Prize for,
so this is the famous one-page paper published in PNAS.
If you look at it, I have no idea what it says.
I mean, he basically just pointed out
that this theorem implies this, implies that-- done.
And so it's good that somebody knew what he was saying,
otherwise we'd be in trouble, all of us.
So what he proved is that such games,
even with more players, more options, and so forth,
they always have such a solution in this sense.
There exists some strategy such as if everybody
were playing in, nobody would have the incentive
to change strategy.
But you have to include so-called probabilistic or
mixed strategies.
And we can draw what this thing is.
So just like always, so everyone else is following A,
then A starts here at 3.
And then it goes to 1.
Whereas the B individuals start at 5, and they go to 0.
So this looks very similar to that,
but they're rather different, in the sense
that in this situation, we had bi-stability.
So if you look at the direction of evolution,
depending upon where you start, you
go to either all B or all A. Whereas
in this situation over here, we have coexistence.
Does not matter where you start.
So long as you have some members of both A and B
in the population, you'll always evolve to the same equilibrium.
Now, the important thing here that's, I think,
interesting is that in a population,
if you have genetic A's and genetic B's that are each
giving birth to their own type, then
you evolve to some coexistence of genotypes.
So here, this is some fraction.
f a star is the equilibrium fraction, fraction
of A in the population.
So this is a case where you have genetic diversity that
leads to phenotypic diversity in the population.
Whereas the mixed Nash equilibrium-- this
is a situation where you have, in principle,
genetic homogeneity.
So this is a single genotype that is implementing
phenotypic heterogeneity.
And indeed, one of the things that we've
been excited about exploring in my group
is this distinction here, where it's known that in many cases,
isogenic populations of microbes can exhibit
a diversity of phenotypes as a result of, for example,
stochastic gene expression and bi-stability.
So that's a molecular mechanism for how
you might get heterogeneity.
Another question is, what is the evolution explanation
for why that behavior might have evolved?
Now in general, we cannot prove why something evolved,
but we can make educated guesses that make experimentally
testable hypotheses.
And for example, in the experiment
that we've been doing, we've been looking at bi-modality
in expression of the galactose genes in yeast.
And that was still a problem set in early-- oh,
no, we removed that one this year.
Well, so experimentally yeast, in some environments,
bimodally or stochastically activate
the genes required to break down the sugar galactose.
And what we've demonstrated is that if you
make the mutants that always turn on or always don't
turn on these genes, then they're actually playing game
where you actually get this exact thing-- where
you get evolution towards coexistence of those two
strategies.
So that's saying that maybe the wild type that
follows this stochastic, mixed strategy-- it
may be implementing the solution of some game that
is a result of such frequency dependence.
There are other possible explanations to this.
In the coming weeks, we'll talk about this idea
of bet hedging-- that given uncertain or fluctuating
environments, it may be advantageous
for clonal populations to have a variety of different strategies
to cope with that uncertainty.
So we'll talk about those models later.
But since we're talking about mixed strategies now,
I wanted to mention that.
Yeah.
AUDIENCE: So f a star, just to be sure,
is going to converge to this probability [INAUDIBLE]?
PROFESSOR: Exactly, so the Nash equilibrium mixed strategy
plays A with probability-- so it's p should
be equal to f a star.
Exactly, so indeed, the heterogeneity there
can be implemented either way.
It's either coexistence of genotypes following
different strategies, or it could
be one genotype implementing both,
or it could be a mixture of those, actually.
Indeed, a characteristic of these situations
is that, let's say that you have a genotype-- a population has
this genotype that is implementing the mixed Nash
equilibrium choosing strategy A with probability p.
That's that equilibrium fraction.
What's interesting is that any individual
in the population following any strategy has the same fitness.
And of course, that's kind of why this was in equilibrium.
This equilibrium is when the two strategies have equal fitness.
But the funny thing is, what that means
is, it doesn't matter what you do at the equilibrium.
Depending on how you look at it, it's either
super deep or super trivial.
But it's a weird thing that if your at the equilibrium,
or if the population or the opponent
is playing this Nash equilibrium in these games,
then it just does not matter what you do.
You can do A. You can do B, actually, in any fraction.
So since A and B have the same fitness,
you can choose between them at any frequency you want,
and you have the same fitness, if the rest of the population
is playing this mixed Nash equilibrium.
And indeed, in this there are nice conditions
for what makes it this Nash equilibrium.
And I'm going to just highlight that you should make sense
of why it means what it is.
So if the payout, the expected fitness or payout--
if you're following the Nash equilibrium
against Nash equilibrium-- is equal to this guy.
So that's what I just said-- that it
doesn't matter what you do.
If everyone else is doing p star,
you have the same fitness.
So that's saying it's a Nash equilibrium.
Whereas there's another interesting kind of statement
here, that--
AUDIENCE: [INAUDIBLE]?
That you can't unilaterally increase your fitness
by switching.
PROFESSOR: Right, it's an equality, which
means it is a Nash equilibrium.
Because it's saying that you don't have the incentive
to change strategy.
It's true that you're not dis-incentivized.
So it's a Nash equilibrium.
It's not a strict Nash equilibrium.
AUDIENCE: [INAUDIBLE].
PROFESSOR: Well, it has to be greater than/equal to,
and it's actually equal to.
The condition it has to be greater than or equal to,
but it's equal to, which means it is a Nash equilibrium.
AUDIENCE: [INAUDIBLE] Nash equilibrium for that situation,
[INAUDIBLE] strategy.
PROFESSOR: That's right.
So this is not the definition of that.
But this thing is true, which means
that it's a Nash equilibrium.
And this other thing that's interesting is that--
so this tells us that it's actually one of these ESS's.
And if you have questions about this, I'm happy to answer it.
It's explained in the book as well.
We are out of time, so I should let you go.
But good luck on the exam next Thursday.
If you have questions and you want to meet with me,
I'm available on Tuesday.
So please let me know.