So by the way, here's a question for you. We can make it a quiz but I'm
not going to make it a quiz. What is the value of the game at p equals 0.4?
>> So I just plug it into those equations, and so negative 10 times
0.4 is like, negative 4 plus 5, is 1. So I'm going to say 1.
>> Mm-hm. So the value here is in fact $0.01. And if you put it
in the other equation you get the
same thing. Yes, right, because they're equal there.
>> That's
actually the answer to the entire kit and kaboodle. If.
>> Sorry, where were the kittens?
>> If A chooses a mixed strategy, where with probability 0.4 he
chooses to be a holder. Notice that it doesn't matter whether B
is a resigner or B is a seer. You will end up
here and there will be a value of plus 1 penny to A.
>> On average.
>> The expected value is.
>> Yeah.
>> plus 1.
>> Neat!
>> Now you might ask yourself, well what if B
decides to be a smarty pants and also do a mixed
strategy? What do you think would happen in that case? So
of course it changes something, but it doesn't change the value
of the game, because B, if B is a resigner, A
is getting one on average. If B is a seer then
A is getting one on average, and an, any average, any
convex average of one and one is going to give us one.
>> That's exactly right.
So in fact if you think about it, if B tries to
do a mixed strategy between these two lines. It's going to have to
be for every single point between the two lines, somewhere the average
is going to have to be somewhere in between there. No matter how you
weight that average. So it's like a bow tie is this space
of possible payoffs. Right and since for any let's if I just,
If B decided to pick some, you know, some values such that
it's in this region of the space. And it's going to end up somewhere,
say between
here that's fine, or here that's fine, or here that's fine,
or here that's fine, or here that's fine. More importantly, right here
it's going to have to be between the two lines. Where those
two lines cross. So, no matter what mixed strategy B chooses. On
average we will end up here, so the value of this
game is the expected value of this game and it is, plus
1 for A. Okay so hang on why is that not
there are other values that can be obtained. There is a minus
5 a plus 5, true. Why is it plus 1? Well its plus
1 here on average, the expected value of this game. Is plus 1.
>> So you say that the strategy for A is to choose .4.
>> Mm-hm.
>> But why? Like, why is that, of all the
different values. What [INAUDIBLE], just because the lines intersect. I
don't understand what makes that a good idea for A.
Like, it's not like it gets any additional payoff for having
things intersect. Well, if a is going to choose a mixed
strategy, and b is going to choose a mixed strategy. This is
the mixed strategy for a, that guarantees an expected value of
plus 1. Meanwhile, let's imagine this is the way you're going to set
it up. So we've been kind of talking in general that
a's going to choose the strategy; b's going to choose a strategy, and they're
going to go. And that, because we know everyone's rational, you already
know what strategy b's going to pick, and b already knows what strategy
a's going to pick Right, because we made this assumption about everyone
is trying to maximize their own utility, there own reward. So in
a mixed strategy, if you know you're going to go for a
mixed strategy, you have exactly the same situation. B can figure out
well what is it that A should choose and A can
figure out what is it that B should choose. So notice that
in this particular version of the game, the way it's setup. Even
if A announced beforehand to B. Listen I am going to choose
to be a holder with probability 0.4. It doesn't matter
what B chooses the expected value is this +1. Right?
>> Yeah.
>> But imagine if A said I am going to pick, I'm going to choose to be
a holder with probability 1. Well then, what
should b do? B should choose to seer.
>> Right.
>> It's exactly the situation we were in before.
>> Okay, but here's the thing I'm having trouble wrapping
my head around. So, it's not special that it's an intersection,
what's special maybe, is it that, you know, if b is always.
Okay; taking what you said before, that a is going to announce a
strategy. So for anything that A can announce B is going to
presumably do what's best for B which is to just minimize right?
>> Mm-hm.
>> So if you look at that triangle at the bottom.
>> Mm-hm.
>> If you think about that as those lines that V, upside down
V shape. Yeah exactly that thing. If you
think about that as being the payoffs, the
payoff function for A, as a function of
the probability P that it chooses to hold.
>> Mm-hm.
>> Then we've chosen the maximum.
>> Right.
>> Right, we're trying to find the, the probability for which A gets
the highest expected value. And it's at that, it's at that peak there.
>> Hm-hm. But notice something, Michael. For any case
where the two lines cross, you're going to have a function of this form.
>> Well, let's be sure of that, because that's,
I wasn't seeing, I was thinking that that was not
true. because basically in this case, we have a, a
line of positive slope and a line of negative slope.
>> Mm-hm.
>> Right? So, what if we, we can have
an intersection between two lines of negative slope. huh. So
again, by doing the same exercise you were doing before,
where you draw the, you sort of take the minimum
of the two lines at all points.
>> Yeah.
>> You'll end up with this. Where do you pick it then?
>> The far left, p equals 0.
>> Yeah.
>> But not the intersection in particular.
>> No, that's true.
>> Okay, alright, I just, wasn't quite
getting that. But so, the intersection is special,
Because in some cases that is where the max is. It can only I guess
the max can only be in those three places, right, it could be far left
far right or the intersection. There' s no other way to have a maximum of lines.
>> Right. And
b by the way there's another case like this where
they never actually cross.
>> Mm.
>> Or like this where they never actually cross. Got it.
>> Right, but in either case, right, it's always going to
be those three points. It's going to be the extreme or
where they cross. If they happen to never cross, then
that point goes away and you just pick the maximum. So
what you could, so, in fact, if you wanted to
think of an algorithm to choose the right thing to
do for A, you basically plot the lines. You take
the min at every point and you find the maximum point.
>> Oh I see, so you could just kind
of discretized it almost.
>> Yeah, and you're done. Seem reasonable?
>> Well discretized seems problematic but like I get
that we're trying to find the probability so that
we maximize. Oh wait we maximize the minimum of
the two other things. So it's like maximin again. Yeah.
>> Except here, so, in fact, it is exactly min
and max or max and min, except, oh, I guess, if
you're thinking from A's point of view. It's min and max
or max and min but in this case, there's this other
parameter, which is the probability. Which is how
you determine how you're doing the min and max.
>> Got it, right. In this bigger space, it's min
and max. I see, yeah, yeah, yeah, that makes sense. What
about so, why is A the one who needs to be
random? Why isn't B the one who needs to be random?
>> It's, you end up with the same, you end up in
the same place. Because again, remember,
you're, you're doing this kind of optimal
notion, where underneath all of that is this belief that both people are
going to be rational. So the only reason to do this, to take
the, you know, maximum minimum, is if you believe
that B is always going to try to minimize what
you do. But B's in exactly the same situation.
Right, from B's point of view, That's what's happening.
>> Huh. It's the same equations?
>> Yeah, it just it gets turned around. That's
actually an interesting exercise. Maybe that's the next homework assignment.
>> Okay.
>> So, did that all make sense to you?
>> [LAUGH] [CROSSTALK] It looks a little
green and scribbley, but yeah. Sure. That's cool.
>> That's how you know
when you're done, Michael. When it's all green and scribbley.
>> [LAUGH] Does this generalize to more than two
options? It seems like it's going to get messy fast.
>> Yes, it does. It generalizes to more than two options. Effectively
now, you're just doing a max over n things, instead of two things.
>> And you have to search for,
there's possibly way more intersections to worry about.
>> Yeah, but all you care about is the minimum.
Think of it as you're always drawing the minimum function.