We live in a complex world with diverse people, firms, and governments whose behaviors aggregate to produce novel, unexpected phenomena. We see political uprisings, market crashes, and a never ending array of social trends. How do we make sense of it? Models. Evidence shows that people who think with models consistently outperform those who don't. And, moreover people who think with lots of models outperform people who use only one. Why do models make us better thinkers? Models help us to better organize information - to make sense of that fire hose or hairball of data (choose your metaphor) available on the Internet. Models improve our abilities to make accurate forecasts. They help us make better decisions and adopt more effective strategies. They even can improve our ability to design institutions and procedures. In this class, I present a starter kit of models: I start with models of tipping points. I move on to cover models explain the wisdom of crowds, models that show why some countries are rich and some are poor, and models that help unpack the strategic decisions of firm and politicians. The models covered in this class provide a foundation for future social science classes, whether they be in economics, political science, business, or sociology. Mastering this material will give you a huge leg up in advanced courses. They also help you in life. Here's how the course will work. For each model, I present a short, easily digestible overview lecture. Then, I'll dig deeper. I'll go into the technical details of the model. Those technical lectures won't require calculus but be prepared for some algebra. For all the lectures, I'll offer some questions and we'll have quizzes and even a final exam. If you decide to do the deep dive, and take all the quizzes and the exam, you'll receive a Course Certificate. If you just decide to follow along for the introductory lectures to gain some exposure that's fine too. It's all free. And it's all here to help make you a better thinker!
Mathematics on Urn Models
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En provenance du cours de University of Michigan
Model Thinking
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À partir de la leçon
Path Dependence & Networks
In this set of lectures, we cover path dependence. We do so using some very simple urn models. The most famous of which is the Polya Process. These models are very simple but they enable us to unpack the logic of what makes a process path dependent. We also relate path dependence to increasing returns and to tipping points. The reading for this lecture is a paper that I wrote that is published in the Quarterly Journal of Political Science
Rencontrer les enseignants
Scott E. PageProfessor of Complex Systems, Political Science, and Economics
Center for the Study of Complex Systems
Hi. The previous lecture I stated two results, regarding the polya process. In
terms of the, each equilibrium is equally likely, and that, any history of R. Red
balls, and B, Blue balls is equally likely. What I want to do in this lecture
is show you why those two proofs are true. Now I'm hoping you prove these on your
own. You don't even have to watch these lectures. You're just watching to see if I
did it the same way you did. But at if you haven't tried them at this point stop the
video and try on your own just by doing some simple examples to see why these
things are true. To get us started I want to remind us how the player process works.
Remember we started with one blue ball one red ball. We pick a ball out, we look at
its color and then we add a ball of the same color of the ball we selected. So
suppose for example that I start out with one red, one blue. The odds of picking a
red ball are one-half. I pick out a red ball. So and I add another red ball. So
now the odds of picking out a red ball are two thirds. Unless suppose I pick up a red
ball again, well then the odds of getting a red ball next period will be three
fourth. So this is how it works, it's straightforward. What we wanna do is see
why these two claims are true. I'm gonna do these claims in opposite orders. The
reason why is, I'm gonna use the second claim to prove the first. So what was the
second claim? Second claim was that any history of B blue balls and R red balls is
equally likely. Let's see why that's true. It's actually really, pretty easy. Let's
take this sequence. So I've got one red followed by three blues, versus one blue
followed by three reds. So when I start this out, I've got one blue, and one red.
So the odds of getting the red ball, Are one-half. Well now after picking the red
ball, I've got one blue, And two reds. So the odds of picking a red, a blue ball, at
this point, are only one third. But after I've picked the blue ball, I've now got
two blue and two red, so the odds of picking a blue ball are two out of four.
And then now for this last one. I've picked a third blue ball. I'm sorry, I
picked a second blue ball. And so the odds, I've got three blue balls, and I've
got two red balls, so the odds of picking a blue ball are three out of five. So I
multiply all this together, I get one times one Times two times three, over two
times three, times four times five. So, it's three times two times one, over two
times three times four times five. Now, let's suppose I'm down here and I pick
blue, blue, blue, red. Well, I start out again with one blue, one red. So the odds
of picking the blue ball are one half. But now what I've got is I've got two blue
balls and only one red ball, so the odds of picking a blue are two thirds. I pick
another blue. So I've got three blue balls and I've only got one red ball, so the
odds of picking the blue Are three out of four. We add another blue ball, because I
pick blue again. So, now I got four blue balls and only one red ball, and the odds
of picking that red ball are just one out of five. Well, if we multiply all these
things together I get one times two, times three times one, over two times three,
times four times five. So, one times two, times three times one is the same as one
times one times two times three, these things are equally likely. Let's do one
more and we'll see why this is the case. Let's suppose I go red blue, red blue. So
I start out red, I've got one blue, one red, so the odds of picking a red. Or one
half and I get a red so now I've got one blue, And two red. [sound]. So the odds of
picking a, a blue ball in this case is just gonna be one out of three. But I do
get a blue ball, so now I've got two blue and I've got two red, so the odds of
picking a red ball are two out of four. So I add another red ball, so I've got two
blue and I've got three reds and so the odds of picking the blue ball are two out
of. Five and So I end up with, is a total of one terms one, terms two, terms two
over two terms three, terms four, terms five. Well That's a good answer for I pick
blue, blue in a red dress, if I pick yesterday with one blue, one red so the
other picking a blue ball on one half. I pick a blue it?s not get a two blue. And
one red. So the odds of picking a blue ball are two-thirds. I do pick a blue so
then I've got three blue and one red. The odds of picking a red ball are one out of
four. But now I've got. Two reds in there, so we got three blue, and two reds, and
the odds of picking a red ball here are two out of five. So, then I multiply all
this out, I get one times two times one times two, over two times three times four
times five. Well, if we look at these two examples, we'll see why this result is
true. Let's see what the claim is. The claim says any history of B blue balls and
R red balls are equally likely. Well, happens in the first case. Well, let's
look at the denominator first. In each case we were picking out of two, three,
four, and five. So, if I wanna know the property of the sequence on the bottom I'm
just gonna get two times three times four times five. That's true here. That's true
here, it will be true in the other two cases. So now I gotta ask, what's the
probability of getting three blue balls and one red ball? Well here I pick the red
ball first, there was one red ball in there. So I get the, a one on the top. Now
I've gotta pick three blue balls. For the first time I pick a blue ball, there is
one Blue ball. Second time I pick a blue ball, there's two blue balls. The one I
picked the first time, plus the original ball. The third time I pick a blue ball,
there's gonna be three. So, the odds are gonna be three out of ever the bottom is.
So, for the one red ball, I'm gonna get a one, and for the three blue balls, I'm
gonna get one, two, and three. Well, the same happens here. I get a one, two, and a
three for the three blue balls, and I get a one for the one red ball. So, here it's
one times one times two times three. Here it's one times two times three times one.
So, if it were the case that I said, blue. Blue, red, blue. That would end up being
one times two times one times three over two times three. Times four, times five.
So what we get is the same probability. Notice here again. Here the case I picked
out two red balls and two blue balls, but the orders were different. Here it went
red, blue, red, blue. Here it went blue, blue, red, red. Again, when I picked the
first red ball, there's only one to pick from. When I picked the second red ball,
there's two. When I picked the first blue ball there's only one to pick. When I pick
this second, there's two. So, I get a one, a one, a two to two. So, here since it
goes red, blue, red, blue, it's one, one, two, two. Here since it goes blue, blue,
red, red, it's blue, first blue, second blue, first red, second red. And because
one times one times two times two equals one times two times one times two, it's
gonna be the same. So, what you see is that result two holds because. The order
[inaudible] matter of some of these get the same numbers one, two, one, two,
multiplied in different orders. So result one, result two, fairly straight-forward.
Let's now go to result one. This is the harder one. Any probability of red balls
is an equilibrium, and it's equally likely. Again, let's do some examples.
Let's suppose I've got all four blue balls, and I wanna know what's the
probability of this happening? Well the odds of picking one blue ball the first
time are one half, then it's one out of, two out of three, and then it's three out
of four, and then it's four out of five. Because what's happening is that, I start
out with one blue ball, one red ball, then I've got two blue balls, one red bull,
then I've got three blue balls, one red bull and then I've got four blue balls and
one red bull, so when I multiply all this [inaudible] I just get. All these things
cancel and I get one over five. Well now suppose I go blue, blue, blue, red. Well
again I start out with one blue one red, so the odds are one half of getting a
blue. Now I've got two blue and one red, so the odds of getting a blue are gonna be
two-thirds. Well now I've got three blue and only one red, so the odds of getting a
blue are gonna be three-fourths. Now I've got four blue. In one reds, the odds of
getting that red ball are gonna be one-fifth so these cancel and I get one
over twenty. So the odds of getting all blue is one over five. The odds of getting
three blues and a red is one over twenty. But notice I could've gotten. Blue, blue,
red, blue. I could have gotten blue, red, blue, blue. Or I could have had red, blue,
blue, blue. So that one red ball. Could have gone anywhere. Right? It could have
gone on the fourth spot, third spot, second pot, spot, or first spot. So, you
gotta multiply this one over twenty times four, and that means there's a one over
five spot of getting exactly one red. So, I've gotta one fifth chance of getting all
blues, a one fifth chance of getting one red. And that also stands to reason, if we
think about this for a second, what are the odds of getting all reds? Well the
odds of getting all reds is gonna be the same as the odds of getting. All blues
which is one-fifth. What are the odds of getting exactly one blue? The odds of
getting one blue is gonna be the same as getting. [sound] One red, which is one
fifth, so the only thing left is two reds, two blues. And what are the odds of that
going to be. Well, they've gotta be one fifth, so there's only one fifth left
over. What we see here is that it looks like this is just sort of magically
working. What we'd like is, we'd like a deeper understanding of what that magic
is. So let's go to a much harder case, let's suppose we've picked out 50, and I
want to know what are the odds of getting, f-, if I've got 50 periods we're getting
all 50 blues, what are the odds of getting a blue in the first period. It's gonna be
one half, second period two thirds, so we're gonna have another blue, third
period. Three fourths. What about in the fiftieth period? Well in the fiftieth
period, I'm gonna have 51 balls in the urn, so the odds are 50 over 51, and in
the forty-ninth period, I'm gonna have 50 balls in the urn, so it's 49 over 50. Well
you can see if I multiplied all these things too, they're all gonna cancel out
and I'm gonna get one over 51. So, there's a one over 51 chance that I'm gonna get 50
blue balls, and that also means there's probably a one over 51 chance that I'm
gonna get 50 red balls. Now you can say what are the odds that I get 49 blue balls
and one red ball? Well let's suppose that we get the red ball last. So what I'm
gonna get is I'm gonna get one over two, two over three, three over four, and so on
to 49 over 50. So, all those things are going to cancel just like before, but in
the last period. I'm only gonna get. One over 51 which is only a one over 51 chance
I'm gonna pick out that red ball. Well this is gonna give me one over 50. Times
51. And now let's think about. Number of previous [inaudible] of previous
[inaudible] say it doesn't, it's equally likely to get the red ball first or last
or second or third or seventeen. So this red ball could have occurred anywhere. It
could have occurred in any one of fifty places. It could have been first, second,
third, fourth. So gotta multiply the probability of getting one red and forty
nine blues, let's multiply that by fifty and I get one over fifty one. So now those
of these things about getting fifty blues and fifty reds is the same as getting one
red and we can sort of start to see the logic. Let's look at the probability of
getting three reds. Okay, well, let's suppose again since it doesn't matter that
the first. Forty seven of end up all being these blue. So we get one over two, two
over three, all the way up to forty seven over forty eight. Now the last three are
gonna be red. So we're going to get one over forty nine cause there's only one red
ball to pick. Now there's two red balls in there, so there's a two out of fifty
chances we'll get a red ball in fifty or the forty nine period. And then the
fiftieth period there's three red balls in there and so it's three over fifty one.
Now again, all these numbers cancel except for the fact that now I'm left with one.
Times two times three over 48 times 49 times 50 times 51. Seems like a double
math. How is this gonna work? Well, let's think about it for a second. I've got
three red balls Those three red balls could have occurred, any one of, a whole
bunch of places. Well how many places? Well there's 50 places the first one could
have occurred. Once I picked where the first one goes there's 49 cases, places
the second one could have gone. And. Then there's 48 places the third could have
gone. But if I pick place seventeen and the sixteen and then twelve, that's the
same as picking twelve, then sixteen and then seventeen. So, you've got to subtract
out or divide by the different orders. Well, for any given three or three
[inaudible]. The different ways they could arrange those three balls are three times
two times one. Right? Because they could have put them in seventeen, sixteen,
twelve, or it could be sixteen, seventeen, twelve, or it could be twelve, seventeen,
sixteen, or I think of all those different orders in which I could put them in spots,
twelve, seventeen, sixteen. There's three places I could choose first. There's two
places I could choose second, and there's one place I could choose third. So we're
gonna divide by three times two times one. But watch, this three times two times one
cancels out with the one, two, three on the top. And this 50, 49, 48 cancels out
with this 48, 49, 50. And I'm left with one over 51. Well just a moment. You know,
says that if I change this 47 and three around to. 46 And four. And have the exact
same logic. What I'm going to get is that there is a one over 51 chance of getting
any number of red balls. So what we've then proven is we've proven claim one.
We've proven that any probabilities of red balls is equally likely. Now it's easy to
show it's equilibrium we've shown is equally likely. Which is sort of
surprising which is way the [inaudible] is so interesting it's highly [inaudible]
dependent. We also should first. Result two, which is that, if I've got five blue
balls and seven red balls, if any order of getting those has the same probability.
Okay, so that was a little fun math on the sign, but it shows us how by constructing
these models, this very simple model, you can then do some mathematics to gain
insights that we never would have expected. So I might have started with a
model that says well, suppose the proportion of, the probability of somebody
buying something is in proportion to the number of previous people that had bought
it, like the blue and red shirts. Okay, so that's a very simple idea, by running on a
formal model we get two really interesting. Results, one is that any
equilibrium is equally likely. And second, we get that any history, with the same
number of Blue shirt purchases and Red shirt purchases, is also equally likely.
Neither of those we would've anticipated and neither of those could be figured out
unless we use the model, to work [inaudible] logic, which is one reason why
we write down these models. Okay, thank you.
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