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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.