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Hi. The previous lecture I stated two results, regarding the polya process. In

Â terms of the, each equilibrium is equally likely, and that, any history of R. Red

Â balls, and B, Blue balls is equally likely. What I want to do in this lecture

Â is show you why those two proofs are true. Now I'm hoping you prove these on your

Â own. You don't even have to watch these lectures. You're just watching to see if I

Â did it the same way you did. But at if you haven't tried them at this point stop the

Â video and try on your own just by doing some simple examples to see why these

Â things are true. To get us started I want to remind us how the player process works.

Â Remember we started with one blue ball one red ball. We pick a ball out, we look at

Â its color and then we add a ball of the same color of the ball we selected. So

Â suppose for example that I start out with one red, one blue. The odds of picking a

Â red ball are one-half. I pick out a red ball. So and I add another red ball. So

Â now the odds of picking out a red ball are two thirds. Unless suppose I pick up a red

Â ball again, well then the odds of getting a red ball next period will be three

Â fourth. So this is how it works, it's straightforward. What we wanna do is see

Â why these two claims are true. I'm gonna do these claims in opposite orders. The

Â reason why is, I'm gonna use the second claim to prove the first. So what was the

Â second claim? Second claim was that any history of B blue balls and R red balls is

Â equally likely. Let's see why that's true. It's actually really, pretty easy. Let's

Â take this sequence. So I've got one red followed by three blues, versus one blue

Â followed by three reds. So when I start this out, I've got one blue, and one red.

Â So the odds of getting the red ball, Are one-half. Well now after picking the red

Â ball, I've got one blue, And two reds. So the odds of picking a red, a blue ball, at

Â this point, are only one third. But after I've picked the blue ball, I've now got

Â two blue and two red, so the odds of picking a blue ball are two out of four.

Â And then now for this last one. I've picked a third blue ball. I'm sorry, I

Â picked a second blue ball. And so the odds, I've got three blue balls, and I've

Â got two red balls, so the odds of picking a blue ball are three out of five. So I

Â multiply all this together, I get one times one Times two times three, over two

Â times three, times four times five. So, it's three times two times one, over two

Â times three times four times five. Now, let's suppose I'm down here and I pick

Â blue, blue, blue, red. Well, I start out again with one blue, one red. So the odds

Â of picking the blue ball are one half. But now what I've got is I've got two blue

Â balls and only one red ball, so the odds of picking a blue are two thirds. I pick

Â another blue. So I've got three blue balls and I've only got one red ball, so the

Â odds of picking the blue Are three out of four. We add another blue ball, because I

Â pick blue again. So, now I got four blue balls and only one red ball, and the odds

Â of picking that red ball are just one out of five. Well, if we multiply all these

Â things together I get one times two, times three times one, over two times three,

Â times four times five. So, one times two, times three times one is the same as one

Â times one times two times three, these things are equally likely. Let's do one

Â more and we'll see why this is the case. Let's suppose I go red blue, red blue. So

Â I start out red, I've got one blue, one red, so the odds of picking a red. Or one

Â half and I get a red so now I've got one blue, And two red. [sound]. So the odds of

Â picking a, a blue ball in this case is just gonna be one out of three. But I do

Â get a blue ball, so now I've got two blue and I've got two red, so the odds of

Â picking a red ball are two out of four. So I add another red ball, so I've got two

Â blue and I've got three reds and so the odds of picking the blue ball are two out

Â of. Five and So I end up with, is a total of one terms one, terms two, terms two

Â over two terms three, terms four, terms five. Well That's a good answer for I pick

Â blue, blue in a red dress, if I pick yesterday with one blue, one red so the

Â other picking a blue ball on one half. I pick a blue it?s not get a two blue. And

Â one red. So the odds of picking a blue ball are two-thirds. I do pick a blue so

Â then I've got three blue and one red. The odds of picking a red ball are one out of

Â four. But now I've got. Two reds in there, so we got three blue, and two reds, and

Â the odds of picking a red ball here are two out of five. So, then I multiply all

Â this out, I get one times two times one times two, over two times three times four

Â times five. Well, if we look at these two examples, we'll see why this result is

Â true. Let's see what the claim is. The claim says any history of B blue balls and

Â R red balls are equally likely. Well, happens in the first case. Well, let's

Â look at the denominator first. In each case we were picking out of two, three,

Â four, and five. So, if I wanna know the property of the sequence on the bottom I'm

Â just gonna get two times three times four times five. That's true here. That's true

Â here, it will be true in the other two cases. So now I gotta ask, what's the

Â probability of getting three blue balls and one red ball? Well here I pick the red

Â ball first, there was one red ball in there. So I get the, a one on the top. Now

Â I've gotta pick three blue balls. For the first time I pick a blue ball, there is

Â one Blue ball. Second time I pick a blue ball, there's two blue balls. The one I

Â picked the first time, plus the original ball. The third time I pick a blue ball,

Â there's gonna be three. So, the odds are gonna be three out of ever the bottom is.

Â So, for the one red ball, I'm gonna get a one, and for the three blue balls, I'm

Â gonna get one, two, and three. Well, the same happens here. I get a one, two, and a

Â three for the three blue balls, and I get a one for the one red ball. So, here it's

Â one times one times two times three. Here it's one times two times three times one.

Â So, if it were the case that I said, blue. Blue, red, blue. That would end up being

Â one times two times one times three over two times three. Times four, times five.

Â So what we get is the same probability. Notice here again. Here the case I picked

Â out two red balls and two blue balls, but the orders were different. Here it went

Â red, blue, red, blue. Here it went blue, blue, red, red. Again, when I picked the

Â first red ball, there's only one to pick from. When I picked the second red ball,

Â there's two. When I picked the first blue ball there's only one to pick. When I pick

Â this second, there's two. So, I get a one, a one, a two to two. So, here since it

Â goes red, blue, red, blue, it's one, one, two, two. Here since it goes blue, blue,

Â red, red, it's blue, first blue, second blue, first red, second red. And because

Â one times one times two times two equals one times two times one times two, it's

Â gonna be the same. So, what you see is that result two holds because. The order

Â [inaudible] matter of some of these get the same numbers one, two, one, two,

Â multiplied in different orders. So result one, result two, fairly straight-forward.

Â Let's now go to result one. This is the harder one. Any probability of red balls

Â is an equilibrium, and it's equally likely. Again, let's do some examples.

Â Let's suppose I've got all four blue balls, and I wanna know what's the

Â probability of this happening? Well the odds of picking one blue ball the first

Â time are one half, then it's one out of, two out of three, and then it's three out

Â of four, and then it's four out of five. Because what's happening is that, I start

Â out with one blue ball, one red ball, then I've got two blue balls, one red bull,

Â then I've got three blue balls, one red bull and then I've got four blue balls and

Â one red bull, so when I multiply all this [inaudible] I just get. All these things

Â cancel and I get one over five. Well now suppose I go blue, blue, blue, red. Well

Â again I start out with one blue one red, so the odds are one half of getting a

Â blue. Now I've got two blue and one red, so the odds of getting a blue are gonna be

Â two-thirds. Well now I've got three blue and only one red, so the odds of getting a

Â blue are gonna be three-fourths. Now I've got four blue. In one reds, the odds of

Â getting that red ball are gonna be one-fifth so these cancel and I get one

Â over twenty. So the odds of getting all blue is one over five. The odds of getting

Â three blues and a red is one over twenty. But notice I could've gotten. Blue, blue,

Â red, blue. I could have gotten blue, red, blue, blue. Or I could have had red, blue,

Â blue, blue. So that one red ball. Could have gone anywhere. Right? It could have

Â gone on the fourth spot, third spot, second pot, spot, or first spot. So, you

Â gotta multiply this one over twenty times four, and that means there's a one over

Â five spot of getting exactly one red. So, I've gotta one fifth chance of getting all

Â blues, a one fifth chance of getting one red. And that also stands to reason, if we

Â think about this for a second, what are the odds of getting all reds? Well the

Â odds of getting all reds is gonna be the same as the odds of getting. All blues

Â which is one-fifth. What are the odds of getting exactly one blue? The odds of

Â getting one blue is gonna be the same as getting. [sound] One red, which is one

Â fifth, so the only thing left is two reds, two blues. And what are the odds of that

Â going to be. Well, they've gotta be one fifth, so there's only one fifth left

Â over. What we see here is that it looks like this is just sort of magically

Â working. What we'd like is, we'd like a deeper understanding of what that magic

Â is. So let's go to a much harder case, let's suppose we've picked out 50, and I

Â want to know what are the odds of getting, f-, if I've got 50 periods we're getting

Â all 50 blues, what are the odds of getting a blue in the first period. It's gonna be

Â one half, second period two thirds, so we're gonna have another blue, third

Â period. Three fourths. What about in the fiftieth period? Well in the fiftieth

Â period, I'm gonna have 51 balls in the urn, so the odds are 50 over 51, and in

Â the forty-ninth period, I'm gonna have 50 balls in the urn, so it's 49 over 50. Well

Â you can see if I multiplied all these things too, they're all gonna cancel out

Â and I'm gonna get one over 51. So, there's a one over 51 chance that I'm gonna get 50

Â blue balls, and that also means there's probably a one over 51 chance that I'm

Â gonna get 50 red balls. Now you can say what are the odds that I get 49 blue balls

Â and one red ball? Well let's suppose that we get the red ball last. So what I'm

Â gonna get is I'm gonna get one over two, two over three, three over four, and so on

Â to 49 over 50. So, all those things are going to cancel just like before, but in

Â the last period. I'm only gonna get. One over 51 which is only a one over 51 chance

Â I'm gonna pick out that red ball. Well this is gonna give me one over 50. Times

Â 51. And now let's think about. Number of previous [inaudible] of previous

Â [inaudible] say it doesn't, it's equally likely to get the red ball first or last

Â or second or third or seventeen. So this red ball could have occurred anywhere. It

Â could have occurred in any one of fifty places. It could have been first, second,

Â third, fourth. So gotta multiply the probability of getting one red and forty

Â nine blues, let's multiply that by fifty and I get one over fifty one. So now those

Â of these things about getting fifty blues and fifty reds is the same as getting one

Â red and we can sort of start to see the logic. Let's look at the probability of

Â getting three reds. Okay, well, let's suppose again since it doesn't matter that

Â the first. Forty seven of end up all being these blue. So we get one over two, two

Â over three, all the way up to forty seven over forty eight. Now the last three are

Â gonna be red. So we're going to get one over forty nine cause there's only one red

Â ball to pick. Now there's two red balls in there, so there's a two out of fifty

Â chances we'll get a red ball in fifty or the forty nine period. And then the

Â fiftieth period there's three red balls in there and so it's three over fifty one.

Â Now again, all these numbers cancel except for the fact that now I'm left with one.

Â Times two times three over 48 times 49 times 50 times 51. Seems like a double

Â math. How is this gonna work? Well, let's think about it for a second. I've got

Â three red balls Those three red balls could have occurred, any one of, a whole

Â bunch of places. Well how many places? Well there's 50 places the first one could

Â have occurred. Once I picked where the first one goes there's 49 cases, places

Â the second one could have gone. And. Then there's 48 places the third could have

Â gone. But if I pick place seventeen and the sixteen and then twelve, that's the

Â same as picking twelve, then sixteen and then seventeen. So, you've got to subtract

Â out or divide by the different orders. Well, for any given three or three

Â [inaudible]. The different ways they could arrange those three balls are three times

Â two times one. Right? Because they could have put them in seventeen, sixteen,

Â twelve, or it could be sixteen, seventeen, twelve, or it could be twelve, seventeen,

Â sixteen, or I think of all those different orders in which I could put them in spots,

Â twelve, seventeen, sixteen. There's three places I could choose first. There's two

Â places I could choose second, and there's one place I could choose third. So we're

Â gonna divide by three times two times one. But watch, this three times two times one

Â cancels out with the one, two, three on the top. And this 50, 49, 48 cancels out

Â with this 48, 49, 50. And I'm left with one over 51. Well just a moment. You know,

Â says that if I change this 47 and three around to. 46 And four. And have the exact

Â same logic. What I'm going to get is that there is a one over 51 chance of getting

Â any number of red balls. So what we've then proven is we've proven claim one.

Â We've proven that any probabilities of red balls is equally likely. Now it's easy to

Â show it's equilibrium we've shown is equally likely. Which is sort of

Â surprising which is way the [inaudible] is so interesting it's highly [inaudible]

Â dependent. We also should first. Result two, which is that, if I've got five blue

Â balls and seven red balls, if any order of getting those has the same probability.

Â Okay, so that was a little fun math on the sign, but it shows us how by constructing

Â these models, this very simple model, you can then do some mathematics to gain

Â insights that we never would have expected. So I might have started with a

Â model that says well, suppose the proportion of, the probability of somebody

Â buying something is in proportion to the number of previous people that had bought

Â it, like the blue and red shirts. Okay, so that's a very simple idea, by running on a

Â formal model we get two really interesting. Results, one is that any

Â equilibrium is equally likely. And second, we get that any history, with the same

Â number of Blue shirt purchases and Red shirt purchases, is also equally likely.

Â Neither of those we would've anticipated and neither of those could be figured out

Â unless we use the model, to work [inaudible] logic, which is one reason why

Â we write down these models. Okay, thank you.

Â