The best thirty two football teams,
soccer teams, in the in the world are
gonna be meeting up in in Russia this
summer. Not only do we obsess on the
World Cup but many of us also have this
obsession about this Panini World Cup
album. So this is the sticker album, I'll
show you the one from four years ago.
This is the one from World Cup 2014 in
Brazil. You get to see all the different
countries that are participating, all the
different players that are participating.
It's kind of sad I couldn't fill it, like
I was not able to get all the stickers
but the idea is that hopefully you are
able to get all of them and you get to
learn a lot about the players and and
just be more engaged with, with what's
happening. So when you just buy it,
like I just did this morning, it looks
like this and it's it's completely empty
and you can see for example Colombia,
this is a team that I'm excited about,
and so these are the names of all the
different players and then you get these
sticker packs. So there's five in a- you
know five little stickers in a little
pack. To anybody who's been doing this
for many years we even know what this
thing's smells like. It's like, you know,
the smell of childhood right here. You'll
be excited about this one, I just got
Massimo Luongo from Australia. You know I
take these stickers and I, and I stick
them on the album. I've been asking
myself how much of my paycheck am I
gonna be spending on this? Because, you
know, I'm a mathematician. I can do the
calculation and I know that this is not
a great idea - gonna be costing a lot of
money. But I kind of want to think about
how much is it gonna cost? The sticker
album this year has 682 stickers,
the last one had 681, but for some reason
they decided that the first one should
have number 0 so there's 682 stickers.
Each one of these packs cost $1.00,
there's 5 stickers in each one, and so
that means that each sticker is 20 cents.
The difficult thing is that I don't get
to choose what's, what's in here. So you
know if I got super lucky and I got all
the stickers that I wanted with no
repeats then there would be 682 stickers,
so when I multiply this this would be
the cost of filling the album - 136
dollars and 40 cents. But of course
there's no way that I'm ever gonna get
that lucky.
Okay I'm gonna, you know, if anybody has
done this before they know that you keep
getting repeats and repeats and repeats.
And so what I'm asking myself is how
many stickers will I need to buy until I
get all of them? Actually this is
something that people have tried to do
before. So the most recent story that
came out talked about how it would cost
about $1,000 to fill the album. And I see
where they get that, but I want to I want
to give us a different way of thinking
about this because the the problem is
that that estimate of a thousand
dollars assumes that you're just buying
and buying and buying and buying
stickers, and that's not really how it
works. How it works is that many people
around, around you are buying stickers
and so whenever you get repeats and they
get repeats then you're able to swap
with each other and that way you're
able to lower the costs. Let's say that
we're trying to find this number n, which
is the number of stickers I expect to
buy before I get every sticker at least
once. This is the lonely sticker
collector version, I'm in, I'm in the
US, I have nobody to trade with and so
I'm just going to imagine that I'm that
I'm buying and buying stickers until I
get them. Let me try to define Ni to
be how many stickers it took me to get
between the ith new sticker and the i
plus one new sticker, okay? So for example
N0 will be how many stickers until
I get the first new one. N1 will be
how many times until I get the second new one.
You can imagine, for example, what is
N200? That means that I've bought a lot of
packs, I have 200 of the players and now I
really want to get the next one.
So I'm looking for for some new one that
I that I don't have in my album yet. And
then I keep I keep buying and buying and
buying stickers until I get one that I
don't have yet. And however many stickers
it takes me, I'm going to call that
number N. The total number of stickers
is going to be basically, you know, how
long did it take me to get the first new
one, plus how long did it take me to get
the second new one, and so on; until I get
the last new one - this is how long it's
going to take me to find the very last
one. And so if I want to find this number
N, how many stickers I need to buy, what
I'm going to do is I'm going to actually
try to find each one of these numbers.
And let's just say we want to compute
N200 for an example. What I want you
to imagine is this: I have 200 stickers
and now I'm going to buy a new one. So
you can imagine I I open a pack, I grab
one of them, and this is this is what I
get: Jonathan dos Santos from
Mexico. And then what I do is that I I
have to look and see, do I already have
Jonathan dos Santos? Is he one of the 200
that I already have or do I not have it
yet? And of course there's there's 682
possibilities of which one this could be.
This is number 461, but you know it could be anything between 0, 1 -
- 681. Let's imagine that I put a circle
around the ones that I already have. So
let's say that I already found sticker
number 1,
I haven't found 0, I haven't found 2;
let's say that I did find 461 already
and I haven't found 681. So I pulled
another sticker out, and I want you to
imagine that this is- - (Brady: Who's this?)
Andreas Christensen from Denmark - number 257. And
now imagine that I open my album and I
find that Andreas Christensen is not
one that I have already, so I get I get
to stick him right away. And so it only
took me one new sticker to get one new
player, okay? So if this was the first one
that I drew, then it would take me one
new sticker. You know, if I happen to draw
this one out then it takes me one
sticker to get a new one - here it takes
me one sticker to get a new one. But now
let's just go back to the reality where
I actually pull Jonathan dos Santos and
I did have Jonathan dos Santos already.
And so now the question is how many
stickers is it going to take me from now
on? Well this was a waste, you know. I
already had Jonathan dos Santos, he's not
useful to me, so I need to draw again. How
many new steps is it going to take me to
get the next one? Well this is precisely
the definition of the number N200. So
I wasted this step and and it's like I
have to start over and it's still going
to take me N200 steps, because that's
what that number means. So if I draw
Jonathan dos Santos it's actually gonna
take me 1 plus N200 stickers before I
get a new one. The same, for example, if I draw
number 1 - it's going to take me one plus
N200. So basically depending on
what I draw here it's either gonna work
right away, I just draw one sticker and
it's a new one, or if I already had it
then it's going to take me 1 plus
N200 stickers. How many stickers did I
already have? I have 200 stickers already.
So in 200 of the draws is going to take
me this many steps to get the next
one. Whereas, for the stickers that I'm
missing, which is a lot, I'm missing
482 of them, is going
to go right away. And so when I average
these numbers over the total number,
which is 682,
well that average is going to be how
long it takes me to get that sticker. So I
get this nice equation; and now if you
look at this equation you're going to
find that 200 plus 482 is equal to 682,
so this equation is just saying this is
equal to 1 plus 200 over 682 times N of
200. What I get is that the number of
stickers that I should expect to buy
before I get the 201st new one is
going to be this. So that's one
computation. Now, as you can imagine the
same argument will work for any one of
these numbers, okay? And so what I want to
do now is generalise and do this for any
one of these numbers; and I think the key
thing that we should realise here is
where did this 482 come from? Well you
saw what happened - it was 680 to minus
200. Ni, which is how many stickers I
need to buy between the ith new sticker and
the i plus first new sticker, is going to
be 682 divided by 682 minus i. And so if
I combine these two what I'm gonna get
is that n is equal to 682 over 682 plus
682 over 681, 682 over 680- - (Brady: You can see)
(when you opened your first pack you were)
(definitely gonna get a sticker you)
(didn't have.) - Exactly. So this is, this is
the number N sub 0, how long is it gonna
take me to get the first new sticker? For
sure no matter what the first one I get
is, that's gonna just gonna come right
away. And as you go on it gets harder and
harder and harder and harder to get a
new one. What is the last one? And so I
get 682 divided by 682 minus 681, which
is equal to 1. - (Brady: That makes sense as well)
(because when you- when there's one sticker)
(you need you've got a 1 in 682 chance of)
(getting it.) - And it's so hard to get it.
And I think it's worth delving on this
point a little bit. This is how long it
takes you to get the very last sticker,
okay? And anybody who's collected this,
who's collected anything, knows that this
gets so much harder and at the very end
if I already got every single sticker,
and I'm only missing one, and I ask
myself how many should I buy until I get
that new one? It's gonna take me a whole
albums worth
just to get that last sticker. So I think
at this point there's two kinds of
people; there's one kind of people who
will be like, oh this is so beautiful
let's explore the mathematics of this,
and there's other people that will say
well what's the answer? Let's talk about
what the answer is first and then we'll
come back to the beautiful mathematics
behind this. What is this number? I
compute, I computed. I went to a computer,
I worked it out that this number is
about 4844.
So this is the number of
stickers that I should expect to buy,
that's a lot of money. I'm not about to
do that, I don't have that kind of
paycheck. What if you use the service
that Panini offers, which is that once
you have 50 stickers that you're missing
then you can just order them from them.
Just send me those and then I'll fill my
album and I don't have to trade it with
anybody. I basically have to do this
computation, except that instead of going
all the way from zero to 682; I want to
do it from 0 to 632.
It's about 1775 stickers, okay. - (Brady: What a saving!)
So we're coming down from a thousand to 355
and I think that's pretty shocking that
to get the first 632 stickers cost you
355, and to get the last fifty it's like
over six hundred dollars. - (Brady: Do we have to)
(pay Panini for those fifty though?) - We do
and I have to be honest that I I think
they do it at face value, but I'm not
totally sure. But that's not fun. I mean
the whole point of this is to buy
stickers and trade it with other people
and so on. And so I think we should go
for a little bit of a more realistic
scenario where we have a bunch of people
that are trying to fill the album
together. So thankfully there's this nice
model that that Donald Newman and Lawrence
Shepp came up with. Let's say that we want
to buy stickers among a bunch of people,
in this case N is 682, that's the number of
stickers, and F is just however many
friends are part of this game.
(Brady: So Federico the model that you're about to)
(talk to me about isn't how hard it will)
(be to fill one album, it's for everyone)
(to fill their album; all the friends will)
(fill their albums?) - Exactly so it's a. it's
a collective goal, everybody wants to
fill the album, we're all gonna help each
other out. And so they came up with this
amazing formula; I should warn you that
is kind of an ugly formula. It's a big
formula, it's messy, and I'm not even
expecting that everybody will understand
it, but the whole point that I want to
make is that there is a formula. Okay so
let me show you that messy formula.
Number of stickers divided by number of
friends, so far this is fine, and now this
is where it gets messy.
So integral from zero to infinity
of some very big expression - it looks
like this: 1 minus, and big parentheses, 1
minus, it's a quotient here, here I get 1 plus
x plus x squared let's say x over 1, x
squared over 1 times 2, x cubed over 1
times 2 times 3, and so on. Until the
number of friends minus 1. And I take all
of that and I divide it by e to the x; I
warned you this was messy.
Now you take a parenthesis here and you
take this whole thing and raise it to
the N, which is the number of stickers,
and then you close the parentheses and
then this is an integral with
variable x and so it's integral here dx.
(Brady: It's a monster!) - It's a monster;
I think is beautiful but you have to
have a taste for these kinds of things.
What I think is beautiful is not that I
see this and I think it's really beautiful,
what I think is beautiful is that I just
need to plug in N for the number of
stickers, F for the number of people and
I'm gonna get the answer for how many
stickers per person we expect to be buying.
(Brady: How much of a saving is it to be a sticker)
(swapper?) - So what we need to do is figure out
how to plug in this thing. One thing I
will say is that if if the number N and
the number F are pretty small, then this
expression is actually not too bad and
if you've taken a course in integral
calculus you will have the tools to
compute this, for N and F small. But the
problem is that we're looking at N is
equal to 682, we want to do this for
various values of F, and so really we're
going to have to ask a computer for help.
And so we're just going to plug
into a computer and we- I did this ahead
of time, and I'll show you some of the
results. So for example for F equal to 1,
I got no friends I'm doing it by myself,
then we should get the old answer that
we got when when we were just buying and
not trading. And you find that the answer
is 4844. 969 dollars. Now let's say that
it's just you and me Brady, and we're
gonna we're gonna do this together and
so now we have two friends. And as you
can imagine is going to be a little a
little bit cheaper. Then the answer that
you get is 3219. So in total we're
gonna buy like 6,400 stickers
but it's 3,200 each. So this is not what
we want to do either, we should probably
find some more friends because I don't
know about you but I'm not about to
spend 644
dollars on this, okay? - (Brady: No? I bet you would.)
No I'm not, I'm not
that crazy. You'll see I've, I've never
filled an album. Let's try five friends.
So if we have five friends the cost
starts getting quite a bit lower, so now
it's going to be two hun-
2050 stickers per person,
which is about 410
dollars per person. - (Brady: Wow so if I- if I'm)
(trading with four other people I'll halve-)
(I will better than halve) what I have to cost?)
Yeah, that's right. And and you can
imagine we probably shouldn't stop there
right? Of course the the the more friends
this is gonna keep going down lower and
lower and lower, okay. And so I actually
just ran this for ten friends, twenty
friends and fifty friends. I should say
that it wasn't me running this. So I got
the wonderful help of my student
Viviana Márquez and I want to give her credit
because she's much better with computers
than me. So that's looking much better I think.
(Brady: Where's this heading towards? You know if)
(I live in Colombia where everyone's)
(obsessed with these stickers and there's)
(a population in the millions like, how)
(low can this go? Does it bottom out)
(anywhere? - So, so there's something called
the law of large numbers which says that
basically once we, once we have enough
people then it kind of costs each of us
essentially the cost of an album. 682, so
if I multiply that by 20 cents I get 136.
Of course it can't be lower than that,
because I need to buy all the stickers.
And so, you know, basically however
whatever price you're willing to pay,
that's kind of how you need to figure
out how much you need to trade. And I
think this is a good argument for really
using the Panini as the social
experience that it is and actually going
out and seeking out friends. And and the
more that you do it the lower that the
price can be, and it can be down to
pretty close to that 136. One thing that
I wanted to do is that, I don't know
about you Brady, but I I find this
formula absolutely amazing. You know, I
sat down and I understood it and I read
the proof and I, and it makes sense to me;
and yet I have trouble believing it it's
so magical. And so one thing that I did
is that I reached out to Viviana Márquez
again and I asked her, hey you're amazing
with programming can you just run some
simulations that will take ten friends
and will buy stickers for ten friends
and figure out how many stickers they
spent, and run it a hundred times so that
we can get some kind of average
of where it'll be. And so I'll show you
Viviana's numbers. I think that's good
that not only do we have a proof of this
that I, that we understand; I think
sometimes it's good to just run some
experiments to make sure that the
formula also holds up in reality and and
and it really does.
Just like I don't
know which stickers are inside this pack,
I might open this in a minute, I also never
know what the problem of the week is
gonna be over at brilliant.org. So let's
have a look today. The basic one says,
every green square is one quarter of the
square it's inscribed in. If the pattern
continues infinitely what fraction of
the total figure will be green? D'you
think you know the answer to that? How
about we have a look at the intermediate.
If a plus b over c equals 6 over 5 and b
plus c over a equals 9 over 2, then what
is the value of a plus c over b? I think
I'm going to need a pen and paper for
that one. But Brilliant is not just about
the problem of the week, although that's
one of my favourite things about it, they
also have an incredible number of
courses you can try covering all sorts
of things: physics, computer science and
mathematics of course. And more and more
people are signing up of all ages and
backgrounds because really this is a
place that encourages you to change the
way you think, think outside the box. How
am I going to solve problems? It really
gets your brain working in different
ways and a lot of people like that, a lot
of people want to be better thinkers.
That's why they're going to Brilliant.
You can go there too, go to
brilliant.org/numberphile so they know you came
from here. And if you use
/numberphile you can also get 20% off a
premium membership at Brilliant. Ooh
I've got the French national team,
special one, oh also got the Colombian
national team! That one's for you
Federico. Anyway brilliant.org/numberphile
for that twenty percent off;
check them out. And thank you to them for
supporting this video.