
Almost all numbers contain the digit “3” [video] - ColinWright
http://www.numberphile.com/videos/three.html
======
Homunculiheaded
I really dislike these numberphile videos as they deliberately construct their
arguments such that the answer makes math seem like a magic trick and that
math is truly confusing, which is the opposite of helping people gain an
intuition around math.

The trick here is that they start with the intuition of counting, in which
each number is itself a single thing, but are actually doing a calculation
based on each number being a string of single digits.

It's much less shocking if you say: Picking a random 10 digit number is the
same as randomly picking 10 single digits. The more digits you pick the more
likely it is that you'll get a '3' somewhere in there. So as your string of
numbers increases in length the less likely it is that you're string won't
contain any given number. If I said to someone "The more dice you throw the
more likely it is you'll get at least one 3", I don't think I would get anyone
who was surprised by that.

The unfortunate thing is that math and number theory in particular are full of
genuinely fascinating observations that don't rely at all on a tricks of
phrasing to be revealed, and rather than making less math literate people feel
'dumb' (as these videos tend to do), spark in interest in exploring math
further.

~~~
japhyr
I teach high school math and science. I had a group of students do a mini-
project around the question, "Is there anything bigger than infinity?" The
core of the project was students watching this numberphile video [0], and
recreating his reasoning on their own.

That project helped them reason about higher-level math in a way they never
thought themselves capable. The idea of "magic" never came up, but they
certainly spoke about their sense of wonder at higher-level math. I really
appreciate the way he presents mathematical concepts.

[0] -
[https://www.youtube.com/watch?v=elvOZm0d4H0](https://www.youtube.com/watch?v=elvOZm0d4H0)

~~~
asdfjkln
I think the quality of numberphile videos vary quite considerably. The one you
linked is a rather honest explanation of the idea of comparing the size of
sets through bijections and a good explanation of Cantors diagonal argument.
As a mathematician I'm quite fine with it.

The one linked in the submission is more of an obvious statement hidden by
obscure use of language. Slightly silly, but not very bad.

The worst offender with "magic" is the one with the Riemann zeta function [1],
which went viral a while ago. The problem here is that they get people started
off on the wrong foot, confusing them with wrong arguments and hidden
definitions. Now, if people really want to understand why this can be made
meaningful, you first have to explain that the better portion of the video is
absolutely wrong, and only then can you explain what is actually going on.

[1] [http://youtu.be/w-I6XTVZXww](http://youtu.be/w-I6XTVZXww)

------
mynegation
This goes against generally accepted mathematical jargon
([http://en.wikipedia.org/wiki/List_of_mathematical_jargon#Des...](http://en.wikipedia.org/wiki/List_of_mathematical_jargon#Descriptive_informalities)).
For countable sets like integers, "almost all" has a meaning "all except a
_finite_ set". Obviously the set of numbers that do _not_ contain 3 (or any
other digit) is an infinite countable subset of all integers. Frequency has
nothing to do with it.

~~~
ColinWright
That turns out not to be the case:

    
    
        > almost all
        > A shorthand term for "all except for a set of measure zero",
        > when there is a measure to speak of.
    

In this case there is no measure to speak of, so it is normal to talk about
taking the ration in a finite initial segment and then letting the size go to
infinity and seeing what happens to the ratio.

And a set of measure zero does not have to be finite, either. Almost all
numbers are transcendental in the sense given above, and yet there are
infinitely many non-transcendental numbers.

So I think you are mistaken, although I would be interested to hear a more
complete statement of what you mean, in case I have misunderstood you.

~~~
mynegation
The same excerpt, starting from "One can also speak" describes how that
"almost all" terminology is applied to integers (or, by extension, to any
countable set, that is already measure zero). The example statement is "almost
all prime numbers are odd", because there is a finite number of primes that
are even (only one number 2).

Put another way, if we stick to the definition of "almost all" as elements not
having the property forming measure zero set, then you do not need the video
to prove the title: all integers are already measure zero. So, for subsets of
integers, "almost all" means something else in jargon.

I could not watch the video, but I found (I believe) equivalent text. And
"almost all" means that as we extend the reach the share of numbers not
containing 3 goes to zero. It is mildly entertaining (albeit a bit too
obvious) result, but my point was that usage of "almost all" is confusing
because _I_ am used to a very specific meaning of it.

~~~
drostie
It's also a little confusing because there are sets which do not have a well-
defined natural density among the integers, like:

    
    
         leastpow2 x = head $ dropWhile (\k -> 2*k <= x)  $ map (2^) [0..]
         p k = k < 3 * (leastpow2 k `div` 2)
         limitless = [k | k <- [0..], k == 0 || p k]
    

which oscillates between attaining 1/2 and then 2/3 for its partial densities.

------
tromp
By the same token, most numbers are more than 8 digits long.

In fact, most are over a billion digits long. Although I've never see a single
one in the wild...

~~~
debacle
5.73x10^1000000000

There you are then. Not that impressive, is it?

~~~
VLM
3 ↑↑ 4

Where ↑ is the knuth uparrow notation, beats your example in only 4 symbols
with about three thousand times as many digits

~~~
ackalker
I'll see you and raise you a © .

Which in my book is c, the speed of light in a vacuum, 299792458 m/s, a mere 9
digits, but inscribed in a circle using Steinhaus–Moser notation[1] is
equivalent to a number so mindbogglingly big that I'm not even going to
attempt to calculate its measure.[2]

[1]:
[http://en.wikipedia.org/wiki/Steinhaus%E2%80%93Moser_notatio...](http://en.wikipedia.org/wiki/Steinhaus%E2%80%93Moser_notation)

[2]: Not that it makes any sense to perform this operation on c, but I think
that use of © in this way is much more interesting than its current usage.

~~~
VLM
I admit defeat, using standard notation I can't think of a better ratio of
numeric length vs symbolic representation. Well played!

Aaronson's got a blog entry tangentially related to this competition

[http://www.scottaaronson.com/writings/bignumbers.html](http://www.scottaaronson.com/writings/bignumbers.html)

"In an old joke, two noblemen vie to name the bigger number. The first, after
ruminating for hours, triumphantly announces "Eighty-three!" The second,
mightily impressed, replies "You win.""

~~~
ackalker
Thank you for a nice challenge, and thanks for the interesting link :-)

As someone who is genuinely interested in AI, I'm very curious if lateral
thinking like we demonstrated can ever be matched by a machine. I'm watching
things unfold with great interest.

------
logfromblammo
The same reasoning applies to every digit. So almost all numbers contain your
favorite decimal digit.

If I showed this video to a friend who particularly enjoys the digit "6", she
would not see the connection, as the video is about "3". No matter how many
times I try to explain that there is nothing special about "6", she won't hear
any of it.

I am quite convinced that the same attitude is rather widespread, and could be
effectively monetized by anyone both charismatic enough and unethical enough
to exploit it.

~~~
tjgq
At the end of the video, he does mention that it works for any digit. A more
honest way to present the argument would be to start with "let x be any
decimal digit..." but I suspect most non-mathematicians wouldn't be able to
follow that.

~~~
timinman
Serious mathematicians will feel his title was misleading or dishonest.
Considering the proportion of mathematicians to the general public, the
teacher is probably not targeting mathematicians - he's a mathematician
targeting the mathematically curious. With that in mind, the title is perfect.

~~~
moron4hire
I don't agree. I think the real power of his videos is that, if he is
demonstrating _anything_ about mathematics, it's that mathematicians aren't
all pedants who can't communicate with anyone outside of their field without
accuracy.

------
smsx
This video was posted on april 1st if no one has noticed.

~~~
ColinWright
I hadn't noticed, but it doesn't matter. The result is correct. Are you
claiming otherwise? Are you claiming that it's false?

~~~
bhousel
Almost all of the news published on April 1st is garbage.

~~~
ikeboy
I don't think this is actually true; regular news sites and newspapers still
go on producing news, and it's only a small percentage that are parodical. I
would bet that less than 10% of purported news articles on April 1st are fake.

Unless you are also claiming that most news published is garbage anyway, in
which case I agree.

~~~
vinceguidry
He was making a joke.

------
kayamon
In real life data sets, Benfords Law shows that the digit '1' occurs with much
higher probability.

~~~
Too
This one is actually true and interesting for real. If you haven't heard of
this before i highly recommend reading up about it.

When measuring random things in nature, _regardless of what unit and base you
use to measure_ (inches, cm, m, feet, etc), measurements with lower first
digit appear more frequent than others! This is due to things in nature
usually being exponentially distributed, when you have an exponential
distribution you will "move away" faster from higher valued digits and reach
the next power of ten faster, there you will restart with a low first digits
which is slow to move away from. For example, assuming a spread of about 50%,
then 1+50%=1.5 (still starting with 1) but 5+50%=7.5 (not starting with same
digit), and 8+50%=12 (starting with 1 again).

------
seccess
For those of you who aren't familiar, Brady Haran, the creator of numberphile,
produces a ton of awesome videos of experts talking about science and math.
Besides numberphile, he does Sixty Symbols (physics), Periodic Videos (chem),
and computerphile (comp sci) [0].

[0] [http://www.bradyharan.com/](http://www.bradyharan.com/)

------
ikawe
Almost all numbers contain the sequence 861523871623871623871263.

------
btbuildem
So in general, almost all numbers contain any digit? Great.

~~~
JadeNB
An interesting consequence (since there are only finitely many digits) is
that, in this sense, almost all numbers contain _every_ digit—a phrasing that,
I think, makes the statement _less_ surprising.

~~~
baddox
Also, almost all numbers contain any given finite sequence of digits.

------
sidcool
In the end he mentions that it's equally true for the number 5. I guess that's
because of the nature of infinity. All numbers would contain almost all other
digits when infinity is involved.

~~~
thekingofspain
Probably a better way to phrase that is almost all numbers contain all the
digits.

------
jnevill
Imagine a game like minecraft, but you only get 10 blocks. Now you will
compose an infinite number of saved games using anywhere from 1 to infinity
number of blocks and each time you place a block in each saved game, you pick
a block randomly.

Some saved game will be a single block, and others will be infinitely vast.
Larger than our real universe, containing every block. It's not a stretch of
the imagination to see that the majority of the games will contain at least 1
of each of the ten block types.

------
smockman36
It's clear that the same principle applies for every digit. But is the
converse true? ie: Almost all numbers DON'T contain the digit "3"

~~~
jlebar
Not unless we equivocate and redefine "almost all" to be something other than
what he used in this video, right?

He showed that the ratio of "whole numbers containing 3" to "whole numbers" is
close to 1. So the ratio of "whole numbers not containing three" to "whole
numbers" is close to 0.

------
ebbv
If you really want to blow your mind, almost all numbers contain all the
digits 0-9. That's a lot more interesting and surprising to me than that they
contain any one digit.

But the reality underlying this is actually really boring; when you start
talking in terms of n -> infinity any fraction will become insignificant as it
self multiplies.

~~~
mikeash
Almost all integers contain the first ten trillion digits of pi, consecutively
and in order.

And yes, it's really not that interesting. Infinity!

------
peterderivaz
Another similar idea is the Kempner series.

If you add 1/n for all natural numbers n, then the series tends towards
infinity.

However, if you leave out any integers that contain a digit 9, then the series
converges (to just under 23). (It would also converge if you left out numbers
containing a digit 3, just to a different limit.)

------
xclojure
Here is a visualization of the number convergence:

(loop [i 1 t [10N] v [1.0]] (if (> i 300) (map (fn [[a b]] (/ a b)) (partition
2 (interleave v t))) (recur (inc i) (conj t (* 10 (last t))) (conj v (+ (* 9
(last v)) (Math/pow 10 i))))))

------
jgwest
Sure, the proportion converges to 1 as the order of magnitude approaches
infinity.

But this convergence is really really slow...

------
oneeyedpigeon
In an infinite universe, almost everyone is a smartass mathematician :)

------
KhalilK
It is an April Fools' video, nothing to see here.

~~~
ColinWright
So, just to be clear, you're claiming that the result is false?

~~~
DINKDINK
It can be a April Fools video and still be correct. Basically it's a moot
video because you can use the same logic to prove that any digit has the same
properties or converse property (that three exists almost NO where because the
probability of a number doesn't have three is P(n)=1-1/10^n and as n->∞
P(n)->0

~~~
ColinWright

      > ... because the probability of a number has
      > three is P(n)=1/10^n and as n->∞ P(n)->0
    

Er, no. The probability of an n digit number having a three is P(n) = 1 -
(9/10)^n, and as n->∞, P(n)->1.

I don't understand why you say it's a moot video. The result as stated in the
video is complete and correct.

~~~
DINKDINK
I meant to say doesn't have 3 and have updated the comment to reflect the
equation I believe is correct.

