A Hairy Problem (and a Feathery Solution) - Numberphile

[@numberphile]

Episode sponsor is MEL Science. Visit melscience.com/sBLS/ and use code Numberphile for 50% discount.

[@KitagumaIgen]

You should put up that other video at Numberphile2 - you usually do!

[@ericgoldman7533]

"Where was this when I was a kid?" A question I ask myself quite often when watching all of these edutainment channels, and seeing all of their sponsors.

[@enbyennui]

Glad Ben mentioned the birthday paradox, as I based my initial guess (almost certain) on having heard of that. Didn't occur to me to consider that it could be 100% though! Small correction, you need a minimum of 367 people to know for certain that at least two must share a birthday (29th Feb babies always throw a spanner in the works).

[@donaldhobson8873]

Once you include a tiny chance that the population of london has suddenly rapidly dropped, then it isn't 100%. Nothing real is certain.

[@spamsac]

@@ketchupmasher If you had 366 they could all have their birthdays on different days, so you do need 367 for 100%.

[@enbyennui]

@@ketchupmasher the pigeonhole proof requires n+1 pigeons :)

[@donaldhobson8873]

@@ketchupmasher 366 days, 366 people. In principle they could have 1 day=1 person.

[@joshyoung1440]

@@ketchupmasher womp womp. Try again lol

[@davidgillies620]

The Pigeonhole Principle is surprisingly powerful in a lot of areas, from compression algorithms to number theory.

[@theanimatedchemist]

As a chemistry educator, I always try to impart the idea of "number sense" or being able to describe a mathematical relationship between properties (say Pressure, Volume, and Temperature with respect to gases). It may not always catch on with everyone but I know some students have fixed mistakes after they get an answer they realize is nonsensical. I am well aware that "gas laws" will not apply in their every day life that often but if I can increase a student's number sense or general confidence with mathematics then I'll take it too!

[@vez3834]

Yep. Even in university a professor of mine for a math course keeps saying that it's important to make a "sanity check". That way you avoid having ridiculous answers and can find what you did wrong.

[@_abdul]

That reminds me, I need a Haircut for my Interview Tomorrow. Numberphile is helping me in ways I couldn't Imagine.

[@thipoktham5164]

I still remember vividly when I read to proof Fermat's little theorem using pigeonhole during my midterm exam. The sad part is, I have completely forgotten the proof, even though I read them right before the exam start

[@guillaumelagueyte1019]

I remember that proof, it's quite elegant but it wouldn't fit in a single KZbin comment.

[@llearch]

@@guillaumelagueyte1019 Margins not big enough? Common problem. ;-]

[@drenz1523]

​@@guillaumelagueyte1019 thats fermats last theorem, not fermats little theorem. nice reference though

[@guillaumelagueyte1019]

@@drenz1523 yeah I realized when writing it but I still wanted to do the joke :D

[@rainerausdemspring894]

Another question: It is easy to see that there are always two different places on earth with the same temperature or two different places with the same air pressure. Of course, we assume that both are continuously dependant on the location. Now estimate the probability that there are two different points on the sphere that have the same temperature AND air pressure.

[@DanatronOne]

100% - it's continuous. If you pick a point on a sphere at random, and another point exactly opposite to it, they're almost certainly going to have a different temperature. One higher, one lower. If you steadily move the points such that they swap places, they'll swap temperatures. But for it to transition from point A being higher to point B being higher (which must happen if they're swapping places), there must be a point during that move where they swapped over, and had the same temperature. This works no matter which path around a sphere you follow, so you'd actually end up with a (very wavy) ring of points that have the exact same temperature as the point on the opposite side of the globe. It has to be a continuous ring, as it forms a barrier; you can't go from one pole to the other without crossing it, as if that were to happen it would imply the numbers could swap without crossing, which is impossible on a continuous scale like temperature. Now if you pick two points on that ring of identical temperatures and instead measure for pressure, you'll find the same thing - a ring between those two points. But since it's on a sphere, and your starting points were on the temperature ring, those two rings of equal measurements must intersect. The intersection point has exactly equal temperature and pressure.

[@mrdavi5064]

100% too ez, in any line on circle, if you have continuous function, there is pair of points that have same y. There must be some line that there are all point have same temperature (if there weren't, you would be able to travel from point A to point B, where point A have greater temperature than B, without traveling through all temperatures between A and B), that mean, on that line there is continuously spread air pressure -> there always exist two points on opposite side of earth such that they have same temperature AND air pressure.

[@kujojotarostandoceanman2641]

Sadly you didn't filter out atmosphere since every atmosphere are extremely similar thus narrowing down the possibility

[@mal2ksc]

@@DanatronOne Except where two cells meet, you don't just get an average. You get a boundary region that constantly flips back and forth between the two extremes and never settles in the middle, because the boundary is constantly moving. The assumption that temperature is continuous is false. Nonetheless, the area where this is happening at any given time is pretty small and doesn't really affect the overall result that yes, there will be two antipodes with the same temperature*. (Neglecting that temperature itself is continuous, thus you could parse it finer and finer until you find a decimal place where they don't meet any more.)

[@rosiefay7283]

Indeed, it's better than that. Take any temperature T which is present at infinitely many places on Earth. Then there are at least two different places with temperature T and equal pressure. Consider one contour line of places at temperature T. The contour is a closed curve. Pressure is continuous on this closed curve. As we go around it, the pressure goes up and down and returns to its starting point, so it must pass every value at least twice. If it doesn't go up and down, it's constant and so the statement is trivially true.

[@MelindaGreen]

Very glad to hear the UK teaches estimation and other highly practical math skills. Also glad to know I nailed the answer, but largely because the pigeonhole problem came up early in my CS training.

[@NoIce33]

10:13 was the first take, wasn't it? :D

[@johnchessant3012]

An example of a "profound" result proved by the pigeonhole principle: Dirichlet's approximation theorem states that any irrational number x is approximated well by infinitely many rational numbers p/q, in the sense that x differs from p/q by at most 1/q^2.

[@wardycruyff]

I think numberphile should do a series in core maths to help average mathematical people like myself be better at using maths in every day life .

[@alexandertownsend3291]

If you want an example of everyday useful math, look up modular arithmetic. Any time you deal with events or phenomena that are cyclicl it is useful. Mod 7 for days of the week mod 4 for the seasons, mod 360 for degrees in a circle, mod 12 for hours on a clock, etc.

[@woody442]

As with anything, if u want to get better - practice. If you are not willing to let math occupy space in your life, it won't fill up space in your head.

[@veggiet2009]

It would be interesting to hear these topics delivered by professors and theoretical mathematicians

[@MegaFonebone]

It could be a whole separate channel.

[@richardsmith3021]

Disagree, I think there’s a ton of other channels that do that quite well.

[@ggtt2547]

Damn that started out as trivial and ended up amazing! Bravo!! Thank you for the video! It truly is a great ad for "core maths" at school. Maybe you should send it to your politicians!

[@dexter9313]

I wish we had something like that in France. I think it's super important that everyone has basic maths skills even if they don't want to follow the scientific path.

[@RegularCody]

Ben is definitely my favourite person on Numberphile

[@Mediumpimpin69]

I completely agree, he's really charismatic.

[@Flavour_Beans]

His stuff always ends up being the stuff I reference in other situations the most.

[@romanski5811]

What's wrong with man buns.

[@Alex-js5lg]

So this is basically the birthday problem, except instead of asking whether two people in a room share a birthday, you're asking the question at a sold out concert.

[@oz_jones]

Pretty much yeah

[@mattp1337]

Never knew the term, but I first encountered the Pigeonhole Principle at age 6 in a book of puzzles. Something along the lines of "if I have 5 colors of socks in my sock drawer, and my bedroom light is burned out, how many socks do I have to grab to ensure I've got at least one matching pair?" That puzzle stuck with me, obvious as it is. But not only has it proven useful on countless occasions (YSWIDT), that particular issue of sock matching in a dark room has come up surprisingly many times in my life :D

[@MushookieMan]

If you have 6 pairs each of 5 colors of socks, how many socks do you have to grab to ensure you have a blue pair? Just put in a new lightbulb!

[@mattp1337]

@@MushookieMan Eleventy-threeve.

[@llearch]

@@MushookieMan Fifty. You'll almost certainly get one earlier than that, but if you got amazingly unlucky, you could pull one blue sock, then every single other sock (in any order - so 6 pairs of four colours each, or 48 socks total) before pulling the second blue sock. I would not want your luck, in that case, but there you go. ;-]

[@EconAtheist]

@@mattp1337 minus one. always minus one.

[@orangelzp]

Here's a cool variation to ponder: Let's say you have 3 different pairs of socks, and your lightbulb is still burnt out. You are also a gambler and mathematician, so you ask yourself "What is the minimum amount of times I have to draw from this drawer to have at least a 50% chance of having a matching pair of socks?" What if you have 5 pairs? 10 pairs? n pairs?

[@neatodd]

Not for the first time, watching this, I wish that I had had access to all these brilliant Numberphile videos when I was a schoolboy

[@ScienceHippie]

The tenth pigeon's embarrassment touched my heart profoundly.

[@Calicoma]

Reading from the beginning, I'm thinking there's got to be a high chance. We're talking about finding any two pairs out of a group of a large number of people. The odds of one pair being correct has to be decently high. There's also an expected range for number of hairs. I'm guessing 70-80%

[@fiver-hoo]

in my high school math(s) club we trained with the book _Consider A Cylindrical Cow_. Highly recommened if you want to bone up on your estimating skill. Take the practice into daily life. Every time you are about to perform a computation, first estimate. Compare your estimation with the actual results. Refine your technique based on what you learned. Later, rinse.

[@PaulMJohnson]

I love that qualification - what a great idea. At work, the ability to sense check numbers, by estimating whether or not something you've been told "feels right", is so useful.

[@WobblycogsUk]

I think it's because the number of hairs on someone's head is quite large, well beyond what people will count. If you asked the same question but used number of arms or number of fingers everyone would get the right answer. An interesting question to ponder is at what point does the number of things get large enough that people stop saying it's a certainty. My guess is it's probably quite low, maybe 50 things.

[@mettanotation]

Fantastic job in choosing the topic and in delivery! The more you learn, the more you learn how easily our intuition can be wrong. Over and over again.

[@douglascodes]

LOL. I had this same problem in my head but with different items. When I worked a restaurant the salad guy would put up salads all night. I wondered how often and what the distribution of lettuce chunks were in each bowl. How often did he make a salad with the same number of pieces? He obviously always tried to make the salads the same size and there were a finite number of pieces. But also this just sounds like a variations of the birthday problem. I haven't finished the video yet, so here's hoping.

[@invictus127]

As soon as you asked this question, I started thinking about it as if it were Fermi's paradox' calculation. So I was VERY much on the mark, and I'm very happy cus this is a rare numberphile video where I was not only right, but on the right track!

[@robertpearce8394]

1 if 0 hairs is valid. Otherwise it sounds like a variant of the birthday problem so I am sticking with 1. Now to watch the video.

[@EnriqueDominguezProfile]

Everybody assumes that you're putting pigeons in holes, but it also works if you're carving holes in pigeons.

[@gigaherz_]

I heard the question and thought "oooh birthday paradox...?" but I wasn't sure about the number of hairs so I didn't commit.

[@N54MyBeloved]

(edited because spoiler comment) . . . . . . Who else knew the answer was 100% chance right away

[@endrawes0]

I wasn't sure if it was the birthday paradox perhaps, but pigeon hole principle makes great sense!

[@lonestarr1490]

@@endrawes0 It's not exactly the birthday paradox, for this asks for the number of people you'll need in order for the probability that two of them share a birthday exceeds 1/2 (thus, it's more likely that it is the case, than that it is not). If I remember correctly, given a normal distribution, the threshold for this is 23.

[@zunaidparker]

Spoiler warning! Lol

[@wagglebutt]

I knew it before the video started because this comment was on top.

[@N54MyBeloved]

@@zunaidparker Don't go looking in the section of the site intended for discussion of a problem-solving video before watching the video 🤔

[@prdoyle]

8:36 - Leap years man!

[@infrabread]

"If you have 366 people, you're guaranteed to share a birthday". You are forgetting about leap years, sir!

[@RibusPQR]

My favorite Ben Sparks hairstyle is when he looks like Denethor.

[@johnchessant3012]

"I'm gonna do what they call in a trade a piece of common sense" great way to introduce the pigeonhole principle

[@asheep7797]

Already knew about this. You've never heard of someone with over 1M hairs. There are over 1M people in London. Pigeonhole principle.

[@shdon]

My first reaction is: 100%. My reasoning is that: the number of hairs on people's heads may cover quite a range, but it is a finite range and the average will likely be in the tens, maybe low hundreds of thousands. There are so many people in London that the range is likely fully covered many times over even if it were perfectly evenly distributed (which it most likely is not). Now... to actually watch the video.

[@fyukfy2366]

This was actually a great breakdown. All around very high quality video and even the average person can feel that "eureka!" feeling I assume mathematicians feel when they figure something out

[@tremkl]

I’d heard the Birthday Paradox before, and while I found it surprising at the time, it did lead me to have a much better gut reaction to this question.

[@N0ahFence]

Ahhh this was the same problem you posed to a bunch of students I took to see you in London! Excellent then and still excellent now.

[@NoIce33]

Gut says "certain". And I think I know what's feathery about the solution :)

[@thisguyispeculiar]

This is similar to asking if in a room full of 367 people, what is the probability of at least 2 people sharing a birthday. Since you can only have 366 different possible birthdays (leap day included)... 367 people will always have at least 1 pair sharing a common birthday.

[@owenthemousey2268]

that's called the pigeonhole principle

[@Voiding210]

If there is m pigeons and n boxes in at least 1 box there is m/n pigeons

[@andrewanyplace]

367 people required to guarantee a shared birthday

[@TrumpeterOnFire]

Very certain was my first impression. Now to watch the rest of the video...

[@Yuusou.]

Actually, you need to have at least 367 people to double up on the birthdays due to the leap year with the 29th Feb added to the year. Other than that great video and explanation.

[@jonatankelu]

You beat me to it.

[@Yuusou.]

@@jonatankelu You'll have a fair chance next time. I promise 🙂

[@lonestarr1490]

I imagine it to be quite frustrating if your birthday only comes up once every 4 years.

[@Marguerite-Rouge]

The question about birthdays and probabilities is indeed really interesting. For example, there is a 50% chance two people share the same birthday in a group of 25 people. It should be a nice numberphile episode to explore that topic!

[@mal2ksc]

@@Marguerite-Rouge Pretty sure they already have.

[@leesweets4110]

My gut reaction actually was nearer to 100%.

[@ColorwaveCraftsCo]

Mine too

[@viliml2763]

my gut reaction was "unlikely" but then I remembered the birthday paradox and I thought "very likely" I didn't consider it could be 100% thought

[@scottcox503]

same, 0.9

[@snowfloofcathug]

Same. Really close to 1. And then I thought about it and realised that London is giant and it’d just straight up be 1

[@CanariasCanariass]

Same here, something about 95%+. I have to say though I knew that we have about 100,000 hairs on our heads and I thought of London having at least 5 million inhabitants so it was kinda easy to guesstimate

[@avihooilan5373]

As soon as he asked the question I knew the answer, not only is it highly likely, it is also certain. Pigeons and holes and all that. The amount of independent hairs we have on our head is around a hundred thousand, yet London has over eight million residents. Also, quick remark, the "City of London" has a few hundred residents, believe it or not, so for that entity it might actually be possible that no two people have the same number of hairs (though, still unlikely as there are probably baldies there).

[@rmsgrey]

The actual population of the City of London (the "square mile") is surprisingly low - under 10,000 - so, while the birthday paradox means the probability is extremely close to 1, it's possible for there to be no matches. Of course, if you count the number of people actually present, you'll get much higher numbers during business hours, with over half a million employed there, though it's not clear how many turn up to the office on any given day.

[@JxH]

0:48 100% Certain. Gut reaction as this sounds a bit like the ponderable about how it's always certain that 'two precisely-antipode points on the Earth having precisely the same temperature and humidity'.

[@JxH]

Correct guess, but for entirely the wrong reason. 🙂

[@protocol6]

8:35 Here he must be using a simplified version of the problem since about a quarter of people are born in leap years, about 1/366th of those (1/1464) are born on leap days, so you aren't absoluetely guaranteed to have a duplicate birthday with 366 people in the room. For that you need 367. There's an extra pidgeon hole and it's a somewhat exclusive club.

[@notdisclosed]

I was hoping that you'd run the birthday paradox equation to see how astonishingly few people it would take to get a duplicate at nearly a 100% chance.

[@ragnkja]

They covered the birthday paradox in one of the first Numberphile videos.

[@gargravarr2]

Here's another puzzle: What's the chance that two people in London have the same number of cells in their body? ... Here's my answer: I googled and found an estimate of 37.2 trillion cells per human body. London has approx. 10 million people. There's an approximation for the generalized birthday paradox where if you have n "people", and n^2 possible "days", there's a ~~50% chance that there's a pair who share a "birthday". (For example 23^2 = 469 which isn't that far away from 366). 10 million squared is 100 trillion. So the chance of two people in London having the same number of cells is probably over 50%, but far from certain.

[@RitaTheCuteFox]

Um no it's far less than 50%. Maybe 0.1%. Unless the margin of people having cells in their body is 37.1 trillion to 37.3 trillion or something but i largely doubt it's that small. The margin is gonna be like just to make it easier to calculate 10 trillion.. so we have 10 million / 10 trillion. Which is like 0.01%. Even worse than my estimation. The chance of having a pair is like n people divided by c amount of people. Thats why 365 people in a room will guaranteed share their birthdate. 365 people / 365 days. = 1. or 100%. Have to take the result times 100 if you want it in percent. At least that's my interpretation of the video and why 50% seems incredible wrong here but idk..

[@donaldhobson8873]

@@RitaTheCuteFox If you have 10^6 numbers, each uniformly chosen from 1 to 10^12, there is a 40% chance two are the same. After all, there are 10^6*(10^6-1)/2 possible pairs of numbers that could be equal. Sure, you need 10^12+1 numbers to be certain of a collision, but by the time you have 10^7 numbers, the chance they are all different is 2*10^-22 (ie basically no chance. You will have a collision.)

[@MRichK]

To add to Donald Hobson said, There aren't 37 trillion different choices for cells. Even baby's have many trillion cells, so the range would even be smaller.

[@BL3446]

Interesting to note here about how the different hairs/cells are being distributed. It wouldn't be like the birthdays where any 2 days are roughly equal in probability. Because someone isn't going to have just 12 cells (or 12 hairs). Even a thousand, or a million, or a billion wouldn't really be possible when speaking on the order of trillions. That small number of cells wouldn't even constitute a person. You'll want to also think about viable ranges too which I think fits well with estimations.

[@colinstu]

My gut reaction right away was 100.000%, and I was stunned to hear that so many think it's the opposite. Like what? It's obvious. Boil down our anatomy, we all average out somewhere for the same count of cells / surface area / etc. That and I've wondered how many feathers are on birds and this has been studied and the counts are fairly consistent between individual members of a species.

[@JohnSmith-nx7zj]

Except that in the case of “number of cells” this logic doesn’t apply. Google suggests there’s ~30 trillion cells in a human body. So even if the number doesn’t vary by much % wise (let’s say everyone has between 29 and 31 trillion) that still means there’s 2 trillion different numbers of possible cells. So it’s clearly possible for all 8 billion humans to have a different number.

[@colinstu]

@@JohnSmith-nx7zj Normal distribution… bell curve. Sure there's stuff at the extremes but, should have a peak somewhere.

[@JohnSmith-nx7zj]

That being said…I would think the chance of two humans having the same number of cells is very high. Of course in practice it’s an unanswerable question since there’s no feasible way to count the cells in a human being to the exact number (not least because they’re constantly dying and dividing).

[@joshyoung1440]

Ben: what's your gut reaction? Me: pretty high Ben: i can name the full decimal expansion. Me: okay so it's 100%, gotcha

[@zwussow]

I'm happy to say my guy reaction was right, but only because I immediately thought of another Numberphile video, about how common it is for two people in a classroom to share a birthday. I would have said it was very likely though, not 100%! 😄

[@BooBaddyBig]

I was thinking 99.999+% because of the birthday party paradox, but yeah, 100% exactly.

[@rmdodsonbills]

You're very very likely to get two people with the same birthday at 366 people, but you're not completely guaranteed until you hit 367.

[@jabbertwardy]

Even though I guessed it right, it's very useful to understand a formal approach to estimating large numbers!

[@Mohamed-lc5wj]

before hearing the answer I believe the possibility is 1.0

[@BryanGorges]

I saw hairy and a bird and thought you had failed with Hairy Woodpeckers here in North America. I was wrong, great video!

[@leppeppel]

My gut reaction is that the probability of someone in London having the same number of hairs as one specific other person in London is basically zero, but the odds of *some* pair of people having the same number is fairly likely. EDIT: I based this guess on my knowledge of the birthday paradox, which states that the likelihood of someone having the same birthday as you or me is ~1/365, but the odds of ANY two people in a group of 23 sharing a birthday is ~50:50

[@Kleyguerth]

I guessed 1 because I was pretty sure that number of hairs is not even close to 1 million, while also being pretty sure London has more than 1 million people living in it. When he said he could tell the answer down to the last decimal digit it became quite obvious it was 1.

[@markkaidy8741]

"366 people in the room" You are definitely NOT "guaranteed to have at least one birthday share".

[@X22GJP]

Being honest I didn’t initially think 100%, but my initial gut reaction was that it feels it should be high just based on the number of people in London versus number of hairs on the head.

[@jogihoogi1]

1, at least 2 people shave

[@samuelrickert6828]

I thought the same thing

[@Alex-js5lg]

Damn, this is like beating a video game in the opening sequence by convincing the antagonist to just move to Tahiti.

[@sweepingtime]

I guessed 1, but it was only gut instinct, I had never heard of the pigeon hole principle before. My gut was telling me that if there are millions of people, the chances of any 2 people with the exact same number of hairs ought to be pretty high. There was a conservative voice telling me to go towards 50%, but in the end I chose 1.

[@HermanVonPetri]

Well, he did say to go with your gut instinct and it worked.

[@ViliamF.]

I thought it was tiny, then I thought about the birthday paradox, and I thought it might be something not so small. But then when he said he knows exactly, a lightbulb went on over my head and I knew too XD

[@marcushendriksen8415]

I remember this example (though it was set in New York as I recall)

[@vick229]

Back again to numberphile

[@PC_Simo]

8:25 That’s, why the password-suggestions often include, like, 15 characters, I suppose 🤔.

[@buttonsjr]

Its 100% for people with Alopecia we all have zero hairs on our head hahaha. My gut reaction was thinking about the birthday paradox, how its a 50% chance 2 people in a room with like 21 or 23 ish people share the same birthday, this seems very similar to that.

[@isacami25]

this is so funny. last year this was one of the questions we set up for our students as warm up, before really starting with the contents of the course.

[@nathangreene3]

Ben reminds me of Russel Crowe. There are several Russel Crowe movies where he has hair like Ben's.

[@ThePiMan0903]

Nice video numberphile!

[@fritsvanzanten3573]

I reasoned the other way around. With 1 million people how many hairs would we need on our head to have a different number each?

[@Xnoob545]

0:57 I'd say Likely to Very Likely

[@BruceDuncan]

Interesting that you mentioned that mathematicians tend to be bad at estimating. As a recovering physicist you nearly caught me out, but I got it pretty quickly. I expect engineers would be even sharper ;)

[@j.thomas1420]

Reminds me that numberphile episode : what's the probability that two people in a auditorium share the same birthday ? Same logic.

[@breakingfree7244]

Mel Science looks amazing

[@DocBree13]

Finally, my stats certificate pays off, lol

[@EconAtheist]

Always fun to see Ben pop up in a new one!

[@CR0SBO]

Instantly a 1. Bound to be more than a single baldy!

[@CR0SBO]

Doh! My gut acted too soon!

[@machineman8920]

adorable drawings and fire cut ben ;]

[@diesdas9400]

The final clip made my day

[@chadjones1266]

Thanks Again

[@milestonechild]

Wonderful explanation

[@tonycolle8699]

At 0:59, I posit 100%. The human head has about 100K hairs and there are more than 100,000 people in London. In fact, there is probably almost certainty that there are two people with zero hairs. Perhaps I did not understand the question but that is my take

[@cerperalpurpose]

The pigeonhole principle didn't seem like common sense in the maths lecture I gatecrashed. The equations make me shivver

[@jeroen94704]

Let's introduce "velocity of nodding" as an official measure of agreement. Not sure about the proper unit. Could be Hz, or perhaps cHz to get nicer numbers.

[@Primalmoon]

Gut reaction was 0. Still experiencing some dissonance because I thought the number of hairs on a head was much much bigger and I'm surprised the number is so low. I think I had confused some other biology facts like number of cells in a body, which is greatly bigger than the population of earth, and subbed that in for the number of hairs on a head.

[@robertpearce8394]

Without doubt the Outnumbered kid.

[@erikbrendel3217]

The outro music sounds a lot like the Factorio theme song "Solar Intervention" - nice :)

[@newburypi]

Loved this one!!! Thanks.

[@nickrodriguez9935]

When he asked the question my immediate reaction was to say 100%, I was thinking about that thing with birthdays and how it was super likely 2 people shared the same birthday. Same applies here I guess

[@GroovingPict]

once he said he knew the exact precise answer, with decimal expansion and all, you knew it had to be either 0% or 100%. And since it cant be 0%, it had to be 100%

[@kevinmcknight5000]

Same.

[@nekogod]

There's 8 something million people in london and I'd wager people have less than 1 million hairs on their heads, so there must be multiple people with the same number of hairs. It also helps Ben says he can give the answer to full precision which he could only do if it was 1 or 0 really.

[@stevefrandsen7897]

I like Ben's videos. Very common senseical, pun intended.

[@wiggles7976]

My gut reaction was that it was guaranteed that there are two people with the same number of hairs on their head, but only because I knew about the pigeonhole principle's application to this problem beforehand.

[@ann_onn]

What you said at 5:30 is kinda wrong. The literal City of London is the square mile, and thus has a population of only about 10,000.

[@sweepingtime]

Yeah, but if you take the pigeon hole principle into account, then part of your estimation needs the number of pigeon holes, hence depending on how big of a London territory you choose your probability scales up or down. It's pretty nifty I think.

[@maarirs12894]

Fantastic video

[@petrospaulos7736]

Classic pigeonhole principle problem!

[@SRangerMtl]

I immediately think of the birthday paradox and guess that the answer is "certain". Gonna watch the video now;) Later : Alright, the solution used another principle!

[@sihingvonfelix4251]

The initial question sounds a lot like the birthday paradoxon with a lot more days and a lot more people.