Lewis Carroll's Pillow Problem - Numberphile

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27 апр 2020

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Комментарии : 1 тыс.

@pleaseenteraname4824 4 года назад

Brady: "Are you having unholy thoughts?" Alex: *nervous laughter*

@disgruntledtoons 4 года назад

And then it shows him in bed with Lewis Carroll...

@babybeel8787 4 года назад

Ahahahaha

@RedRad1990 4 года назад

A way of dealing with impure thoughts in bed? Nowadays we call it 'incognito mode'

@Albimar17 4 года назад

i bet one of these problems gotta be, how many armies could i fit in in between my top tier tooths? XD

@JarodM 3 года назад

@@Albimar17 Ten Roman Legions ⚔

@jakethesnake17 4 года назад

Gf: He’s probably thinking about other girls. Bf: *_Pillow Problems_*

@raffimolero64 4 года назад

don't lie to me, you're having B l a s p h e m o u s T h o u g h t s .

@elnico5623 3 года назад

Unholy Thoughts

@schizoframia4874 3 года назад

Body pillow

@Triantalex 6 месяцев назад

??.

@arkajyotijha906 4 года назад

"Before you get to bed, you have sceptical, blasphemous, and unholy thoughts" My entire day: *Well actually...*

@gulugul78 4 года назад

I do that at least three times a week using my own ball bag

@hamiltonianpathondodecahed5236 4 года назад

@Richard Groller Alice wants BoB

@ostrich_dog 4 года назад

@@hamiltonianpathondodecahed5236 I appreciate

@seabassthegamer6644 4 года назад

I mean that's technically still before bed

@livedandletdie 4 года назад

@@hamiltonianpathondodecahed5236 Bob's your uncle.

@esotericVideos 4 года назад

Therapist: "Are you having any sceptical thoughts?" Lewis Carroll: *Takes drag from cigarette* "All I have are sceptical thoughts."

@darreljones8645 4 года назад

Why did they misspell the word "skeptical"?

@esotericVideos 4 года назад

@@darreljones8645 From google: "Skeptic is the preferred spelling in American and Canadian English, and sceptic is preferred in the main varieties of English from outside North America." So presumably it's a UK thing.

@billywhizz09 4 года назад

It can be spelled like that

@AdityaKumar-ij5ok 4 года назад

esotericVideos jokes

@robertlewis5439 4 года назад

@@billywhizz09 I'm sckeptical about that.

@pafnutiytheartist 4 года назад

To be honest it's much more intuitive than the Monty Hall paradox. This one took me much less time to re-frame in a sensible way.

@alephnull4044 4 года назад

pafnutiytheartist Chebychev is that you?

@yensteel 4 года назад

Totally agree!

@olmostgudinaf8100 4 года назад

What clinched it for me was the realization that the red ball he took out was not necessarily the same one he put in. The rest follows.

@user-rv9vk8by5i 4 года назад

My first realisation was that if he kept repeating it and kept pulling out a red ball, the chance that there's no other colour approaches 100%. So of course, after just one iteration, the probabilities will change.

@charlytaylor1748 4 года назад

@@user-rv9vk8by5i don't bring infinity into this!

@cupass6179 4 года назад

i'm amazed at how badly i wanna see a green ball right now. this is a weird feeling

@noslowerdna 4 года назад

truth

@toblerusseta 4 года назад

Is that because you're a dog ?

@qoakoa 3 года назад

Happening

@proloycodes 2 года назад

saaame

@kalla103 2 года назад

yess i was waiting for him to take a green out

@erbro 4 года назад

The strange thing is that these puzzles often only seem difficult with small numbers. With big numbers they can become obvious. If you pick a random ball 300 times and put it back, and it comes up red every time, most people will conclude that the one in the bag is probably red.

@RandySpaulding 4 года назад

This episode of Numberphile sponsored by Red Ball energy drinks

@charadremur333 4 года назад

Took me a seconds.

@rosuav 4 года назад

You put a red bull into the bag, and pull a red bull out of the bag. What is the probability that someone drank it in between? Very low if there were any cans of V Blue in there - it tastes way better. Wait, I might be missing the point here.

@tiyenin 4 года назад

Nobody's talking about the fact that if the ball you pull out is green, then there is a 100% probability that the remaining ball is red.

@nathaniliescu4597 4 года назад

It was yes.

@meghanshu7424 3 года назад

@@nathaniliescu4597 replying a reply Reply paradox

@arisontube 4 года назад

Another way to solve it is with Bayes Theorem: The probability that the first bead is red: PA = 0.5 The probability that the first bead is green: PA- = 1 - PA = 0.5 The probability to pick a red bead given that the first bead is red: P(B|A) = 1 The probability to pick a red bead given that the first bead is green: P(B|A-) = 0.5 Therefore, the total probability that we pull a red bead is: PB = P(B|A) * PA + P(B|A-) * (PA-) = 0.5 * (1) + (1-0.5) * 0.5 = 0.75 If we apply Bayer Theorem, the probability that the initial bead is red given that we pulled a red bead is: P (A|B) = P(A) * P(B|A) / P(B) = 0.5 * 1.0 / 0.75 = 0.66 If we perform the experiment several times, each time updating PA and PB as P(A)_next = P (A|B) _previous and PB_next = P(B|A) * PA_next + P(B|A-) * (PA-)_next we get the following values for P (A|B) for each successful attempt: 1 0.666 2 0.800 3 0.888 4 0.941 5 0.969 etc.

@nathanbell6962 4 года назад

I don't know what your talking about but I agree with you wholeheartedly because you must be smarter than me.

@EwingTaiwan 4 года назад

This is what I'm thinking about, this is nicely done. A little mistake: " The probability to pick a red bead given that the first bead is green: *P(B|A-) = 0.5* " In plain text, that should be "the probability of B given ( A- ), not ( A ). Also, it should be "Bayes" not "Bayer" (unless there exist other aliases that I'm not aware of)

@leadnitrate2194 4 года назад

@@nathanbell6962 if you're interested, I'd suggest to you the 3blue1brown video(s) on Bayes' theorem. Amazing.

@Elmaxo1989 4 года назад

Also those probabilities of red remaining, after each additional successful drawing of a red ball, are (2^[n])/(2^[n]+1)

@MarkWiemer 4 года назад

@@Elmaxo1989 Yep! My favorite part is that this works with n=0 as well

@NoriMori1992 4 года назад

>"I know exactly the time it was invented." >proceeds to give the date but not the time 😝

@larrykuenning5754 4 года назад

Probably before #72 (invented the same night but crazier).

@chinareds54 4 года назад

@M N Well, more accurate than that because it was at Carroll's bedtime.

@NoriMori1992 4 года назад

@M N Yes, I realize that, but usually when you say you know the "time" something happened, one expects to be told a _time_ (as in "5 o'clock" or "8:47 am"), not a date. Of course, I'm sure you realize that as well, and are merely nitpicking my nitpick. 😛

@thisrandomdude_ 3 месяца назад

pfffft, bedtime obviously

@williamaitken7533 4 года назад

I was thinking to myself how this was similar to the Monte Hall problem before Alex brought it up!

@wingracer1614 4 года назад

Same here. It's a different version of Monte Hall

@nikediva1 4 года назад

Yes exactly

@LordPrometheous 4 года назад

Yeah, this very thing was mentioned in the movie "21" and they called it variable change. I immediately thought of the movie.

@Croccifixo 4 года назад

Was actually coming to the comments to ask if this wasn't the same, then he mentioned it

@sleepinflame1288 4 года назад

I thought so as soon as he said “red or green”. It took me so long to get the intuition behind Monty Hall, I’m very aware about the value of the additional information

@chinareds54 4 года назад

The interesting thing is although in the case of one iteration, it may seem like the probability is unchanged, if you do the experiment 100 times and 100 times in a row you pull out a red ball, anyone would agree that it's most likely the original ball hidden ball was also red. So therefore the probability must change each time you do it. Incidentally, this is also one of the ways of explaining the Monty Hall problem, by increasing the number of decoy doors.

@Jodabomb24 4 года назад

When you say "given that we drew a red ball out, [question]", what you're really talking about is something called "post-selection". It has ties to Bayesian statistics and conditional probabilities and all that kind of stuff. There are really interesting things being done with this kind of mathematics in the context of quantum mechanics, and especially in the context of so-called "weak measurements".

@zerid0 4 года назад

The probability is 100%. There's no way he would have taken the chance of it failing by drawing green and having to reshoot the video.

@gustavgnoettgen 4 года назад

I still wonder if he even has a green ball there. 🤔

@tinynewtman 4 года назад

@@gustavgnoettgen If he did, it would probably be a fuzzy tennis ball so he could know not to pick it by texture alone.

@billowytrots8366 4 года назад

@@gustavgnoettgen I think it might have been blue.

@gustavgnoettgen 4 года назад

@@tinynewtman but it's also just a fairly small bag (OR SHOULD I SAY 'PILLOW'????🤣) so mixing them up isn't easy in the first place.

@gulugul78 4 года назад

@@gustavgnoettgen he had one green ball in his bag... unfortunately it was removed by way of orchidectomy😳

@norbi275275 4 года назад

You can "easily" get it using Bayes theorem: P(A) - pulling red = 3/4 P(A^B) - pulling red and red is inside = 2/4 then P(B) - red being inside = (2/4)/(3/4) = 2/3

@alephnull4044 4 года назад

Yes, it's conditional probability at work here.

@slurpleslixie 4 года назад

Yeah that's how I did it, seemed like an obvious case of bayes' theorem

@ZygfrydJelenieRogi 4 года назад

Thanks! This explains more than the video tbh

@danmarino900 4 года назад

this isn’t bayes theorem /:

@danmarino900 4 года назад

you’re just using the definition of conditional probability; bayes theorem is a consequence of this, not vice versa

@Zizzily 4 года назад

If you're red-green colorblind, does that make it 100% probability?

@sprsmalstegn5911 4 года назад

yes

@Ganliard 4 года назад

No, then the ball just has a hidden property

@aryamankejriwal5959 4 года назад

😂

@jeremydavis3631 4 года назад

I know this might have been a joke, but the probability actually wouldn't change at all. It would stay at 50%. If you can't tell whether the ball you've pulled out is red or green, you gain no information by doing that, so you can't rule out any of the four equally likely possibilities. The only way to get 100% would be to pull out a green ball (and recognize it as green), since that would prove that the red ball you'd put in the bag was still inside.

@magichands135 4 года назад

If you have dyscalculia it's probably 109%

@4ltrz555 4 года назад

Pillow problems are the mathematical equivalent of the brain talking meme template.

@gustavgnoettgen 4 года назад

Who would win: The most advanced computing organ known to itself, able to construct and operate vehicles to other planets and surgery on itself to mention only two things Or A soft boi

@knifeninja200000 4 года назад

As well as shower thoughts

@oldcowbb 4 года назад

someone please make it

@Triantalex 6 месяцев назад

false.

@Vodnuth 4 года назад

"Are you having any unholy thoughts?" "All I have are unholy thoughts"

@Endothermia 4 года назад

"Is it just me, or is it getting unholier out there?"

@JarodM 3 года назад

🤡

@TheKopakah 4 года назад

I didn't know Michael Sheen was into puzzles

@estherscholz8400 4 года назад

Doing math to avoid unwanted thoughts is relateable.

@johannesbragelmann6629 4 года назад

Me at the end of the video: NOW TELL ME: IS THERE A GREEN BALL?

@recklessroges 4 года назад

aah its 50/50 ;-)

@VoidFame 4 года назад

6:44 you can see the red bleed through.

@bokkenka 4 года назад

@@recklessroges -- No, he said it's 2/3%

@katrinareads 4 года назад

@@bokkenka You mean 2/3 or approximately 67%. 2/3% is 0.006666 repeating, or approximately 0.67%. Very different.

@olmostgudinaf8100 4 года назад

@@katrinareads a friend of mine talking about his acquaintance: "I don't know why other businessmen complain about small margins. I buy a widget for 5 and sell it for 8 and I can live on those 3% quite comfortably."

@mjswart73 4 года назад

The animator is on fire with this one

@bentoth9555 4 года назад

Me, thinking: It's the Monty Hall problem, basically. Alex, a minute later: If you do the archaeology of the Monty Hall problem, this is where it all began.

@riccardogilardi3124 4 года назад

"Before you get to bed, you have sceptical, blasphemous, and unholy thoughts" Well, I read too many Caroll biographies not to be scared of what he might have thought

@yashbijlani6652 4 года назад

Normal people: money, relationship, etc problems Mathematicians: pillow problems

@julienbongars4287 4 года назад

Software Engineers: Javascript

@Ian.Murray 4 года назад

I hate when I think about et cetera problems...

@olmostgudinaf8100 4 года назад

@@julienbongars4287 You consider Javascript "software engineering"? Bah!

@julienbongars4287 4 года назад

@@olmostgudinaf8100 r/gatekeeping

@olmostgudinaf8100 4 года назад

@@julienbongars4287 ;) was implied. I would have made it explicit for a C or C++ guy, but I assumed that a JavaScript guy would be familiar with implicit types ;)

@ten.seconds 4 года назад

I can turn this into a more intuitive version. Say if Alex repeat the process 100 times. I think most people would think that the other ball is definitely not green since the effect compounds. It's less likely that we're in the green ball universe even if we only do it once and the red ball is drawn out. Recall the modified monty hall problem where there's 100 doors, you pick one and the host open 98 doors with goats behind them. It's the same idea.

@leecoldsmoke 4 года назад

I was surprised they didn't mention this in the video.

@Albimar17 4 года назад

what about monty python?

@111fernandovg222 4 года назад

Mathematicians: pillow problems Me: I want to watch the green ball!

@Fatabuna 4 года назад

lol, there must a be a poodle somewhere amongst your ancestors

@KatzRool 4 года назад

Imagine being so legendary that you ponder complex mathematical issues to stifle your dark wandering mind.

@Pembolog 4 года назад

He was a well respected Mathematician is his own right, he was a lecturer at Oxford

@NortheastGamer 4 года назад

You don't have to be legendary, for example: I think about math and logic stuff all the time to distract myself from stuff and I'm totally average in every way ;)

@jacobscrackers98 4 года назад

It's not that complex really.

@arpitdas4263 4 года назад

Yeah if i was thinking of diddling kids, I'd probably turn to riddles as well

@Sauspreme 4 года назад

it the "Randomly selected" interaction that changes the probability.

@tsgsjeremy 4 года назад

I'll have you know that all those rabbits during the thinking time had me thinking many many unholy thoughts.

@WG55 4 года назад

When I was studying for my degree in mathematics, I came upon a copy of Lewis Carroll's _Pillow Problems_ in the library, and remember the infamous Problem no. 72 from "Trancendental Probabilities": "A bag contains 2 counters, as to which nothing is known except that each is either black or white. Ascertain their colours without taking them out of the bag. (8/9/87)" He deduces from tortured and ridiculous logic that one must be white, and one must be black. "To the casual reader it may seem abnormal, and even paradoxical; but I would have such a reader ask himself, candidly, the question 'Is Life not itself a Paradox?'

@MushookieMan 4 года назад

That was deduced by Dodgson in 1893, but he made several assumptions. Obviously nothing can be deduced otherwise.

@WG55 4 года назад

@@MushookieMan Yes, he was obviously making a joke. 😆

@NoriMori1992 4 года назад

Between that and the Alice books, I get the impression he had an incredibly offbeat sense of humour 😂

@larrykuenning5754 4 года назад

I didn't see this comment until I'd posted about problem #72 myself. Yes, I think it's a joke, funny only to mathematicians. Did you notice it was invented the same night as the problem in the video?

@bigjimrand 4 года назад

I was asked a question like this on the first or second day of my philosophy degree and I got it wrong and it felt really exciting! With all these "counterintuitive" probability puzzles, it instantly becomes much saner if instead of asking, 'what's the probability that x?', you ask, 'what world could I be in?' and properly imagine being in them all, even if it's some dry equation - when I do that it loosens any attachment I had to one specific outcome, and it reminds me to exhaustively check all the options. I know now that as soon as I feel that signature Monty Hall brain-ache, I must have missed a world and I just need to loop back and visit it.

@wolffang21burgers 4 года назад

If you do it n times: (drop a red ball in, take a red ball out) Probability is (2^n) / (2^n + 1) (as there is a 1/(2^n) chance of picking a Red if you are in the Green ball universe).

@inakibolivar664 4 года назад

That is the conclusion I got to and I was honestly looking for a comment to confirm my theory

@DavidBeaumont 4 года назад

If you do it n-times and don't pull out the green, then yes, it's a 1/2^n chance. But that's not the actual puzzle. The puzzle says "you randomly take out a ball *and it's red* " which is possible, but not certain. So it trims down the space of all possibilities by removing the case where a green ever gets taken out. That's the *sneaky* bit.

@inakibolivar664 4 года назад

That's the case when n = 1, I don't understand what you are saying that hasn't been said yet, it's not sneaky at all, its what the original comment said

@wolffang21burgers 4 года назад

@@DavidBeaumont Yes sorry, I though that would be inferred. But also my wording wasn't great. So if you are in the Green universe: there is a 1/2^n chance. If you are in the Red universe: there is a 2^n/2^n chance. Hence, 2^n / (1+2^n)

@inakibolivar664 4 года назад

@@wolffang21burgers Your wording was perfect

@uraldamasis6887 4 года назад

03:35 I said to myself "The probability the other ball is red is 100%. Because if he put a green ball in there, there is a 50% chance his explanation would be ruined and he would look like a fool."

@holyknightthatpwns 4 года назад

Or he's lying to us about "randomly selecting" the ball

@maxberan3897 4 года назад

@@holyknightthatpwns Surely that is misjudging the situation. He wasn't performing an experiment, he was play-acting the "given" in the puzzle. And what is "given" by the terms of the puzzle is that the ball removed was red. And we were asked, given that piece of information, what is the probability that the other ball was red.

@user-xl2kf9rr7c 4 года назад

@@maxberan3897 Oh, noooo! Are you sure?!

@Bartooc 4 года назад

Or the other ball was actually yellow and he's fooling with all of us.

@alexliu2221 4 года назад

actually, he might have held on to the red ball while his hand was inside the bag

@Agresiel 4 года назад

I am so proud I knew the answer to this and connected this to the Monty hall problem before Alex said. THANK YOU NUMBERPHILE! you are teaching me things that is committed to long term memory!

@George4943 4 года назад

Strangely similar: I randomly throw two dice where I cannot see. My interlocutor says, "I see a 6." What is the probability that the other die is also 6?

@wildBillMunson 4 года назад

Sample space is: {(1,6), (2,6), (3,6), (4,6), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)}. Therefore, only a 1/11 chance the other die is also 6.

@benweieneth1103 4 года назад

I'd say it depends on the algorithm of your interlocutor. If it's "look at one of the dice and say what number is showing", then it's 1/6. If it's "say whether a six is showing on either die", then it is indeed 1/11.

@hip-notized8635 4 года назад

Finally, a perfect video, my-2 a.m-watch list will be legendary

@CaptainSpock1701 4 года назад

Almost there. Reading this comment at 1h07 in the morning!

@sterby1 4 года назад

By applying physics and properties of light through a fine mesh of material I can deduce there was no green ball in the bag... ...with 66% probability

@arpitdas4263 4 года назад

Excellent. A fellow man of culture

@Daniel-qc2tl 4 года назад

Since this is a video either the red ball wasn't picked randomly or there was never a green ball. We also don't even know if there was another ball

@King0Mir 4 года назад

This and Bertrand's box paradox are actually different from the Monty Hall problem in a particular way: In Monty Hall, the host knowingly picks one of the goats, whereas in this case the subject still randomly picks the red ball (or gold coin).

@REDBULLHEADiphone 4 года назад

Absolutely delightful. Thanks for sharing.

@NoriMori1992 4 года назад

I thought this felt very Monty Hall-ish, nice to see the connection confirmed 😊 I'm pretty proud of myself for guessing the answer was 2/3 - even if Monty Hall helped me do it!

@mikec4390 Год назад

Except the answer was 1/3. He was asking the probability of it being green.

@cigmorfil4101 Год назад

Curiously, the Monty Hall priblem is subtly different. Because the host *knows* where a booby prize is, he can always show a booby prize, the initial random separation into two groups of 1 and 2 items keeps the probabilities for each group when he shows a booby prize from the 2 group. If the host did *not* know where the star prize was and _randomly_ selected one of the two remaining doors to open and showed a booby prize, then the probabilities *do* change and swapping is no better than keeping, as they are now both 50% chance!

@gustavgnoettgen 4 года назад

Is it also a blasphemous thought to imagine a sceptic thinker in bed with Lewis Carroll?

@autolykos9822 4 года назад

Hearing that problem just screamed "Bayes' Theorem" at me (try it, it gives the same result). As a bonus, once you've worked it out, it trivially gives you the probability for the ball still being red after repeating the experiment N times, and always pulling out a red one.

@yueshijoorya601 3 года назад

Today is the 28th of November 2020. I just saw this video switch from "6 months ago" to "7 months ago", at 8:56 pm.

@aleschudarek4672 4 года назад

I dont know why, but I really want to drink RedBull right now :D

@TheRealGuywithoutaMustache 4 года назад

So it's a probability issue, I knew it sounded familiar at first, then I realized it was similar to the topic I learned 3 years ago in class

@alanwolf313 4 года назад

Oh f***, didn't expect to see you here XD

@casualbeluga2724 4 года назад

@@alanwolf313 it's not the og one

@MechMK1 4 года назад

If you had a mustache, perhaps you would have remembered sooner.

@TheHyruleCitizen 4 года назад

It's similar to the Monty Hall problem, I believe.

@alanwolf313 4 года назад

@@casualbeluga2724 I know, but i still see him in a lot videos

@barmanitan 4 года назад

Man, can't catch a break from Monty Hall. It's everywhere!

@jameswest8280 Год назад

I love counterintuitive stuff like this. The Monty Hall problem blew my mind, even though I read about how it works, it's still kind of a mystery.

@hasko_not_the_pirate 4 года назад

9:00 Why does he consider it “smoke and mirrors”? It’s just a posteriori knowledge that we’re given. There’s no lie in it. The ball was picked at random and we just happen to know the result of that picking.

@Swiftclaw123 4 года назад

This is literally a basic application of Bayes’ Rule

@mina86 4 года назад

Yes, was just about to comment P(2 red | red out) = P(red out | 2 red) P(2 red) / P(red out) = ½ / ¾ = ⅔

@kanjurer 4 года назад

mina86 yeah, eazy peazy lemon squeazy

@WideMouth 4 года назад

It took me 10 seconds and I’d never even heard of Bayes’ Rule.

@mattbox87 4 года назад

@@WideMouth well done! You get the fundamentals of probability I don't mind saying it took me a while I think I was wrangling equations too much and not pondering "pillow problems" like our friend Lewis

@OlliWilkman 4 года назад

@@mina86 I thought about it slightly differently (maybe in a more complicated way), framing it as P(red in | red out) = P(red in) P(red out | red in) / (P(red in) P(red out | red in) + P(green in) P(red out | green in)) = ¾ ⅔ / (¾ ⅔ + ¼ 1), but the answer is the same of course.

@dejremi8190 4 года назад

Oh yes even during lockdown numberphile videos keep being releasing what a pleasure

@sudheerthunga2155 4 года назад

8:15 Exactly randomly! It gives us information on the distribution.

@zerid0 4 года назад

This is very similar to the mounty hall problem I think. Imagine each door being a ball, 2 reds and a green. You choose one randomly but don't look at the colour, then we show you a red ball that you haven't picked and you have to find out the probability that your pick is red. For these sort of problem, the fact that the probability changes becomes more obvious if you change the scale. Imagine having a ball that is 50/50 red or green. Then you add 99 red balls into the bag, draw 99 balls. If all the balls drawn are red, what is the probability of the final one being green? It would be very unlikely to draw 99 reds if the initial ball was green. So the probability of that must be much lower than it being red.

@Pembolog 4 года назад

>This is very similar to the mounty hall problem I think. If only the mentioned that in the video

@pbp6741 4 года назад

Good video. I do wish he had expanded the discussion to repeated sampling.

@mattc3581 Год назад

Initially the probability is 1/2, after randomly pulling a red ball it is 2/3, if you replace it and randomly pull a red ball again it is now 4/5, then 8/9 then 16/17 and so on. Since the number of ways you can draw the red from a red/red scenario is twice the number of ways you can draw a red from the red/green scenario there always remains only one way of the hidden ball being green but the number of ways it can be red doubles each time.

@Jaojao_puzzlesolver 4 года назад

I love that thinking time animation.

@thesos320 4 года назад

One of the few times I got it right! I didnt think of it the same way though. It was mostly intuitively. Simple but interesting!

@ErulianADRaghath 4 года назад

It is unfortunately 6am here, and I really should get some rest.

@l.3ok 4 года назад

I interpreted the problem in two ways: I)Well, if you take a random ball of the bag, considering that it can be green, we have that the probability of the ball inside the bag being red is 75%. II) On the other side, if the ball that I take of the bag needs to be red, we have that the probability of the ball inside the bag being red is of 66.66...%. This problem is very similar to the Monty Hall problem, once you solve one of them, you can solve almost immediately the other (English is not my native language, so I may have made some grammatical mistakes).

@torreyrg42 4 года назад

I also wondered if this is just a restatement of the Monty hall problem, or vise versa or at least related somehow based on what you do know, what you don’t know, and what the probabilities are. Edit... should have watched 10 more seconds into the video before commenting where the progression of this problem into the Monty hall problem is properly explained.

@pryan22 4 года назад

Well if he puts a red ball in and then randomly pulls out a green, then the probability that the ball in the bag is red is 100%....right? Isn't that why it would be pointless if he pulled a green out of the bag?

@tinnitusthenight5545 4 года назад

I love how he never shows the other ball

@mpalin11 4 года назад

Very nice explanation!

@azdarksonal 4 года назад

Oh wow this is my first time seeing the man behind the camera, I always assumed he’d be clean shaven for some reason. :p

@CeeJMantis 4 года назад

He has several other channels, but there's one called Objectivity where he is frequently on camera so he can interact with the objects

@jzieba0204 4 года назад

Then you dont watch the videos till the end, cuz he nearly always sponsors something at the end.

@Bartooc 4 года назад

Welcome, you must be new to channel then.

@jerry3790 4 года назад

I find it disappointing that people rarely talk about Carol’s contributions to mathematics. A lot of them are just as interesting as his writing!

@bsharpmajorscale 4 года назад

Maybe because a lot of it is that complicated logic stuff. :P

@ArieteArmsRAMLITE 4 года назад

I met him in a pub in Camden once.

@marcognudi664 3 года назад

Always a pleasure watching Michael Sheen as his puzzle-solving alter ego!

@EternalDensity 4 года назад

"The surprising thing is that it changes the probability." The surprising thing is finding that surprising.

@dennis.geurts 4 года назад

a nice addition might have been to actually have 'randomly' selected the green ball: Then everyone would immediately have felt intuitively that now the probability that the ball in the bag of being red had increased to 100%. Thus proving that adding a ball and then randomly picking one out does change probabilities.

@Trias805 4 года назад

6:30 67%

@MultiSteveB 2 года назад

1:49 Interesting how one pinwheel is two pieces, but the other pinwheel (which is rotating in the opposite direction) is a single piece. :D

@wompastompa3692 4 года назад

I just draw trees when dealing with probabilities.

@auferen 4 года назад

Soution: -Be colorblind -whatever you pick, you won't be able to see the difference so you go to sleep peacefully

@prkhrsrvstv1 3 дня назад

Very interesting artwork on the wall behind.

@thatmcgamer3106 4 года назад

This is cool, I started seeing the connection to the Monty Hall, before you mentioned it.

@redsalmon9966 4 года назад

"One red in, one red out" The easy way to get yourself out of this mindset that got you tricked is that the one in doesn't have to be the one out.

@hamiltonianpathondodecahed5236 4 года назад

one red Doraemon in one red Doraemon out

@yuvalne 4 года назад

This problem is equivalent to the Monty Hall problem, isn't it?

@ancientswordrage 4 года назад

They do say that towards the end.

@caseygreyson4178 4 года назад

Yuval Nehemia if you watched the full video, you would realize that they say this was the original inspiration behind that problem.

@bergerniklas6647 4 года назад

I mean he mentions it in the Video, so yeah...

@ElZafro_ 4 года назад

Yep, 7:10 he says it

@Syrange13 4 года назад

How? Can someone explain please?

@philjamieson5572 4 года назад

I enjoyed this. Thanks.

@cuteypetz 4 года назад

weirdly, the monty hall problem always takes me some time to wrap my head around, but this puzzle (even if a similar premise) was much simpler to grasp and understand the maths (as in, I got to the answer before it was shown). I wonder if that's because this only involves two objects, rather than three, which pares the maths down a bit? 🤔

@proloycodes 2 года назад

same

@y1521t21b5 4 года назад

0:16 What you revealed was a date. Still waiting for the actual time ;-)

@Mystery_Biscuits 4 года назад

0:08 “But before that, I want to tell you about...” TODAY’S SPONSOR: RAID: SHADOW LEGENDS

@jeepien 4 года назад

Stating that a red ball is removed is analogous, in the Monty Hall problem, to Monty having the secret knowledge of where the prize is, and always showing you a goat. Instead of leaving the green-ball footage on the cutting-room floor, just have a miniature Monty in the bag who always pushes a red ball into your hand.

@WomenCallYouMoid Год назад

0:43: yes, I know, the thoughts come streaming in.

@rosuav 4 года назад

This actually looks like a great lead-in to Bayes Theorem. You have a prior probability ("is the one in the bag green"), and a stated event ("the one I removed was red"), and the calculation is P(Red seen if Green present) * P(Green present) / P(Red seen overall), or 50% * 50% / 75%. The key here, as you showed, is that the probability of drawing Red out is actually 3 in 4.

@ca-ke9493 4 года назад

The multiple ways to reframe the monty hall problem is so fascinating. I really like this particular set up for the monty hall problem, as I can imagine how as you put in a red ball and randomly pull out a red ball, we become more and more certain that the original ball is red and not green as it is really unlikely that you'd not get a green ball with more tries, if it was green. One way which conditional probability really gets so confusing - its not intuitive by common sense or by rigourous math to me.

@bwill325 4 года назад

I'm proud of myself for recognizing it was a form of the Monty Hall problem before it was revealed.

@caterpillow 4 года назад

these were made for me

@jj.wahlberg 4 года назад

I have a probability midterm on Friday so this is great timing

@theRealPlaidRabbit 4 года назад

I like the thinking time image. Reminds me of my friend Leonardo of Pisa

@borissokol2958 4 месяца назад

I find this very intuitive because the question makes it clear and almost explicit that the inside ball could have changed (if it couldnt, none of it would make sense) and it could only be changed into red.

@waltercisneros9535 4 года назад

Amazing way to wake up and have my breakfast, greetings from Colombia 🇨🇴

@codyheiner3636 4 года назад

I'd recommend Bayes' rule for this one, eliminates the weird mental trickery and makes it very straight forward. P(other green | picked red) = (1)(1/4)/(3/4) = 1/3.

@cauchym9883 4 года назад

The way the problem is presented reminds me a bit of the capture-recapture method to estimate population sizes. Perhaps you could make a video on this sometime?

@LoaFrz 4 года назад

I think the difficult thing about this problem is the explanation of the setup. (At least for me) I miss interpreted the part where we start with a known red ball OUT of the bag and an unknown ball IN the bag. I had to rewatch the beginning to listen to the setup again. Fun puzzle!

@user-iy7yy8er3r 2 месяца назад

Out of all red picks, 1/3 are cases when the picked red is new but old one is also red. Also, 1/3 of red picks is when initial ball was red and now is picked. So in total 2/3 chance that the initial ball was red

@PAUNOMOLUSCO 4 года назад

2:56 Randomly in the same sense that a host of these 3 doors games opens a door and asks you if you want to switch doors after you picked one, but before, of course, it’s revealed what is behind your door.

@otakuribo 4 года назад

"sceptical, blasphemous, and unholy thoughts" Alex you just sold me your new book

@ChrisStavros 2 года назад

It's a little different from the Monty Hall problem, because it that, the host reveals an incorrect door with the knowledge that it's incorrect, whereas here we've revealed a non-green ball and it's baked into the probability of the question from the outset, while we're supposed to assume that it *could* have been green.

@mydearfriend007 4 года назад

Amazing video

@qwertyTRiG 4 года назад

I managed to work this one out fairly handily. It may be the ancestor of the Monty Hall problem, but it's a lot simpler.

@Dangles1989 4 года назад

The conditional probability formula can help. Pr(A given B) = Pr(A and B)/Pr(B) where B is that the selected ball is red and A is that the nonselected ball is red.