Lewis Carroll's Pillow Problem

Lewis Carroll's Pillow Problem - Numberphile

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4 жыл бұрын

Featuring Alex Bellos.
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Пікірлер: 1 000

@pleaseenteraname4824 4 жыл бұрын

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

@disgruntledtoons 4 жыл бұрын

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

@babybeel8787 4 жыл бұрын

Ahahahaha

@RedRad1990 4 жыл бұрын

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

@Albimar17 4 жыл бұрын

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

@JarodM 3 жыл бұрын

@@Albimar17 Ten Roman Legions ⚔

@jakethesnake17 4 жыл бұрын

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

@raffimolero64 4 жыл бұрын

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 жыл бұрын

Unholy Thoughts

@schizoframia4874 3 жыл бұрын

Body pillow

@Triantalex 6 ай бұрын

??.

@esotericVideos 4 жыл бұрын

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

@darreljones8645 4 жыл бұрын

Why did they misspell the word "skeptical"?

@esotericVideos 4 жыл бұрын

@@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 жыл бұрын

It can be spelled like that

@AdityaKumar-ij5ok 4 жыл бұрын

esotericVideos jokes

@robertlewis5439 4 жыл бұрын

@@billywhizz09 I'm sckeptical about that.

@arkajyotijha906 4 жыл бұрын

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

@gulugul78 4 жыл бұрын

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

@hamiltonianpathondodecahed5236 4 жыл бұрын

@Richard Groller Alice wants BoB

@ostrich_dog 4 жыл бұрын

@@hamiltonianpathondodecahed5236 I appreciate

@seabassthegamer6644 4 жыл бұрын

I mean that's technically still before bed

@livedandletdie 4 жыл бұрын

@@hamiltonianpathondodecahed5236 Bob's your uncle.

@pafnutiytheartist 4 жыл бұрын

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 жыл бұрын

pafnutiytheartist Chebychev is that you?

@yensteel 4 жыл бұрын

Totally agree!

@olmostgudinaf8100 4 жыл бұрын

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 жыл бұрын

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 жыл бұрын

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

@arisontube 4 жыл бұрын

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 жыл бұрын

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

@EwingTaiwan 4 жыл бұрын

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 жыл бұрын

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

@Elmaxo1989 4 жыл бұрын

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

@MarkWiemer 4 жыл бұрын

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

@RandySpaulding 4 жыл бұрын

This episode of Numberphile sponsored by Red Ball energy drinks

@charadremur333 4 жыл бұрын

Took me a seconds.

@rosuav 4 жыл бұрын

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 жыл бұрын

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 3 жыл бұрын

It was yes.

@meghanshu7424 3 жыл бұрын

@@nathaniliescu4597 replying a reply Reply paradox

@cupass6179 4 жыл бұрын

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

@noslowerdna 4 жыл бұрын

truth

@toblerusseta 4 жыл бұрын

Is that because you're a dog ?

@qoakoa 3 жыл бұрын

Happening

@proloycodes 2 жыл бұрын

saaame

@kalla103 Жыл бұрын

yess i was waiting for him to take a green out

@zerid0 4 жыл бұрын

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 жыл бұрын

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

@tinynewtman 4 жыл бұрын

@@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 жыл бұрын

@@gustavgnoettgen I think it might have been blue.

@gustavgnoettgen 4 жыл бұрын

@@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 жыл бұрын

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

@erbro 4 жыл бұрын

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.

@NoriMori1992 4 жыл бұрын

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

@larrykuenning5754 4 жыл бұрын

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

@chinareds54 4 жыл бұрын

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

@NoriMori1992 4 жыл бұрын

@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_ 2 ай бұрын

pfffft, bedtime obviously

@norbi275275 4 жыл бұрын

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 жыл бұрын

Yes, it's conditional probability at work here.

@slurpleslixie 4 жыл бұрын

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

@ZygfrydJelenieRogi 4 жыл бұрын

Thanks! This explains more than the video tbh

@danmarino900 4 жыл бұрын

this isn’t bayes theorem /:

@danmarino900 4 жыл бұрын

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

@chinareds54 4 жыл бұрын

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.

@Zizzily 4 жыл бұрын

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

@sprsmalstegn5911 4 жыл бұрын

yes

@Ganliard 4 жыл бұрын

No, then the ball just has a hidden property

@aryamankejriwal5959 4 жыл бұрын

😂

@jeremydavis3631 4 жыл бұрын

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 жыл бұрын

If you have dyscalculia it's probably 109%

@williamaitken7533 4 жыл бұрын

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

@wingracer1614 4 жыл бұрын

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

@nikediva1 4 жыл бұрын

Yes exactly

@LordPrometheous 4 жыл бұрын

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

@Croccifixo 4 жыл бұрын

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

@sleepinflame1288 4 жыл бұрын

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

@4ltrz555 4 жыл бұрын

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

@gustavgnoettgen 4 жыл бұрын

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 жыл бұрын

As well as shower thoughts

@oldcowbb 4 жыл бұрын

someone please make it

@Triantalex 6 ай бұрын

false.

@Jodabomb24 4 жыл бұрын

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".

@Vodnuth 4 жыл бұрын

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

@Endothermia 4 жыл бұрын

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

@JarodM 3 жыл бұрын

🤡

@johannesbragelmann6629 4 жыл бұрын

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

@recklessroges 4 жыл бұрын

aah its 50/50 ;-)

@VoidFame 4 жыл бұрын

6:44 you can see the red bleed through.

@bokkenka 4 жыл бұрын

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

@katrinareads 4 жыл бұрын

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

@olmostgudinaf8100 4 жыл бұрын

@@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."

@bentoth9555 4 жыл бұрын

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.

@yashbijlani6652 4 жыл бұрын

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

@julienbongars4287 4 жыл бұрын

Software Engineers: Javascript

@Ian.Murray 4 жыл бұрын

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

@olmostgudinaf8100 4 жыл бұрын

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

@julienbongars4287 4 жыл бұрын

@@olmostgudinaf8100 r/gatekeeping

@olmostgudinaf8100 4 жыл бұрын

@@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 ;)

@riccardogilardi3124 4 жыл бұрын

"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

@WG55 4 жыл бұрын

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 жыл бұрын

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

@WG55 4 жыл бұрын

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

@NoriMori1992 4 жыл бұрын

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

@larrykuenning5754 4 жыл бұрын

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?

@ten.seconds 4 жыл бұрын

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 жыл бұрын

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

@Albimar17 4 жыл бұрын

what about monty python?

@uraldamasis6887 4 жыл бұрын

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 жыл бұрын

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

@maxberan3897 4 жыл бұрын

@@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 жыл бұрын

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

@Bartooc 4 жыл бұрын

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

@alexliu2221 4 жыл бұрын

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

@KatzRool 4 жыл бұрын

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

@Pembolog 4 жыл бұрын

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

@NortheastGamer 4 жыл бұрын

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 жыл бұрын

It's not that complex really.

@arpitdas4263 4 жыл бұрын

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

@wolffang21burgers 4 жыл бұрын

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 жыл бұрын

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

@DavidBeaumont 4 жыл бұрын

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 жыл бұрын

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 жыл бұрын

@@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 жыл бұрын

@@wolffang21burgers Your wording was perfect

@TheKopakah 4 жыл бұрын

I didn't know Michael Sheen was into puzzles

@mjswart73 4 жыл бұрын

The animator is on fire with this one

@estherscholz8400 4 жыл бұрын

Doing math to avoid unwanted thoughts is relateable.

@bigjimrand 4 жыл бұрын

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.

@Agresiel 4 жыл бұрын

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!

@111fernandovg222 4 жыл бұрын

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

@Fatabuna 4 жыл бұрын

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

@l.3ok 4 жыл бұрын

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 жыл бұрын

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 жыл бұрын

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?

@REDBULLHEADiphone 4 жыл бұрын

Absolutely delightful. Thanks for sharing.

@sterby1 4 жыл бұрын

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 жыл бұрын

Excellent. A fellow man of culture

@Daniel-qc2tl 4 жыл бұрын

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

@George4943 4 жыл бұрын

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 жыл бұрын

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 жыл бұрын

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.

@tsgsjeremy 4 жыл бұрын

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

@sudheerthunga2155 4 жыл бұрын

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

@autolykos9822 4 жыл бұрын

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.

@NoriMori1992 4 жыл бұрын

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 Жыл бұрын

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

@cigmorfil4101 Жыл бұрын

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!

@hasko_not_the_pirate 4 жыл бұрын

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.

@jameswest8280 Жыл бұрын

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.

@King0Mir 4 жыл бұрын

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).

@Sauspreme 4 жыл бұрын

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

@hip-notized8635 4 жыл бұрын

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

@CaptainSpock1701 4 жыл бұрын

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

@thesos320 4 жыл бұрын

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!

@dejremi8190 4 жыл бұрын

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

@gustavgnoettgen 4 жыл бұрын

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

@Swiftclaw123 4 жыл бұрын

This is literally a basic application of Bayes’ Rule

@mina86 4 жыл бұрын

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 жыл бұрын

mina86 yeah, eazy peazy lemon squeazy

@WideMouth 4 жыл бұрын

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

@mattbox87 4 жыл бұрын

@@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 жыл бұрын

@@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.

@jeepien 4 жыл бұрын

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.

@yueshijoorya601 3 жыл бұрын

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.

@zerid0 4 жыл бұрын

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 жыл бұрын

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

@aleschudarek4672 4 жыл бұрын

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

@Dangles1989 4 жыл бұрын

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.

@mpalin11 4 жыл бұрын

Very nice explanation!

@pbp6741 4 жыл бұрын

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

@mattc3581 Жыл бұрын

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.

@azdarksonal 4 жыл бұрын

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 жыл бұрын

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 жыл бұрын

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

@Bartooc 4 жыл бұрын

Welcome, you must be new to channel then.

@barmanitan 4 жыл бұрын

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

@marcognudi664 3 жыл бұрын

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

@jerry3790 4 жыл бұрын

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 жыл бұрын

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

@ErulianADRaghath 4 жыл бұрын

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

@Jaojao_puzzlesolver 4 жыл бұрын

I love that thinking time animation.

@KanishkMalkan 4 жыл бұрын

i see it this way....initially we don't know which ball is inside(R/G) and then we add a red ball and pick a red one...to find out the probability, we can try finding which case it was(RR/RG) or basically find out the colour of the ball initially....every time we pick a red ball from the bag containing 2 balls, we near towards the case that has red ball in the starting and if we do this long enough probability -->1 or certainly there are 2 red balls in the bag...every red ball pick changes the probability as (n+1)/(n+2) where n is the no. of picks

@TheRealGuywithoutaMustache 4 жыл бұрын

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 жыл бұрын

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

@casualbeluga2724 4 жыл бұрын

@@alanwolf313 it's not the og one

@MechMK1 4 жыл бұрын

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

@TheHyruleCitizen 4 жыл бұрын

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

@alanwolf313 4 жыл бұрын

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

@dennis.geurts 4 жыл бұрын

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 жыл бұрын

6:30 67%

@MultiSteveB 2 жыл бұрын

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

@EternalDensity 4 жыл бұрын

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

@wompastompa3692 4 жыл бұрын

I just draw trees when dealing with probabilities.

@thatmcgamer3106 4 жыл бұрын

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

@rosuav 4 жыл бұрын

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.

@redsalmon9966 4 жыл бұрын

"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 жыл бұрын

one red Doraemon in one red Doraemon out

@auferen 4 жыл бұрын

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

@borissokol2958 3 ай бұрын

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.

@Parax77 4 жыл бұрын

I find this much easier to continue the dual bag point and to phrase the question such: When I draw a random ball, and it is red, which bag did I have: Bag1(RR) or Bag2(RG)? The odds are then 2 out of 3 red balls were in Bag1(RR) and one of three Red balls was in Bag2(RG) and the probability is exactly that; 2/3 that you have Bag1(R Remaining) and 1/3 that you have Bag2(G Remaining) so it becomes evident that the probability of the remaining ball is 2/3R and 1/3G.

@yuvalne 4 жыл бұрын

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

@ancientswordrage 4 жыл бұрын

They do say that towards the end.

@caseygreyson4178 4 жыл бұрын

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

@bergerniklas6647 4 жыл бұрын

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

@ElZafro_ 4 жыл бұрын

Yep, 7:10 he says it

@Syrange13 4 жыл бұрын

How? Can someone explain please?

@y1521t21b5 4 жыл бұрын

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

@codyheiner3636 4 жыл бұрын

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.

@Mike-lx9qn Жыл бұрын

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

@Mystery_Biscuits 4 жыл бұрын

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

@julianrosenfeld7177 4 жыл бұрын

If a relatively simple probability puzzle like this is so easily misinterpreted, no wonder it can be so easy for most people to misinterpret statistics where there are more variables and the stakes are higher. I think statistics and probability should be a mandatory course for all schools. It’s just too important for it to be misunderstood by so many people, which is dangerous as we have definitely seen as of late.

@doublenegation7923 4 жыл бұрын

No joke, I'm 13 and figured it out like seconds upon the riddle itself was presented. It's easy Lol.

@julianrosenfeld7177 4 жыл бұрын

conan gray yes me as well. I immediately recognized it as being similar to the Monty hall problem and it wasn’t too difficult for me to just reason through it intuitively. Unfortunately many people don’t know the first thing about probability.

@philjamieson5572 4 жыл бұрын

I enjoyed this. Thanks.

@user-iy7yy8er3r Ай бұрын

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

@cuteypetz 4 жыл бұрын

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 жыл бұрын

same

@xn85d2 Жыл бұрын

The reason the probability changes is that you've taken a ball out of the bag 'randomly' but then thrown away every occasion where you pulled out the green ball. In effect, you're pretending that the green ball being removed from the bag didn't happen but still using the probabilities including the fact that it did. So if you had put the red ball in, deliberately picked a red ball and taken it back out, the probability would not change at all, which is what people would think intuitively.

@Kamis2244 Жыл бұрын

It's like giving a input to the system (you throw a ball with known colour) and then (you take the same colour ball) feedback is obtained giving you more certanity about what can happen (after this perturbation we know with 66,(6)% chance there is a ball with exact colour as our input).

@LucaCrosta 4 жыл бұрын

Define p(0) as the initial probability of the ball inside being red, repeat the same process n times (put inside the red ball, randomly extract a ball which results being red). The probability p(n) of the ball still inside at this point being red, can be written as p(n)=(p(0)*2^n)/(1+p(0)*(2^n -1)). For the dimostration: the hard part is finding that (2^n)/(1+2^n) is a solution for p(n+1)=2*p(n)/(1+p(n))

@LoaFrz 4 жыл бұрын

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!

@Lodrik_Bardric 4 жыл бұрын

IT would have been nice to show that you can repeat this and every time you take out randomly a red Ball you go on reducing the probability of a green ball remaining inside, but you'll never get it to zero. 1 pick is 1/3, 2 picks is 1/5, 3 picks is 1/9, and so on... That means the probability of n picks of a red ball for a green ball remaining is 1/(2n+1)

@ca-ke9493 4 жыл бұрын

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.

@theRealPlaidRabbit 4 жыл бұрын

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

@ashtonlewis4814 4 жыл бұрын

An excerpt from another of Alex's books, "Can You Solve My Problems?" It's packed with age-old problems concerning all sorts of things! - A napkin ring is the object that remains after a sphere has been you drill a cylindrical hole through a sphere, where the center of the hole passes through the center of the sphere. - A certain napkin ring is 6 cm deep (i.e. the height of the remaining shape is 6 inches). **What is its volume?**

4 жыл бұрын

What I like to think to make this more intuitive for my mathematically challenged brain is: "Well, if there actually IS a green ball in the bag, isn't it strange that I keep pulling out only red ones?" so each time you randomly pick a red one it becomes less and less likely that there is a green ball there.

@ChrisStavros 2 жыл бұрын

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.

@leif1075 4 жыл бұрын

Can you please tell me who wrote that lovely white rabbit run interlude psychedelic music at around 4:41? Thanks. It's nice.

@kikones34 4 жыл бұрын

I think that ultimately the only way to make sense of this problem or any of its many variants is to understand conditional probabilities. If Rb = The original ball in the bag was red, and Rt = The ball he took out is red, the problem is asking the question "What's the probability of Rb given Rt?". The "given Rt" part is limiting the possibilities to a particular universe in which Rt has happened, it's removing the uncertainty of Rt by making it a known outcome. So, while P(Rb) is still 0.5 no matter what, P(Rb | Rt) incorporates the information that Rt has occurred. After understanding this, you can either think it through or use Bayes' formula to solve it.

@ArieteArmsRAMLITE 4 жыл бұрын

I met him in a pub in Camden once.

@kevinbahr5954 4 жыл бұрын

I’m sure someone else has said this, but this is a great situation to apply Bayes’ Theorem. If we take the ball in the bag being red as “Event A” and pulling out a red ball is “Event B”: The original probability of a red ball in the bag P(a)=0.5 The probability of removing a red ball given that the remaining ball in the bag is red P(b|a)=1.0 The total probability of removing a red ball is the sum of the probability of pulling out red if the original ball was red, plus the probability of pulling out red if the original ball was green. P(b)=(0.5*1.0)+(0.5*0.5)=0.75 And apply Bayes’ Theorem: P(a|b)=P(b|a)*P(a)/P(b)=1.0*0.5/0.75=0.667

@cauchym9883 4 жыл бұрын

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?