@@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.
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.
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.
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)
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.
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.
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.
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
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".
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.
@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. 😛
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.
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
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.
"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
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
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.
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).
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.
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
@@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)
@@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."
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?'
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?
@@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 ;)
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 ;)
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 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.
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.
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?
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.
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).
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.
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?
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.
4:00 If I ignore that you seem to actually be doing this randomly and not cheating, and that removing a non-red ball would ruin the prop value (which together mean that the other ball is also red), 2/3. And then 4/5, and 8/9, and 16/17ths, if you keep doing it.
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.
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.
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.
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)
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).
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.
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!
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!
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
03:34 For those who are confused. People may question "what if we pull out a green ball?" It's actually obvious: If we pull out the green ball, there is 100% probability the ball inside the bag is red, as the red was the one we've put into the bag. That's why the puzzle statement mentiones ONLY pulling out the red ball.
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!
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.
It's almost easier to think of the case of many repetitions. If you keep putting in a red and then randomly pulling out a red, eventually you accept that there must definitely be two red and zero green. Therefore the probability of 2 red must be going up each time, not staying 50:50
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.
Yeah...went off the rails there a bit at the end. It was pretty easy to determine that putting a red ball in and pulling a red ball out meant that the other ball only had a 33% chance of being green, but that absolutely doesn't mean that putting the previously drawn ball in the bag (red to start with) and pulling out a random one would result in red being pulled out 3/4 of the time ( 8:27 ). If the other ball does turn out to be green, then randomly drawing a significant number of times would come close to 1/2 red and 1/2 green (of course, truly random could potentially result in every draw being red, even with a green ball, but, over a significant amount, the probability should balance it out if it isn't being tampered with). And, obviously, if the other ball turns out to be red, red will be drawn 100% of the time (although, while you can find out the other ball is green, by drawing a green ball, you can never actually state the other ball isn't green, regardless of how many reds you draw, when you only see one random ball at a time; it simply lowers the chance of it being green for each red draw).
3:31 I don't understand how it is surprising. It obviously changes the probability. It is not that you put a red ball in and take a red ball out. You don't know whether you took the old ball or the red one you just put. It should easily be intuitive that the probability changes, but maybe not that intuitive to what it changes to
@@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
@@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.
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))
In my first maths test at uni xxx years ago, we had this question: You've got three cards. The first is white on both sides, the second is white on one side and red on the other and the third is red on both sides. You pick a card at random and look at one side, which is red. What's the probablilty the other side is red as well? Yet abother version of Monty Hall,. I remember I struggled some but finally got it right.
I think the easiest way to solve this problem is, that instead of seeing it as having three cards, you have six sides, all equally probable from start. When you look at your card and it is red the probability for one of the white sides go to zero while the probability for each read side becomes 1/3. For two of the red sides the other side is also red , thus the answer for the question is 2/3.
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.
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.
It isn't excluding the remove-green case that "increases the probability" - on the contrary, if you just remove one ball ignoring the color, the remaining one is 75% likely to be red. I'm not sure why this problem would seem counterintuitive in any case. One way to think about it is that inserting the red ball increases the "redness" of the possible outcomes by more than removing a red (which may or may not be the newly inserted one) decreases it.
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.
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.
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?**
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.
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? 🤔
So, the math seems ok. It’s the words I think you are having trouble with. If you have a bag with a ball that can be either red or green and you add a red ball, then the probability of randomly removing a red ball is 2 out of 3. That’s because you are choosing from either (R,R) or (R, G). No problem: 2/3. But once you remove a red ball, that’s done. When you then go to remove another ball you will be choosing from (_,R) or (_,G). There are now only two choices. That the odds of choosing a red ball used to be 2/3 is of no consequence. You used the wrong words. The chance of the second ball NOW being red is 1/2.
I think this is actually the opposite of the Monty Hall problem in terms of the error in reasoning that leads to the intuitive choice being wrong. In the MH problem, most people update their prior after being shown a goat, despite this providing no new information about the original choice. (If the host would randomly choose which of the remaining 2 doors to open, at times showing a car, then seeing a goat should lead one to update their prior about the original choice to 50%). Here the problem is the opposite. Seeing a red ball pulled out the bag is a signal which should lead to Bayesian updating: you should revise your prior from 1/2 to 2/3. So in the MH problem, people are paying attention to a signal when they shouldn't be, here they're ignoring a signal that they should pay attention to.
4:36 My intuition is that the chance of the ball being green after doing this is 25% (or possibly 1/3 ). Because half the time it's 100% chance of being red, and the other half it's 50% and 150%/2 is 75%. Or something like that.
I thought the exact same thing. Oops. At least it being higher than 50% was immediately clear to me this time, when I first encountered the Monty Hall problem it was a lot more confusing.
Having watched to 3:07, here's my guess. If you assume that there's equal chance of pulling the 100% or the 50% ball out, then there's a 50% chance of there being a 50% chance of the remaining ball being red, plus a 50% chance of a 100% chance of the remaining ball being red. 50% * 50% + 50% * 100% = 75% that the remaining ball is red. Of course, if you pull a green ball out (and are allowed to look at it), you know there's a 100% chance that the remaining ball is red. Edit: I misunderstood the question. Taking a green ball out is not allowed.
There is no change in the probability. It does not change from 50% to 66% percent. The first situation is that the single ball in the bag is either red or blue, 50% either way. That is the initial ASSERTION which we presume to be true. But after a red ball is placed in the bag and a red ball is removed from the bag, we do not know whether what's left in the bag is the same ball that was originally in the bag. So the question becomes, What is the probability that the ball left in the bag is the SAME ball that was originally in the bag? THAT probability must be incorporated into the calculation of the probability of that remaining (not necessarily the original) ball in the bag being red.
So if you do this and get a red- and put it back in and pull out a random ball and again its red a 2nd time- does the probability of there being a green in there go down even more? Im guessing it does.
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
4 года назад+1
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.
I understand why Brady calls it a “sleight of hand” (because the problem provides additional information that is not random), but really want to emphasize seeing results and working out the probabilities back is *how it usually works in real life.* Not everything in science is set up so you know the probabilities beforehand: more often than not, you have observations, some *partial* knowledge of the probabilities in the processes that lead to the observations, and it’s a matter of working backward to figure out how likely that the observations are so.
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).
The term _postselection_ refers to conditioning a probability space on a given event, turning an ordinary probability P(A) into a conditional probability P(A | B). In this video, they are _postselecting_ so that you only see the samples in which the ball taken out of the bag is red. It's fun to imagine a magical postselection button that, when pressed, retroactively _destroys_ _the_ _timeline_ in which it was pressed. If you decide ahead of time that you will press the button any time you draw a green ball, then you can randomly sample from the bag and always get a red ball, because the timelines in which you get a green ball don't exist. Actually, it would be more correct to say that you would never see the button get pressed... but that might also be because you draw a green ball and then decide not to press the button after all, or draw a green ball and then have a heart attack before you can press the button, etc.
So for each time you input and take out it the probability is 1/Xvn = 1/(2(Xv(n-1)-1)+1) = 1/ I'm using v to say subscript as the opposite of ^ Xvn is the probability on try n. n is the number of inputs/outputs. Xv1=3 Xv0=2
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.
P = 1 / (2^T + 1) T = times you take the red ball out P = probability that the ball left in is green In fact, the general form is: P = 1 / ((1/G-1)*2^T + 1) G = first probability of the ball being green (in the case of this video, G is 0.5)
i imagine if i repeat this process -- every time i put a red one in, a red one comes out then the chance of the bag containing a green ball tends to 0 so if the bag starts with 50% green ball & ends with 0% green ball then doing this process once will make the chance lies somewhere in between 50% to 0%
Would I be right in saying that if you don’t look at the colour of the ball you draw at random, the probability that the remaining ball is red rises to 0.75? Only when you look (and see it’s red), does the probability then fall to 0.67. 😊
Brady: "Are you having unholy thoughts?"
Alex: *nervous laughter*
And then it shows him in bed with Lewis Carroll...
Ahahahaha
A way of dealing with impure thoughts in bed? Nowadays we call it 'incognito mode'
i bet one of these problems gotta be, how many armies could i fit in in between my top tier tooths? XD
@@Albimar17 Ten Roman Legions ⚔
Therapist: "Are you having any sceptical thoughts?"
Lewis Carroll: *Takes drag from cigarette* "All I have are sceptical thoughts."
Why did they misspell the word "skeptical"?
@@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.
It can be spelled like that
esotericVideos jokes
@@billywhizz09 I'm sckeptical about that.
Gf: He’s probably thinking about other girls.
Bf: *_Pillow Problems_*
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 .
Unholy Thoughts
Body pillow
??.
"Before you get to bed, you have sceptical, blasphemous, and unholy thoughts"
My entire day: *Well actually...*
I do that at least three times a week using my own ball bag
@Richard Groller Alice wants BoB
@@hamiltonianpathondodecahed5236 I appreciate
I mean that's technically still before bed
@@hamiltonianpathondodecahed5236 Bob's your uncle.
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.
pafnutiytheartist Chebychev is that you?
Totally agree!
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.
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.
@@Seb135-e1i don't bring infinity into this!
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.
I don't know what your talking about but I agree with you wholeheartedly because you must be smarter than me.
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)
@@nathanbell6962 if you're interested, I'd suggest to you the 3blue1brown video(s) on Bayes' theorem. Amazing.
Also those probabilities of red remaining, after each additional successful drawing of a red ball, are (2^[n])/(2^[n]+1)
@@Elmaxo1989 Yep! My favorite part is that this works with n=0 as well
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.
This episode of Numberphile sponsored by Red Ball energy drinks
Took me a seconds.
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.
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.
It was yes.
@@nathaniliescu4597 replying a reply
Reply paradox
i'm amazed at how badly i wanna see a green ball right now. this is a weird feeling
truth
Is that because you're a dog ?
Happening
saaame
yess i was waiting for him to take a green out
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.
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
Yes, it's conditional probability at work here.
Yeah that's how I did it, seemed like an obvious case of bayes' theorem
Thanks! This explains more than the video tbh
this isn’t bayes theorem /:
you’re just using the definition of conditional probability; bayes theorem is a consequence of this, not vice versa
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".
If you're red-green colorblind, does that make it 100% probability?
yes
No, then the ball just has a hidden property
😂
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.
If you have dyscalculia it's probably 109%
>"I know exactly the time it was invented."
>proceeds to give the date but not the time 😝
Probably before #72 (invented the same night but crazier).
@M N Well, more accurate than that because it was at Carroll's bedtime.
@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. 😛
pfffft, bedtime obviously
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.
I still wonder if he even has a green ball there. 🤔
@@gustavgnoettgen If he did, it would probably be a fuzzy tennis ball so he could know not to pick it by texture alone.
@@gustavgnoettgen I think it might have been blue.
@@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.
@@gustavgnoettgen he had one green ball in his bag... unfortunately it was removed by way of orchidectomy😳
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.
Pillow problems are the mathematical equivalent of the brain talking meme template.
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
As well as shower thoughts
someone please make it
false.
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.
"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
I was thinking to myself how this was similar to the Monte Hall problem before Alex brought it up!
Same here. It's a different version of Monte Hall
Yes exactly
Yeah, this very thing was mentioned in the movie "21" and they called it variable change. I immediately thought of the movie.
Was actually coming to the comments to ask if this wasn't the same, then he mentioned it
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
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.
I was surprised they didn't mention this in the video.
what about monty python?
Doing math to avoid unwanted thoughts is relateable.
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).
That is the conclusion I got to and I was honestly looking for a comment to confirm my theory
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.
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
@@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)
@@wolffang21burgers Your wording was perfect
The animator is on fire with this one
Me at the end of the video:
NOW TELL ME: IS THERE A GREEN BALL?
aah its 50/50 ;-)
6:44 you can see the red bleed through.
@@recklessroges -- No, he said it's 2/3%
@@bokkenka You mean 2/3 or approximately 67%. 2/3% is 0.006666 repeating, or approximately 0.67%. Very different.
@@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."
"Are you having any unholy thoughts?"
"All I have are unholy thoughts"
"Is it just me, or is it getting unholier out there?"
🤡
I didn't know Michael Sheen was into puzzles
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?'
That was deduced by Dodgson in 1893, but he made several assumptions. Obviously nothing can be deduced otherwise.
@@MushookieMan Yes, he was obviously making a joke. 😆
Between that and the Alice books, I get the impression he had an incredibly offbeat sense of humour 😂
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?
Normal people: money, relationship, etc problems
Mathematicians: pillow problems
Software Engineers: Javascript
I hate when I think about et cetera problems...
@@julienbongars4287 You consider Javascript "software engineering"? Bah!
@@olmostgudinaf8100 r/gatekeeping
@@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 ;)
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.
Imagine being so legendary that you ponder complex mathematical issues to stifle your dark wandering mind.
He was a well respected Mathematician is his own right, he was a lecturer at Oxford
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 ;)
It's not that complex really.
Yeah if i was thinking of diddling kids, I'd probably turn to riddles as well
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."
Or he's lying to us about "randomly selecting" the ball
@@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.
@@maxberan3897 Oh, noooo! Are you sure?!
Or the other ball was actually yellow and he's fooling with all of us.
actually, he might have held on to the red ball while his hand was inside the bag
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.
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?
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.
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.
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).
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.
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?
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
I'll have you know that all those rabbits during the thinking time had me thinking many many unholy thoughts.
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.
it the "Randomly selected" interaction that changes the probability.
4:00 If I ignore that you seem to actually be doing this randomly and not cheating, and that removing a non-red ball would ruin the prop value (which together mean that the other ball is also red), 2/3. And then 4/5, and 8/9, and 16/17ths, if you keep doing it.
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
Excellent. A fellow man of culture
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
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.
Mathematicians: pillow problems
Me: I want to watch the green ball!
lol, there must a be a poodle somewhere amongst your ancestors
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.
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.
>This is very similar to the mounty hall problem I think. If only the mentioned that in the video
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)
Is it also a blasphemous thought to imagine a sceptic thinker in bed with Lewis Carroll?
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).
Finally, a perfect video, my-2 a.m-watch list will be legendary
Almost there. Reading this comment at 1h07 in the morning!
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.
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!
Except the answer was 1/3. He was asking the probability of it being green.
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!
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
I dont know why, but I really want to drink RedBull right now :D
03:34 For those who are confused.
People may question "what if we pull out a green ball?"
It's actually obvious:
If we pull out the green ball, there is 100% probability the ball inside the bag is red, as the red was the one we've put into the bag.
That's why the puzzle statement mentiones ONLY pulling out the red ball.
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
Oh f***, didn't expect to see you here XD
@@alanwolf313 it's not the og one
If you had a mustache, perhaps you would have remembered sooner.
It's similar to the Monty Hall problem, I believe.
@@casualbeluga2724 I know, but i still see him in a lot videos
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!
Good video. I do wish he had expanded the discussion to repeated sampling.
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.
It's almost easier to think of the case of many repetitions. If you keep putting in a red and then randomly pulling out a red, eventually you accept that there must definitely be two red and zero green. Therefore the probability of 2 red must be going up each time, not staying 50:50
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!
Maybe because a lot of it is that complicated logic stuff. :P
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.
6:30
67%
Yeah...went off the rails there a bit at the end. It was pretty easy to determine that putting a red ball in and pulling a red ball out meant that the other ball only had a 33% chance of being green, but that absolutely doesn't mean that putting the previously drawn ball in the bag (red to start with) and pulling out a random one would result in red being pulled out 3/4 of the time ( 8:27 ). If the other ball does turn out to be green, then randomly drawing a significant number of times would come close to 1/2 red and 1/2 green (of course, truly random could potentially result in every draw being red, even with a green ball, but, over a significant amount, the probability should balance it out if it isn't being tampered with). And, obviously, if the other ball turns out to be red, red will be drawn 100% of the time (although, while you can find out the other ball is green, by drawing a green ball, you can never actually state the other ball isn't green, regardless of how many reds you draw, when you only see one random ball at a time; it simply lowers the chance of it being green for each red draw).
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
He has several other channels, but there's one called Objectivity where he is frequently on camera so he can interact with the objects
Then you dont watch the videos till the end, cuz he nearly always sponsors something at the end.
Welcome, you must be new to channel then.
3:31 I don't understand how it is surprising. It obviously changes the probability. It is not that you put a red ball in and take a red ball out. You don't know whether you took the old ball or the red one you just put. It should easily be intuitive that the probability changes, but maybe not that intuitive to what it changes to
This is literally a basic application of Bayes’ Rule
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
mina86 yeah, eazy peazy lemon squeazy
It took me 10 seconds and I’d never even heard of Bayes’ Rule.
@@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
@@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.
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))
It is unfortunately 6am here, and I really should get some rest.
In my first maths test at uni xxx years ago, we had this question:
You've got three cards. The first is white on both sides, the second is white on one side and red on the other and the third is red on both sides. You pick a card at random and look at one side, which is red. What's the probablilty the other side is red as well?
Yet abother version of Monty Hall,. I remember I struggled some but finally got it right.
Isn't that just... 1/3? The question is asking what the probability of picking the card with two red sides is, right?
2/3. It's like the exact same problem as the one in the video.
is it 50%?
I think the easiest way to solve this problem is, that instead of seeing it as having three cards, you have six sides, all equally probable from start. When you look at your card and it is red the probability for one of the white sides go to zero while the probability for each read side becomes 1/3. For two of the red sides the other side is also red , thus the answer for the question is 2/3.
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.
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.
"The surprising thing is that it changes the probability."
The surprising thing is finding that surprising.
It isn't excluding the remove-green case that "increases the probability" - on the contrary, if you just remove one ball ignoring the color, the remaining one is 75% likely to be red.
I'm not sure why this problem would seem counterintuitive in any case. One way to think about it is that inserting the red ball increases the "redness" of the possible outcomes by more than removing a red (which may or may not be the newly inserted one) decreases it.
I just draw trees when dealing with probabilities.
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.
No joke, I'm 13 and figured it out like seconds upon the riddle itself was presented. It's easy Lol.
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.
0:16 What you revealed was a date. Still waiting for the actual time ;-)
8:15 Exactly randomly! It gives us information on the distribution.
"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.
one red Doraemon in one red Doraemon out
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?**
0:08 “But before that, I want to tell you about...”
TODAY’S SPONSOR: RAID: SHADOW LEGENDS
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.
Soution:
-Be colorblind
-whatever you pick, you won't be able to see the difference so you go to sleep peacefully
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? 🤔
same
This problem is equivalent to the Monty Hall problem, isn't it?
They do say that towards the end.
Yuval Nehemia if you watched the full video, you would realize that they say this was the original inspiration behind that problem.
I mean he mentions it in the Video, so yeah...
Yep, 7:10 he says it
How? Can someone explain please?
So, the math seems ok. It’s the words I think you are having trouble with.
If you have a bag with a ball that can be either red or green and you add a red ball, then the probability of randomly removing a red ball is 2 out of 3. That’s because you are choosing from either (R,R) or (R, G). No problem: 2/3.
But once you remove a red ball, that’s done. When you then go to remove another ball you will be choosing from (_,R) or (_,G). There are now only two choices. That the odds of choosing a red ball used to be 2/3 is of no consequence. You used the wrong words. The chance of the second ball NOW being red is 1/2.
came to the same conclusion, i wrote a little program that simulates the riddle 10 million times, it always results in about 50% red
I think this is actually the opposite of the Monty Hall problem in terms of the error in reasoning that leads to the intuitive choice being wrong. In the MH problem, most people update their prior after being shown a goat, despite this providing no new information about the original choice. (If the host would randomly choose which of the remaining 2 doors to open, at times showing a car, then seeing a goat should lead one to update their prior about the original choice to 50%). Here the problem is the opposite. Seeing a red ball pulled out the bag is a signal which should lead to Bayesian updating: you should revise your prior from 1/2 to 2/3. So in the MH problem, people are paying attention to a signal when they shouldn't be, here they're ignoring a signal that they should pay attention to.
4:36 My intuition is that the chance of the ball being green after doing this is 25%
(or possibly 1/3 ).
Because half the time it's 100% chance of being red, and the other half it's 50% and 150%/2 is 75%. Or something like that.
Yes, I knew that 1/3 would somehow show up...
I thought the exact same thing. Oops. At least it being higher than 50% was immediately clear to me this time, when I first encountered the Monty Hall problem it was a lot more confusing.
Having watched to 3:07, here's my guess. If you assume that there's equal chance of pulling the 100% or the 50% ball out, then there's a 50% chance of there being a 50% chance of the remaining ball being red, plus a 50% chance of a 100% chance of the remaining ball being red. 50% * 50% + 50% * 100% = 75% that the remaining ball is red. Of course, if you pull a green ball out (and are allowed to look at it), you know there's a 100% chance that the remaining ball is red.
Edit: I misunderstood the question. Taking a green ball out is not allowed.
There is no change in the probability. It does not change from 50% to 66% percent. The first situation is that the single ball in the bag is either red or blue, 50% either way. That is the initial ASSERTION which we presume to be true. But after a red ball is placed in the bag and a red ball is removed from the bag, we do not know whether what's left in the bag is the same ball that was originally in the bag. So the question becomes, What is the probability that the ball left in the bag is the SAME ball that was originally in the bag? THAT probability must be incorporated into the calculation of the probability of that remaining (not necessarily the original) ball in the bag being red.
So if you do this and get a red- and put it back in and pull out a random ball and again its red a 2nd time- does the probability of there being a green in there go down even more?
Im guessing it does.
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
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.
I understand why Brady calls it a “sleight of hand” (because the problem provides additional information that is not random), but really want to emphasize seeing results and working out the probabilities back is *how it usually works in real life.* Not everything in science is set up so you know the probabilities beforehand: more often than not, you have observations, some *partial* knowledge of the probabilities in the processes that lead to the observations, and it’s a matter of working backward to figure out how likely that the observations are so.
Alex: The probability is equal of it being red or green.
Shrodinger: The ball is both red and green.
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).
The term _postselection_ refers to conditioning a probability space on a given event, turning an ordinary probability P(A) into a conditional probability P(A | B). In this video, they are _postselecting_ so that you only see the samples in which the ball taken out of the bag is red.
It's fun to imagine a magical postselection button that, when pressed, retroactively _destroys_ _the_ _timeline_ in which it was pressed. If you decide ahead of time that you will press the button any time you draw a green ball, then you can randomly sample from the bag and always get a red ball, because the timelines in which you get a green ball don't exist. Actually, it would be more correct to say that you would never see the button get pressed... but that might also be because you draw a green ball and then decide not to press the button after all, or draw a green ball and then have a heart attack before you can press the button, etc.
Imagine finding bugs by collapsing timelines where a piece of code doesn't crash
So for each time you input and take out it the probability is 1/Xvn = 1/(2(Xv(n-1)-1)+1) = 1/
I'm using v to say subscript as the opposite of ^
Xvn is the probability on try n.
n is the number of inputs/outputs.
Xv1=3
Xv0=2
That leads to some strange conlcusions like Xv-1 = 1.5
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.
P = 1 / (2^T + 1)
T = times you take the red ball out
P = probability that the ball left in is green
In fact, the general form is:
P = 1 / ((1/G-1)*2^T + 1)
G = first probability of the ball being green (in the case of this video, G is 0.5)
i imagine if i repeat this process -- every time i put a red one in, a red one comes out
then the chance of the bag containing a green ball tends to 0
so if the bag starts with 50% green ball & ends with 0% green ball
then doing this process once will make the chance lies somewhere in between 50% to 0%
yeah, the green ball odds go like 1/2, 1/3, 1/5, 1/9...
Would I be right in saying that if you don’t look at the colour of the ball you draw at random, the probability that the remaining ball is red rises to 0.75? Only when you look (and see it’s red), does the probability then fall to 0.67. 😊