That is because you are looking at the problem from a single person perspective. The odds you are imagining is if you pick one person and ask what is the probability that somebody in the room matches THEIR birthday. That is a different question than if ANY two people in the room have the same birthday. The same fallacy arises in the analysis of DNA evidence. An analysis usually only looks at a few dozen markers in the DNA. From this DNA matches usually quoted to have a really low number of people matching your DNA. That's true; but, it isn't the complete picture. If you instead consider how many people in large population will have matches, the number is quite a bit higher.
It's not counterintuitive, you're just thinking about it wrong. Think about it in terms of graph theory, where there is a fully connected graph with each person as a node. The probabilities that a birthday will be shared is proportional to the edge count in this graph.
This is kind of like the infinite sum that adds up to -1/12. At no point in a finite summation would it be equal to that, but we ASSIGN the value to it. So I think it's kind of questionable to say that it EQUALS said value.
Don’t worry, the whole point is that p-adic numbers don’t agree with the standard number system, it is a completely different set of rules. In our normal rules that sum would in fact equal infinity, which makes the statements given there misleading.
@@juliusjakas6858 When it comes to calculating machines there is no such thing as infinite. (Okay IEEE-754 does have infinities; but, that is just a signal and not useful for actual calculation.) A computer will never be able to do an infinite calculation because it will always have a finite amount of memory. This limitation can be used as an advantage. The hardware necessary to subtract numbers is also more complicated than the hardware necessary to add numbers. Adding such functionality to an adding machine or a microprocessor would take up space and require more material -- so lots of adding machines and lots of microprocessor's neglect having any subtraction hardware at all. The solution is to leverage the fixed word size of a microprocessor or the fixed number of digits of the adding machine along with the modular rollover when the size of that fix number is exceeded. Adding machine operators could do subtraction by adding ten's complement numbers and computers can subtraction by adding two's complement numbers. For a four digit ten's complement adding machine, the number -1 would be represented by 9999. If I wanted to subtract 1 from 5309, I would add the 5309 + 9999 to get (1)5308. The one that would normally be carried over to the 5th digit is simply discarded because there is no hardware to store the 5th digit. So the correct result displayed is 5308. So the joke is that if the universe is being simulated inside a computer, then any attempt to use an infinite number results in a rollover leading to glitchy results inside of the universe. Imagine a kid put ten toys into a bag and then took them out again one by one. But after he had already taken all ten toys out, he attempts to take out another toy and instead of there being no toys he takes one out and there are ...9999 toys left in the bag afterward because whoever wrote the simulation forgot to check for underruns of the counter determining how many toys were in the bag. We will not even discuss what would happen if the counter was used to index an array of toys in the bag... human sacrifice, dogs and cats living together, mass hysteria.
...9999 isn't equal to -1 because the two are different types of numbers Using p-adic numbers with real numbers allows you to falsely “prove” any ...nnnn = -1 conclusion, where “n” stands for a single digit. For example in the 2-adic system where numbers are written with the digits 0 and 1 (e.g. 10 = 2, 11 = 3, 100 = 4) you could show: ...1111 + 1 = 0 => ...1111 = -1 This too would be false because p-adic numbers don’t follow real number arithmetic. Rather, the problem should be written as: ...9999 + ...0001 = ...0000 ...0000 != 0
The way you actually solve this problem is by saying that the p-adic numbers are different from the real and complex numbers. Zero is just defined as the additive identity of a set, so ...0000 = 0. Similarly the p-adic numbers are allowed to contain one. The statement ...1111 = -1 is true in the 2-adic numbers. It's not true in the other p-adic numbers and it doesn't make sense in the reals.
THANK YOU, its clickbait for people who don't know what p-adic numbers are. A more sensible answer would be 100000..., but really it just makes no sense in the context of real numbers
I don't really understand, why would the dry mass stay the same weight? It's doubled now, so would it not be 2kg? then again, I'm currently brain dead tired so idk
@@EvanUnknownas dry mass they mean the weight the potatoes would have with no water. the total weight is always the dry mass + the water. The real statement is to find the total weight so that the total water is 98% of the total weight, not that the potatoes are 98% hydrated. It's a bit confusing (because wrongly stated) and it plays on that to look counterintuitive
not only did you give a short introduction of proof of 0.9…=1 in just shy of 30 second, but you also explained in the RIGOROUS manner, I am impressed already.
@@nutronstar45pretty sure something simple can work. Assume we have 0. To get a number with one digit after the decimal point we have to add a number from 1 to 9 and multiply by 0.1. To get as close to 1 as possible we have to use the number 9. Now we have 0.9, 0
Absolutely beautiful how you just get to the point without wasting everyones time. If more youtubers were like that the plattform would be so much better
I feel like it’s a little too brief, but that’s only because I’m very interested in how the weird probabilities actually work, but I also realize that those explanations might require higher levels of math.
i mean, you’re not *wrong* in the sense that there is surely stuff we don’t know, but all of this is super consistent with math. there are no contradictions here (in the mathematical sense).
With the concept of infinity and probability there’s many different ways people could look at it, I do believe there is only one correct answer for everything, but that’s assuming we are all answering the same question, but since he’s missing some explanations I’m not sure what view he’s looking at it from exactly.
@@penguincute3564 Hexadecimal or decimal, all the numbers in computer are stored as binary. For 0xffffffff, all 32 bits in the memory is 1. If you interpret it as signed integer, then it is -1.
only 300 views? before i saw the views i really thought it has hundreds of views because the animations are just like those other good math videos with a lot of views... how can this channel also only has less than 2k subs??? it deserves more than hundreds of thousands...
@@xazii There are only 366 possible birthdays. With 366 people you may or may not have at least two sharing a birthday, but when you add person 367 there must be at least two people sharing a birthday according to the pigeonhole principle.
@@xazii With a meetup of everyone with the same birthday, I think it's safe to say that they will have the same birthday 💀 I understand the trip up, but this counter example is hilarious because literally everyone would have the same birthday.
@@YT7mc yea i totally get that hahha, I prolly didnt thoroughly read the comment when I wrote that, I probably thought he said something else hahahah, that was some of the dumbest shit ive ever said tho
2:55 i also got 2/3 but i think the core of the problem is understanding the question. the question asks for the probability of the second ball is red WHEN the first ball is red. so the first ball can not be green this is different than a question saying: out of these 2 boxes, what is the possibility of getting two reds? then the probability wouldn't be independent of each other and would be 1/2 at least i think idk
I got messed up on this. If we choose a random ball and we know beforehand the ball has to be red then it is 2/3, but if you choose a random ball and it just happens to be red then it's 1/2. Its like with the Monty hall problem and the 3 doors. You switch only because you knew Monty couldn't choose the door with the prize. Though, I think I may have just repeated your answer.
Oh wait, no, I was wrong! Picking a random box then a random ball is the exact same as picking just a random ball. So if you got a red ball there's a 1/3 chance you're in the green ball box and a 2/3 chance you're in the red ball box. You have 4 equally probably events: Choose green in box A Choose red in box A Choose red in box B Choose other red in box B Doesn't matter if the green ball could be picked or not (unless both boxes were equally probable so the red ball in box A was 50% and not 25%)
The probability problem is all based on the wording, both 50% and 66% are correct depending on the wording. "Let's take the ball out of the random box, the ball is red. What is the probability that the 2nd ball in this box is also red?" This kind of setup leaves the question open to ambiguity around whether or not the ball was chosen randomly, or if just the box was. "the random box" implies the box is what was random, and 'take the ball' implies it was specific. Assuming it's all random, there a 50% chance to pick A or B. A red ball from A is 50%*50%=25%. A red from box B is 50%*100%=50%. A 75% chance total to pick a red, 25% from A, 50% from b, or 2/3, 66% that you picked box B, which will result in double reds. However, if you assume just the box was random and we chose to take the red ball first. The ball left in the box would either be Red or Not. If this ball was picked from A, there is a 0% chance the remaining ball is red. If it was picked from B, there is a 100% chance the remaining ball is red. This results in a 50% chance that the ball was from A or B. This shows the importance of being extremely clear with and direct with probability. Most of the confusion in probability comes from wording, not the maths itself.
2/3 is actually never right for that problem. You ALWAYS have a red ball first, as specified in the problem, so your 25% from A 50% from B is inaccurate. It’s a given picked a red ball from box A if you at all picked from box A, so it ALWAYS has to be a green ball left. It’s a given you picked a red ball from box B if you picked from box B, so there will ALWAYS be a red ball left. The 2/3 only comes in if all balls are in one box, and you picked a red ball first. Then you now have 2/3 chance to pick another red ball, because two of the three balls remaining to be picked are red.
@@theeraphatsunthornwit6266 You have a situation where every time you look at this problem, the ball is red. It's a given. If using this form of thinking, the 50% would apply. There is also no clear distinction that the ball is random, only the box. I'm not saying I disagree with you, just that it's very important to make probability absolutely clear with no room for false interpretation, otherwise we end up with problems like this
@@theeraphatsunthornwit6266 well at least for me. I misinterpreted it exactly like in scenario two and was very confused about if I had made a mistake.
The most common result would be the mode, no? And I think it is $1. The median would be a value that splits the possibilities in half: Getting a bigger payout than the median or a smaller payout both have probabability >= 0.5
The last one is tricky because in the real world you cannot expect to win more than 100 million, so the actual value is more like 26€. After enough coin flips the lottery would just go bankrupt
@@shadowpenguin3482It will take a long time for the lottery to go bankrupt though, they can probably make it sustainable if the payment required is like greater than $5 and no one is sitting there playing 10 billion times
Did I just see the right triangle of combinations at 7:10 which also relates to creating an input counter in circuit maker 2?! Amazing! Now rotate the x and y values so that n is the height-map for each combination and include the xor values for if numbers are only single integers. 0 would be in its own plane because we are only counting 1's here.
I feel like the ball problem may have been under specified. If I assume fungibility of the balls (eg. I don't care if I have your $1 bill or my $1 bill it's $1) then when I return to the box I'll either see red or green: Scenario 1: Balls: RG Remaining: G Scenario 2: RR Remaining: R Options: RG Probability: 50% (frequency of R/total possibilities) The 66% case would kick in if balls were not interchangeable (eg. red ball 1 , red ball 2 and red ball 3 exist) or I have the ability to pick from the other box (eg. I can pick from Box 1 even though my first pick was Box 2). I may be off here but I feel like it depends on the parameters of the problem
Just because the two balls are the same color doesnt mean they are the same ball. There are three red balls each equally likely to be picked. Picking one of the two in the same box leads to an outcome where the other ball in the box is red.
@@michaelshaffer6717it only asked if the next ball would be red, not if it would be a specific red ball or anything. There are only two outcomes, red or green, and each outcome has a 50% chance of being true. 2/3 would only be true if the balls were all in the same box.
@@axelarnesson5066 But the odds about which box you picked from aren't 50-50, is the tricky bit. The fact you got a red ball means it's twice as likely you picked from the red-red box, because if you'd chosen from the red-green box, it's half likely you would have gotten a green ball. At first, there's a 50-50 chance you picked from either box, but getting a red ball is more information, which changes the odds. What you're thinking is that it's equally likely you picked from one of two boxes; what you really know is that it's equally likely you've found one of three red balls, and two of the red balls share a box with another red ball.
What you are missing is that if you randomly pick the GR box, you have 50% chance to get the wrong ball, and then you might try from the ither box Start : Randomly pick option 1 or 2 Option 1 : You randomly pick the RR box (50%), you get a R ball (100%) and the other is R. So 50% to get R Option 2 : You randomly pick the RG box (50%), then you have 50% to get the R one and see that the other is G (so a total of 25%), but also 50% to get the G one, then you go back to Start So you have 50% to get RR and the other is R, 25% to get RG and randomly pick the R one and the other one is G, but on the remaining 25% you restart with a non-zero probability to then pick the RR one and get a R. So it has to be more likely to get R as the other ball Try it yourself, get 2 dices (but you have to know which one you roll) and a coin then foip the coin to pick a dice. For the first dice, whatever you roll, you note 1. For the 2nd dice, you note 6 if and only if you get a 6. Otherwise you do the coin flip again. Now, i’m pretty sure that you see how it’s rigged. It’s exaclty the same thing for the 2 balls problem
They do and they frequently make bad decisions because they don't do the math. There are lots of psychological fallacies that lead to everything from bad purchases to burning money for bad investments to voting for the wrong leaders that result because of it.
The periodic numbers one isn't actually true, as it is actually just a semantic argument that relies entirely on the ways we write numbers rather than the actual value itself. Sure, if you add it by continuosly getting a zero and carrying the one, you will get an infinite amount of zeros, but you'll also never finish this calculation. It'd be like if you added 1 to a hundred digits of 9s, but stopped halfway and claimed it equaled 0. If you just finished the calculation, you'd eventually get a 1 in the first digit. Same with the infinite one, just that you can never finish the calculation. Not only that, but this trick only works because we use a base 10 system, which is an arbitrary choice. If you converted the infinite 9s number to hexadecimal and added 1 to it, you would no longer get an infinite amount of 0s and would instead get something entirely different which obviously does not equal -1. Instead, this trick would work with an infinite amount of Fs in hexadecimal.
I see what you mean but not at all the case, first they are called p-adic not periodic the p in p-adic stands for prime as these are usually used in a base with prime numbers so the number changes based on what base your in ...9999999 in 10-adic is different to ...9999 in hexadecimal. just as 10 in base 10 is different to 10 in hexadecimal. and no you wouldn't always take the result as true but certain parts of math have uses for these p-adic numbers and using these results
The p-adic numbers are just different mathematical objects. Veritasium did a video on them and showed how they provided a shortcut to an answer to a practical problem about the regular old real numbers that would be hard to find otherwise. I've never studied p-adics, but the idea is that you introduce a different metric on p-adics than the one you are used to so that a sequence like 9, 99, 999, 9999, 99999, ... actually has a limit just like the sequence 0.9, 0.99, 0.999, ... has a limit in the real numbers with the Euclidean metric.
@@oLouis53 @JustinBA007, you are both right in a sense. Whilst the actual value does indeed not equal -1 in any base for that matter if you consider it differently (like was done in the video) you can make an argument for it converging on something and it does have an actual use in mathematics. It simply depends on what rules you are using for your allowed operations. It's kind of like Fairey addition in that it isn't mathematically correct but still useful.
@@oLouis53ok, that makes sense, but that is not made clear at all in this video. The way it's presented in this video makes it sound like he is saying that an infinite number of all 9s is actually equal to -1, which is just not true in normal math. Also, not to mention that the chapter of the video is titled "periodic numbers," which only further confuses things.
This principal is actually fundamental to how computers perform arithmetic, namely how they handle subtraction by adding: Negative numbers in a computer are actually written this way, i.e. -1 is 0b11111111 for an 8-bit binary number. This means it's trivial to negate numbers (just flip all the bits and add 1) and you can just treat them like all other numbers. To give a decimal example, we can represent 1000 values with a three digit decimal number, 000 - 999. this effectively defines [501, 999] as [-499, -001] in a way that preserves addition and subtraction. For instance, 209 - 057 becomes 209 + 943, which is 1,152. Since we're stuck with three digit numbers, we drop that leading one and end up with 152 (i.e. 209 - 57)!
For last one it doesnt only matter how much money you are willing to pay but also how many times you are willing to play, looking at numbers best way is to play 1000 times and max money for that is 5,6$
The idea that an infinite string of 9s, written as "...999999999" (with an infinite number of 9s going off to the left), is equal to -1 is a speculative and controversial concept in number theory.
I learned way more interesting math than in school! You need more subs! You are so underrated! Please make more videos I really enjoy them! I just subbed to you and turned on all notifications!
This is incredible. It took my teacher 4 weeks to stumble over this and you nailed it in 8 minutes. I wish professors in universities would teach like you! Just amazing!
The additive persistence one is not that surprising, as the sum of the digits of a number has an additive persistance of that number -1, so for example the sum of the smallest number with an additive persistance of 3 (199) 's digits is just 19, the smallest number with an additive persistence of 2. This is true for 4 and 5, and i assume it's also true for the others as well.
I'm trying to get my head around the last one. I set up a simulation in excel where it randomly picks a 0 or a 1 (my head or tail), if it's a 1 then it pays out based on the number of 0s that came beforehand. I then add up the total winnings and divide by the number of games played and it does this a million times (well slightly more: the number of rows permitted in excel). The average win per game seems to settle at around 4-5, but there are spikes where it gets really big. There's every possibility that I'm not calculating this correctly, but also I am aware that each time I'm doing this I'm working out the average win for that particular set of games rather than what happens generally. There are a ridiculous number of ways to put 0s and 1s together when you have a non-fixed number of either of them, so I guess every now and then there might be a combination that averages vastly more than what I'm seeing here. The general trend is upwards though, but when I plot game# against average win so far I see big spikes (huge wins) followed by decays.
Yeah well basically you are never going to see the massive upside because the average wins per round grows at a logarithmic rate. You really have to look at it analytically: Expected value is the sum of all possible outcomes each multiplied by their payout. if i bet 2 dollars on a coin flip coming up heads, my expected value is (2*0.5)+(0*0.5), which is 1 dollar. Each possible outcome in our game, one more tails than the previous one, is a (1/2)^n chance of happening times a 2^n payout .Round one is a 50% chance of making 2 dollars, so that's an expected payout of 1. round two is a 25% chance of making 4 dollars, which is also one. Do you see what is happening? every round has an expected payout of 1 dollar, so when we sum them together we get a result of an infinite expected payout. Remember, the values he shows in the chart is average wins _per round_. how much money you would make is actually that times the number of rounds played. vsauce2 has a great video on this: czcams.com/video/RBf1s4TassI/video.html
I've also simulated this in python and for a million games played it only tends towards 6$ if I'm understanding the rules of the game correctly. Not sure what's wrong.
@@copalo343Sounds like you're potentially getting a similar result to me . Some of my outputs tended to around 6, but I got different results each time I ran the simulation because you do get different outcomes as every one of the 1 million coin flips I do was calculated each time I ran it. As I understand it, the longer it takes before you flip a win, the more money you get, but I'm not sure if we're considering what we ought to pay for each flip, or what we pay to flip the coin until it wins. It's possible that maybe what you need to do is consider a run of (e.g.) four losses and then a win to be one "game" and do that a million times rather than just do one million coin flips. You end up with a significantly larger dataset with a different number of coin flips each time. I'm not sure how to handle that in excel but I think I could do it in matlab as that can handle changing matrix sizes quite happily, plus I could log multiple results and see what sort of distribution the final outputs have. I have very little experience in python. This might be a task for tomorrow or wednesday.
@@copalo343I'll try to remember to do that if I'm successful. I'm a bit too tired to do it tonight (plus I need to install matlab for this computer) and I have a band rehearsal on tuesday night so I might not get chance to do much on it then. Realistically Wednesday is the earliest I'm going to get a chance to start work on it and I will definitely need some time to re-acquaint myself with matlab commands as I've not used it for a few years.
this is one of my favorite math things - facts that are super unintuitive but are still, well, facts. my favorite thing about math is how weird it can be while still being correct!!
The one with the red and green balls is like Betrands Box paradox, where that version has a 3rd box consisting of 2 green balls. Also wouldve been cool if you did the Multiplicative Persistance as well as Additive
and also kinda like the monty hall paradox, where you always change because monty hall will never reveal the door with the money, so if you change, there is 2/3 chance you win, if you dont, its 1/3, because its 1/3 chance you guessed it right first, and 2/3 you guessed wrong
@@atepomarok9339yeah your right, they did the math wrong, the probability of pulling a red ball after one red ball is 100%… not 200%, multiply, not add.
You're wrong on a couple of these, at least as your presented them. The way your providing the information is not the same as wherever you took these problems from.
@@williamwalsh4743 I typed out a whole explanation already, and then accidentally deleted it. The potatoe one seemed off, but I'm not sure about that one. The red ball green ball one is wrong fs. Without further specification, it makes most sense that you would be randomly choosing a box, and then randomly selecting a ball from within. The way he calculated assumed that you have an equal probability of drawing every ball from either box, which is not true.
@@maniacpwnageking1. The potato one is fine. The dry mass stays the same, so it should always have 1 kg, work from there. 2. The ball one isn’t wrong. He just said you randomly choose a red ball, why would you assume that he first randomly chooses a box? That’s your assumption which you have chosen to make.
Sometimes(p=1/4), you will get green ball and skip that. When you get red(p=3/4) you, obviously, have 1 outcome to get next green and 2 to get next red.
Why would you make implicit reference to the Archimedean property for showing 0.999 repeating is the same as 1 for the reals, and then immediately change to the 999... p-adic example, not even mentioning that the p-adics are a completely different set of numbers that don't have the Archimedean property?
I may be wrong, because I'm not the best at math, but in the first paradox (the one that explains that ...9999999 = -1) the problem here is that you're assuming that this number has an end (because then you can't add 1 using the arithmetic operations, in this case addition), which is impossible because you said it yourself, the numbers are infinite. Now, EVEN if this was correct (which, like I said, it's mostly not) the moment you get at the first numbers (which are 99999...) when you add the carried 1 to the nine at the most left, you get 100000... . However, like I said, this is probably impossible, because then this number would no longer be infinite, because in the last example I gave, there IS an end.
@@ctanimations8488actually after learning about p-adic numbers, it turns out ...999=-1 is TRUE. It's not made clear in the video, but p-adic numbers use an entirely different math system where addition operations like that in the video are valid. in the math system we're used to, it's not valid, but in the p-adic number system, it is valid.
@@shieldgenerator7 Thanks for the clarification, like I said, I'm not the best at math so this makes it much clearer. Maybe I'll investigate later about p-adic numbers, once again, thanks
Great vid! However, i don't get why there must be something between two numbers for these numbers to count as "different". Is that some axiom i'm not aware of?
Agreed. The point of saying greater or less than means you know the values of the numbers and you compare them. If I had a number line with just the integers then could I say that because I cannot fit a number between 4 and 5 then they are the same number? No. By this logic every number would have the same value.
This is a fun video and that's great but I do need to point out ...999 does not infact =-1 For anyone who is interested you cannot simply add one to an infinitely large number (well you actually can using ordinals but not in this case) thus to understand the value of such a number you would need to approximate it using something like an infinite series which you would find diverges to infinity.
The ball one doesn't make sense though. There are 3 outcomes but 2 of the outcomes are combined as 1 since they're both red balls. (Nvm I understand now)
I wouldn't say combined but I agree with you because the 3 outcomes which he mentioned (and thus used to conlude a probability of 1/3) are not equiprobable if he chose a random *box*
you may be combining them, but that is actually wrong. Imagine you also number the balls. Clearly, you have a 25% chance for pulling each ball. Keeping the numbers, there are 3 numbers that result in pulling a red (let's say 2,3,4). Each of these is equally likely, so 1/3 each. Now, of the set (2,3,4), how many are from box B?
Nvm I was wrong. Here is the python code that I wrote, feel free to point out any logical errors import random redcount = 0 greencount = 0 remainderoftrials = 0 for i in range(10000): initialboxchoice = random.randint(1,2) box1 = ['g','r'] box2 = ['r','r'] if initialboxchoice == 1: removed = box1.pop(random.randint(0,1)) chosenbox = box1 else: removed = box2.pop(random.randint(0,1)) chosenbox = box2 if removed != 'g': if chosenbox == ['r']: redcount+=1 elif chosenbox == ['g']: greencount+=1 else: print('ERROR') else: remainderoftrials+=1
print(f"Other ball was green: {greencount}") print(f"Other ball was red: {redcount}") print(f"Remainder of trials, i.e. those where the first ball was green: {remainderoftrials}") print(f"Experimental probability that the second ball in that box was also red: {redcount/(redcount+greencount)}") P.S. Ik that the code could be written much better, I just wanted to make super sure that it was error free Conclusion: Probability = 2/3
Hmm...no. The question ask.. what is the chance of another ball in the box you just chosen is red. It does not ask you to choose a ball from the remaining 3 balls.
Thanks for once again showing that infinity is not a number but rather an idea. In the real (finite) world, one wouldn't have enough time to continually play the "hope I don't flip heads yet" game for anything more than a few dollars.
Actually, the problem with the 2 boxes is misleading. The probability of either box is equal, therefore the probability that the other one is green is 50%. The other box has a 50% chance so there are 2 possibilities of being a red ball at 25%. So to sum up: Pulling Red ball from A: 50% Pulling Red ball A from B: 25% Pulling Red ball B from B: 25% Still 50/50
The probability of box 1 is 1/3 as it is said a green ball is never drawn first, and there are 3 total red balls. With 3 possible starts there is box 1 (1/3 chance) with ball 2 being green, box 2 (1/3 chance) with ball 2 being red, and box 2 (1/3 chance) with ball 2 being red. 2/3 chance of ball 2 being red, and 1/3 chance of ball 2 being green
The probability of picking either box is equal, but you have been told that a red ball has been selected. You can draw a probability tree to visualise it. There are four possible balls at the start and after you select the first red ball, there are 2 red balls and 1 green ball remaining so the probability of picking a red ball is 2/3 since you still don’t know which box you are in.
Edit: Nevermind, I misunderstood the problem. Original comment: I suppose it depends on whether we assume that we have to choose the box first and then the ball or that we go straight to choosing the ball. If we have to choose the box first: 1. box A (50%) --> red ball (100%) --> 50% × 100% = 50% 2. box B (50%) --> red ball A (50%) --> 50% × 50% = 25% 3. box B (50%) --> red ball B (50%) --> 50% × 50% = 25% 1 results in green ball, so 50% chance of choosing a green ball. 2 and 3 result in red ball, so 25% + 25% = 50% chance of choosing a red ball. If we choose the ball immediately: 1. red ball from box A (33⅓%) --> 33⅓% for green ball 2. red ball A (33⅓%) --> 33⅓% for red ball 3. red ball B (33⅓%) --> 33⅓% for red ball 1 results in green ball, so 33⅓% chance of choosing a green ball. 2 and 3 result in red ball, so 33⅓% + 33⅓% = 66⅔% chance of choosing a red ball. (Sorry for any spelling or grammar mistakes I might have made, English is not my first language)
@@endengineer2441 for the first situation, you said the probability of picking a red ball out of box A is 100%, but it's not; there's still a green ball. it's 50%. so since the probabilities of each of them are equal (50%, 50%, 50%), then it's out of three, not out of four.
Actually if you have 9999... repeating and add one it doesn't give you 0 since there is (at least in some sense) 1. The fact that you cannot reach the end (or like in this case beginning) of it doesn't mean it doesn't exist so its both logical and mathematical error
The 2/3 comes from if the balls were all in the same box and you picked one from that one box. Separating into two boxes creates a separate problem with the constraint that you must pick from the same box. It’s a given that the first ball is red, so you either have picked from the first box (a 0% chance of the next ball being red), or from the second box (a 100% chance that the next ball is red). Since there are two boxes to pick from, and each box can only give a single outcome (positive or negative respectively), the answer is still 50%. The question becomes 2/3 if you have them all in the same box, because if you have picked a red, then you could either pick the second red ball, the third red ball, or the green ball. That’s 2 of 3, or a 2/3 chance.
Let's analyse the problem with the rules of conditional probability to find the definite answer. For two random events A and B, the probability of A given B, written as P(A|B) is equal to the probability of A and B occurring together (P(A and B)) divided by the probability of B occurring at all (P(B)). In this problem we have two boxes, A and B. A contains one red ball and one green ball, B contains two red balls. We pick a box at random, then pick a ball at random from that box and note that its colour is red. We then want to know the probability that the other ball is also red. This probability is precisely the probability that we picked Box B - if we picked Box A, then the other ball is definitely green, whereas if we picked Box B, all balls are red. For the sake of brevity, we will call "picking Box B at random" B and "Picking a red ball out of a box first" R. We want to know the probability that we picked box B given that we picked a red ball out of a box first. This is written P(B|R). By the law of conditional probability we know that P(B|R) = P(B and R)/P(R). So we want to know P(B and R) and P(R). P(R) is the probability of picking a red ball out of a box first. There are 4 possible scenarios with equal probability: We pick Box A and pull the red ball out first, we pick Box A and pull the green ball out first, we pick Box B and pull red ball 1 out first or we pick Box B and pull red ball 2 out first. As you can see, in 3 of these scenarios we pull a red ball out first, so P(R) = 3/4. But what is P(B and R)? Well, using the law of conditional probability again, we can note that P(R|B) = P(R and B)/P(B). How does this help? Well, we can rearrange this equation to get P(R and B) = P(R|B) * P(B). We can also note that P(B and R) = P(R and B), as both are the probability of events R and B both occurring. P(B) = 1/2, as we have a 50:50 chance of picking either box at random at the start. P(R|B) is also very simple: if we know that we picked Box B to start, then our chances of pulling a red ball out first are 100%, as all balls in Box B are red. So P(B and R) = P(R and B) = P(R|B) * P(B) = 1 * 1/2 = 1/2. Putting all this together, we know that P(B|R) = P(B and R)/P(R) = 1/2 / (3/4) = 1/2 * 4/3 = 2/3. This might seem counter intuitive. The important takeaway is that knowing the outcome of the first pull informs you about the probability that you are in a world where you picked Box B earlier. It's sort of similar to the famous Monty Hall problem.
The counter-intuitive point is that, if for the case we pull a red ball out, there's only 1/3 chance for the box to be A, and a 2/3 chance for the box to be B
Top commenter is right, if there is an equal chance of pulling from each *box*, it doesn't matter how many red ones are in the box on the right. The probability only becomes 2/3 if each *ball*, not each box, has the same probability of being chosen, in which case the separate boxes are just a misdirect and have nothing to do with the problem.
2:54 if you randomly choose a box with 50/50% chances, then the probability is 50%. You're talking about choosing a random ball with some magic, but to actually take a ball you need to choose a box before
for the probability problem, It stated that you're taking a ball out of a random box and that the ball is red. If you took the red ball from box A then the next ball is green. If you took the red ball from box B the next ball from the same box is red so it should be 50-50. You stated that we take a red ball out of a random box. The two options are taking a red ball from box A or from box B since you're randomly choosing the box that you choose. I'm kinda confused tbh. edit: I now realize where I was wrong
It's conditional probability, doesn't make much sense at first but if you want to learn more, you can check bayes' theorem. For this problem, the question is "what is the probability that the other ball is red if the first is also red" which can be written as P(R2 | R1) with R1 : "the first ball is red" and R2 : "the second ball is red". You can calculate each probability separately but using a tree like shown in the video is way more intuitive and easy for this case (correct me if i'm wrong)
I agree. It should be 50%. He said ”a random box” so it’s 50/50 which box he chooses. If he had said ”a random ball from these two boxes” it’s another story. It’s super important to be clear about these things when handling probability.
@@dominobuilder100It's 66% because of the part where you already took one ball out and know that it's red. The 2 red box is more likely than 1 red 1 green to get you a red ball first, hence getting red first makes it probable that you picked the 2 red box.
No each ball in box b is 25% box b has only red balls so that's 100% which is 50%. Of both boxes. There is only 1 red ball in box a and the chance that you picked it first assuming you got a red ball first and chose box a is 100% and the other ball in there is 100% green so 0% that both balls in box a are red and 50% + 0% is 50% boom.
@@loganblakely3448 2/4 chance that you pick the box with 2 reds and get a red ball, and 1/4 chance that you pick the 1 green 1 red box and get a red ball, and 1/4 chance of getting the green ball. You got a red ball, so the case where you get a green is now excluded. You're left with the 2/4 of the 2 reds box, and the 1/4 of the red green box. That's how we get the 2:1 chance, or 66%
something similar is that the percent of children who are single children is significantly lower than the percent of parents who have a single child (provided the avg children per parents is above 1)
Intersting. Never heard this b4 Oh i get it why. Even a pair of parents is counted as one. Imagine an exaggerated village with 2 pair of parent One parent has 1 children One parent has 100 children When computed as ratio, you are one child in a 101 child that is single
the potato paradox is wrong, if 1% dry mass = 1kg, 2% dry mass isn t 1kg, because 1% of water evaporates, it means it s more dry mass, because if it t less water, the potate is more "dryer"
The term "dry mass" is refering to the mass of the object that is made out of molecules other than H2O. That is not going to change as the water evaporates.
Infinite 9s repeating plus one isn't just 0, there is a carry of one after all infinite nines, so I would describe it as 1 with infinite 0s afterwards, no different than 1.0 repeating multiplied by ten to the power of infinity.
While I do agree with you, I think the point was to say that since you can never reach the last 9 to make a 1 (maybe calling it the first 9 would be more accurate), it's like putting a bunch of 0's until the end of time and beyond. Therefore, the sum is 0, making the number -1. To me it's just a matter of how you look at it. Because you can think about it the way you did and make it exactly what you said "1.0 repeating multiplied by ten to the power of infinity" I personally find it more logical "your" way
This is not actually the case, because you can't actually represent a number that way, because the 1 has no position. Any digit you ask about is 0. So the result is actually 0. It forms a completely coherent number system for it to work like this even if it's a little counter intuitive.
This is why I hate when we put an answer down for an equation that can never be completed, 1/3 can't equal 0.3 recurring, because we can't calculate forever and saying it is 0.3 recurring is just an estimation, just like it is true for ...99999. we can't assume that adding 1 to this equals zero as we have an infinitely large 1 waiting to be put on the end of this equation, the true number it equals in 1 ...0000. as this is the best representation for this
The reason for Potato paradox to be even a paradox is because real potato doesn't have 99% of its content as water. It's hard to find some everyday food item to have 99% water, maybe diet coke.
The probability that the second ball is red: Case 1: you took it from box A 0% chance Case 2: you took it from box B 100% chance "What is the probability that the second ball in this box is also red?" 50%
Case 1: you took the only red ball from box A Case 2: you took one of the red balls from box B Case 3: you took the other red ball from box B You can get actual boxes and balls and do the experiment yourself (have someone else randomize them or use your own system to,) and the other ball will be red 2/3rds of the time. I tried it myself.
Before you pick, the chance of picking a box is 50:50, but after you get a red, now with this new information, the chance has changed. Same like you can teleport randomly anywhere in the world with equal chance, now u teleport, you see an arab, what is the chance that u teleport to an arab country
The ...999 example doesn't convince me at all. If you add 1 to it, you'd have an 1 after the infinite 000s. So it'd be more a 1...000 thing rather than a 0, and if there's no mathematical notation for it, blame mathematicians, not me. Next time you're gonna tell me that summing all integers gives us -1/12.
I can answer that for you, but still take in count I could be wrong you may say that that number is infinity since that number is the sum of infinite non-converge geomtric series. (9+90+900+9000...) which makes sense with why you can't just simply add one into it and ask what is the number's representation in decimal base. from the statement of the video that this number is equall to -1 ( if this statement was true..) we could construct a proof by contradiction that this number does not exist by saying that a number is equall to -1, we already asumed that he does exist we can easily say that 9+90+900 is bigger that 1 we get x>1 and x=-1 , a contradiction therefore x does not exist.
Also there is a similar thing going on with the sum of all integers gives us -1/12 (if it was true) we already know that 1+2+3... is bigger than one which gives us a contradiction but I have a question is infinity a number? or can we say infinity exists?
@@ronflypotato4242 9 plus any number of positive ints will remain greater than 0, -1 is less than 0, (aka ...999 > 0 > -1), ...999 = -1 contradicts this. This is the same as your proof, right?
Yeah, the argument in the video seems like just semantics imo. Like, sure, if you do the addition manually and carry the one you get an infinite number of 0s, but that's just a consequence of the way we write numbers. That doesn't make the actual value -1.
yeah i used to believe "adding the natural numbers gives us -1/12" but it's more like "adding the natural numbers would diverge and not have a value, but if we extrapolated from other results we'd get -1/12"
The last one is weird. If there is a fixed expected value why would the average win grow with attempts? I mean it was like looking for the average win of a lotto ticket and the same thing happens. OK it's at night and I am super tired. Maybe I look at it again tomorrow.
Because it only goes to infinity when you take into account extremely rare events, 20 flips until heads for example gives 2^20 dollars but also has a probability of 1/2^20, and the more simulations the more likely to get those.
2:12 only if you do not know the box you chose from, because it really does not matter which red you chose, but yeah if you took one from box then if you swap boxes aronud, 66-33 , but if you choose and it is red, choosing next is 50% if you scramble it will act like if it is one box anyway
Probability Problems like box one can be fixed quite easily with Bayesian thinking! Although its not very natural to think of them when sample space is very small... So, it gets very paradoxical.. I think the ball problem is equivalent to monty hall problem?
@@RuthvenMurgatroyd P-adics can be treated as a different branch of numbers seperate from real numbers. This sum, people try to put into natural numbers set and thats just not true.
the anwser to the probability problem is 50%, the balls from the second box cannot travel into your chosen box via osmosis, so you only have a single ball that is either red or green, not three balls that are red, red, and green
Před 8 měsíci+3
You picked a ball randomly and it was red. Given this, three cases are possible: • You picked the red ball from box A. (The next ball will not be red) • You picked ball 1 from box B. (The next ball will be red) • You picked ball 2 from box B. (The next ball will be red) 2/3 of the cases are valid for our conditions. Probability is 2/3. It is not 50% because the probability of picking a red ball from box B is higher than that of box A.
@@quentind1924 Yes they do. P-adic numbers have no issue with decimals. The only thing you have to avoid is _infinite_ decimals. The reason they tend of a avoid decimals is that when p is prime, the decimals are unnecessary, but when p is not prime, you need the decimal part to be able to represent all the numbers.
I don't understand the last game. When does it end? What is the choice I have to make? How much I I bet per coinflip? Or per total game until it ends? Do I have to say how many times I want to flip until I get heads? Because if we play once until heads comes up, I don't think it would be wise to bet infinite money. Or do I have to say at which flip heads comes up and if I'm right I win? I really don't understand what I bet on and how long it is played aka the winning conditions and what choice I make that I bet on.
The game ends immediately when you get a head for the first time, if tails , continue playing. You get the money according to how many times the coin is flipped in a run of a particular game, according to the table, the longer the run the bigger the money. The problem ask what is the maximum money you SHOULD pay to participate in this game., as an entrance fee to play a round.
Here's a fun demonstration. For a number x of length k, the periodic number 0.(x) = x/99...9 where there are k values of 9. Try it yourself. Therefore 0.(9) = 9/9 = 1. P.S. A demonstration for that formula is that 10^k * 0.(x) = x.(x) = x + 0.(x) So (10^k - 1) * 0.(x) = x 0.(x) = x/(10^k -1) = x/99...9 where 9 is repeated k times.
To say that 0.999...=1, you have to be a little more careful. It's easy to show that the largest number that is possibly smaller than 1 is 0.999..., and hence either 0.999...=1 xor there are no numbers between 0.999... and 1 (and 0.999... is smaller than 1). But to show that it must be 1, you still have to use e.g. that the set of real numbers is dense - there cannot be no numbers between any two real numbers. Of course, you can also easily sum the infintie series of 0.9+0.09+0.009+... as it is a geometric series. After seeing the last expected outcome game, I noticed another way to find 0.999...=1 using probability calculus. Note that { the sum over all positive integers n of [the probability of getting the first head at the nth toss] } must be 1, since in each game [you must get the first head at the nth time for some positive integer n if you do end up getting a head] AND [the chance that you never get a head is limit n goes to infinity of (1/2)^n, which is 0], and summing over the probabilities of all [the possibilities of non-zero probabilities] must yield 1. And this probability sum is also a geometric series 1/2 + (1/2)^2+(1/2)^3+..., so this example tells us that you can make an exact analogy between the probability sum and the geometric series. For the series 0.9+0.99+0.999+... = the sum over all positive integers n of 9/10^n, let's consider a random number generator that generates all numbers from 0 to 9 with equal probability - let x be an arbitrary number from 0 to 9, then the probability of getting [any number that isn't x] for the first time after using the generator exactly n times would be 9/10^n. As before in the expected outcome game, { the sum over all positive integers n of the probability of getting [any number that isn't x] for the first time after using the generator exactly n times] } must be 1 by probability calculus, and must also be equal to the sum over all positive integers n of 9/10^n, and hence we obtain 0.999... =1.
That potato one isn't surprising; it's just weirdly worded. I wouldn't call reducing the amount of water by a small amount (basically half) dehydration. It's a semantic trick question. What is actually happening is you are taking away 50 kg of water (so 50/99 % of the water, or basically half), and yet leaving the 1 kg of non-water, so of course, the result is half of 100 kg which is 50 kg; as you took away 50 kg of water, but specified this in a deceptive way. Of course this is a deliberate trick question, which is why it is worded that way.
I don’t agree with the first demonstration because I don’t believe 9 repeating is a number, therefore 9 repeating + 1 would not be zero, it would be undefined. Feel free to correct me if I’m wrong on that. Very fun video!
It’s a different number system. In this "mathematical world", ...999 exists. But if in real situation someone asks you what ...999 is equal to, say that this number doesn’t make any sense unless you are in the P-adic system
MY POINT EXACTLY, YOU CAN DO THAT INFINITELY FOR EVERY POSSIBLE CONCIEVABLE NUMBER SO AT SOME POINT IT STOPS MATTERING AND YOU CAN JUST SAY THAT EVERY NUMBER IS BASICALLY THE SAME NUMBER, ITS RIDICULOUS
The birthday paradox is one of my favorite paradoxes because it feels so wrong, yet it's mathematically sound.
That is because you are looking at the problem from a single person perspective. The odds you are imagining is if you pick one person and ask what is the probability that somebody in the room matches THEIR birthday. That is a different question than if ANY two people in the room have the same birthday.
The same fallacy arises in the analysis of DNA evidence. An analysis usually only looks at a few dozen markers in the DNA. From this DNA matches usually quoted to have a really low number of people matching your DNA. That's true; but, it isn't the complete picture. If you instead consider how many people in large population will have matches, the number is quite a bit higher.
Reminds me of when I told it to my mom after I learned it in school and she straight up said "That's not true, that can't be true."
not really a paradox though is it?
It's not counterintuitive, you're just thinking about it wrong. Think about it in terms of graph theory, where there is a fully connected graph with each person as a node. The probabilities that a birthday will be shared is proportional to the edge count in this graph.
@@The9thDoctor Doesn't counterintuitive literally mean differing from how the average person would think about it ?
Every fiber of my being disagrees with p-adic numbers
As a computer scientist every fiber of my being does. :D
This is kind of like the infinite sum that adds up to -1/12. At no point in a finite summation would it be equal to that, but we ASSIGN the value to it. So I think it's kind of questionable to say that it EQUALS said value.
@@seejoshrun1761Except that =-1/12 just doesn't work at all, the math behind it is deeply flawed. It just doesn't make sense
@@seejoshrun1761channel mathologer just disporve that like 7 years ago😂
Don’t worry, the whole point is that p-adic numbers don’t agree with the standard number system, it is a completely different set of rules. In our normal rules that sum would in fact equal infinity, which makes the statements given there misleading.
It's crazy how barely anyone has seen this
now that's counterintuitive math
*this masterpiece
Ikr
No. That is the *third* most viewed video of the channel.
severely underrated
99999... = -1 because the simulation running our universe is on a twos-complement machine.
But is it a proof or only assumption? Why adding +1 you cannot write it as 9999...+1=10000.... making it 'meta' infinite? 'Surplus that goes beyond'
When the number is too big there’s an integer overflow!
How can infinite number be "too big". I believe in video mentioned "proof" denies that ...9999 is infinite.
That was exactly my thought seeing the thumbnail.
@@juliusjakas6858 When it comes to calculating machines there is no such thing as infinite. (Okay IEEE-754 does have infinities; but, that is just a signal and not useful for actual calculation.) A computer will never be able to do an infinite calculation because it will always have a finite amount of memory. This limitation can be used as an advantage.
The hardware necessary to subtract numbers is also more complicated than the hardware necessary to add numbers. Adding such functionality to an adding machine or a microprocessor would take up space and require more material -- so lots of adding machines and lots of microprocessor's neglect having any subtraction hardware at all. The solution is to leverage the fixed word size of a microprocessor or the fixed number of digits of the adding machine along with the modular rollover when the size of that fix number is exceeded. Adding machine operators could do subtraction by adding ten's complement numbers and computers can subtraction by adding two's complement numbers.
For a four digit ten's complement adding machine, the number -1 would be represented by 9999. If I wanted to subtract 1 from 5309, I would add the 5309 + 9999 to get (1)5308. The one that would normally be carried over to the 5th digit is simply discarded because there is no hardware to store the 5th digit. So the correct result displayed is 5308.
So the joke is that if the universe is being simulated inside a computer, then any attempt to use an infinite number results in a rollover leading to glitchy results inside of the universe. Imagine a kid put ten toys into a bag and then took them out again one by one. But after he had already taken all ten toys out, he attempts to take out another toy and instead of there being no toys he takes one out and there are ...9999 toys left in the bag afterward because whoever wrote the simulation forgot to check for underruns of the counter determining how many toys were in the bag. We will not even discuss what would happen if the counter was used to index an array of toys in the bag... human sacrifice, dogs and cats living together, mass hysteria.
...9999 isn't equal to -1 because the two are different types of numbers
Using p-adic numbers with real numbers allows you to falsely “prove” any ...nnnn = -1 conclusion, where “n” stands for a single digit.
For example in the 2-adic system where numbers are written with the digits 0 and 1 (e.g. 10 = 2, 11 = 3, 100 = 4) you could show:
...1111 + 1 = 0 => ...1111 = -1
This too would be false because p-adic numbers don’t follow real number arithmetic. Rather, the problem should be written as:
...9999 + ...0001 = ...0000
...0000 != 0
This is the most rational answer Ive seen to this.
What's the difference between ...0001 and 1?
Meh, ...0000 = 0 , lol
The way you actually solve this problem is by saying that the p-adic numbers are different from the real and complex numbers. Zero is just defined as the additive identity of a set, so ...0000 = 0. Similarly the p-adic numbers are allowed to contain one. The statement ...1111 = -1 is true in the 2-adic numbers. It's not true in the other p-adic numbers and it doesn't make sense in the reals.
THANK YOU, its clickbait for people who don't know what p-adic numbers are. A more sensible answer would be 100000..., but really it just makes no sense in the context of real numbers
I mean a 99% water-content potato is just a glass of water with some starch sprinkled on it.
So a 50% change to dry-weight really isn't too much
Lmao fr
I always used melons in exactly this exercise: they consist of 92% water, that's close enough ;-).
@@sonobox-lu6mr still, a -12.5% dry weight change from 92% to 91% sounds a lot more normal than the hypothetical 99% percent water potato.
I don't really understand, why would the dry mass stay the same weight? It's doubled now, so would it not be 2kg? then again, I'm currently brain dead tired so idk
@@EvanUnknownas dry mass they mean the weight the potatoes would have with no water. the total weight is always the dry mass + the water. The real statement is to find the total weight so that the total water is 98% of the total weight, not that the potatoes are 98% hydrated. It's a bit confusing (because wrongly stated) and it plays on that to look counterintuitive
not only did you give a short introduction of proof of 0.9…=1 in just shy of 30 second, but you also explained in the RIGOROUS manner, I am impressed already.
jesus died for you
and then completely botched the ...999 + 1 = 0 proof by making a false equivalency between p-adic numbers and real numbers.
it's not rigorous, they did not prove that there are no real numbers between 0.9 repeating and 1
@@allergictobs8261 unrelated
@@nutronstar45pretty sure something simple can work. Assume we have 0. To get a number with one digit after the decimal point we have to add a number from 1 to 9 and multiply by 0.1. To get as close to 1 as possible we have to use the number 9. Now we have 0.9, 0
Missing an explanation for some of these
Absolutely beautiful how you just get to the point without wasting everyones time. If more youtubers were like that the plattform would be so much better
I feel like it’s a little too brief, but that’s only because I’m very interested in how the weird probabilities actually work, but I also realize that those explanations might require higher levels of math.
yes
this is true too@@BinaryHedgehog1
The first one was a bit off but the rest this is good
engk wrong
These don't feel like "proofs", it just feels like our standard mathematical system has core deficiencies.
i mean, you’re not *wrong* in the sense that there is surely stuff we don’t know, but all of this is super consistent with math. there are no contradictions here (in the mathematical sense).
and that’s because these aren’t proofs. these are just facts or theorems. if you want to see proofs they’ll be a little bit more involved than this
With the concept of infinity and probability there’s many different ways people could look at it, I do believe there is only one correct answer for everything, but that’s assuming we are all answering the same question, but since he’s missing some explanations I’m not sure what view he’s looking at it from exactly.
Yes, mathematics has flaws, and mathematicians know those flaws _far_ better than you do.
The way we express different functions and operations, yes.
Math in general? No.
I was thinking 0xffffffff is -1 because it's a signed integer.😂
I think that's actually not a coincidence at all
Isn’t that hexadecimal?
Thats actually a very similar concept.
@@penguincute3564 Hexadecimal or decimal, all the numbers in computer are stored as binary. For 0xffffffff, all 32 bits in the memory is 1. If you interpret it as signed integer, then it is -1.
That's precisely what I though when I saw the thumbnail.
only 300 views? before i saw the views i really thought it has hundreds of views because the animations are just like those other good math videos with a lot of views... how can this channel also only has less than 2k subs??? it deserves more than hundreds of thousands...
Yeah
You were surprised it had only 300 because you were expecting hundreds? You do know that 300 is in fact hundreds, right?
In a room of 367 people, there is a 100% chance that someone has the same birthday
nope, what if it was a meetup for people with the same birthday? It can never be 100%
@@xazii There are only 366 possible birthdays. With 366 people you may or may not have at least two sharing a birthday, but when you add person 367 there must be at least two people sharing a birthday according to the pigeonhole principle.
@@mwickholm dang that was dumb of me. You're obviously right. That is embarrassing, I was not thinking straight when I wrote that lmao
@@xazii With a meetup of everyone with the same birthday, I think it's safe to say that they will have the same birthday 💀
I understand the trip up, but this counter example is hilarious because literally everyone would have the same birthday.
@@YT7mc yea i totally get that hahha, I prolly didnt thoroughly read the comment when I wrote that, I probably thought he said something else hahahah, that was some of the dumbest shit ive ever said tho
2:55
i also got 2/3
but i think the core of the problem is understanding the question. the question asks for the probability of the second ball is red WHEN the first ball is red. so the first ball can not be green
this is different than a question saying: out of these 2 boxes, what is the possibility of getting two reds? then the probability wouldn't be independent of each other and would be 1/2
at least i think idk
I got messed up on this. If we choose a random ball and we know beforehand the ball has to be red then it is 2/3, but if you choose a random ball and it just happens to be red then it's 1/2.
Its like with the Monty hall problem and the 3 doors. You switch only because you knew Monty couldn't choose the door with the prize.
Though, I think I may have just repeated your answer.
Oh wait, no, I was wrong! Picking a random box then a random ball is the exact same as picking just a random ball. So if you got a red ball there's a 1/3 chance you're in the green ball box and a 2/3 chance you're in the red ball box.
You have 4 equally probably events:
Choose green in box A
Choose red in box A
Choose red in box B
Choose other red in box B
Doesn't matter if the green ball could be picked or not (unless both boxes were equally probable so the red ball in box A was 50% and not 25%)
The probability problem is all based on the wording, both 50% and 66% are correct depending on the wording. "Let's take the ball out of the random box, the ball is red. What is the probability that the 2nd ball in this box is also red?" This kind of setup leaves the question open to ambiguity around whether or not the ball was chosen randomly, or if just the box was. "the random box" implies the box is what was random, and 'take the ball' implies it was specific.
Assuming it's all random, there a 50% chance to pick A or B. A red ball from A is 50%*50%=25%. A red from box B is 50%*100%=50%. A 75% chance total to pick a red, 25% from A, 50% from b, or 2/3, 66% that you picked box B, which will result in double reds.
However, if you assume just the box was random and we chose to take the red ball first. The ball left in the box would either be Red or Not. If this ball was picked from A, there is a 0% chance the remaining ball is red. If it was picked from B, there is a 100% chance the remaining ball is red. This results in a 50% chance that the ball was from A or B.
This shows the importance of being extremely clear with and direct with probability. Most of the confusion in probability comes from wording, not the maths itself.
thanks for explaining this, i knew there was some funny business going on in this problem
2/3 is actually never right for that problem. You ALWAYS have a red ball first, as specified in the problem, so your 25% from A 50% from B is inaccurate. It’s a given picked a red ball from box A if you at all picked from box A, so it ALWAYS has to be a green ball left. It’s a given you picked a red ball from box B if you picked from box B, so there will ALWAYS be a red ball left.
The 2/3 only comes in if all balls are in one box, and you picked a red ball first. Then you now have 2/3 chance to pick another red ball, because two of the three balls remaining to be picked are red.
I think u should not interpret it as the second scenario...i think taking a ball out of a random box is clear enough.
@@theeraphatsunthornwit6266 You have a situation where every time you look at this problem, the ball is red. It's a given. If using this form of thinking, the 50% would apply. There is also no clear distinction that the ball is random, only the box.
I'm not saying I disagree with you, just that it's very important to make probability absolutely clear with no room for false interpretation, otherwise we end up with problems like this
@@theeraphatsunthornwit6266 well at least for me. I misinterpreted it exactly like in scenario two and was very confused about if I had made a mistake.
The last one is tricky since the median i.e. the most common result is very different from the average if I remember correctly.
The most common result would be the mode, no? And I think it is $1.
The median would be a value that splits the possibilities in half: Getting a bigger payout than the median or a smaller payout both have probabability >= 0.5
The last one is tricky because in the real world you cannot expect to win more than 100 million, so the actual value is more like 26€. After enough coin flips the lottery would just go bankrupt
@@shadowpenguin3482It will take a long time for the lottery to go bankrupt though, they can probably make it sustainable if the payment required is like greater than $5 and no one is sitting there playing 10 billion times
Actually, if you played that game an infinite number of times, you would lose exactly $1/12
You would die well before that. So there is indeed a finite bound to any expected reward.
This is a joke/sarcasm to the mistaken proof that infinity = -1/12
I loved this video, the graphics and the subtitles really elevate the script. Youve gained a subscriber!!
Did I just see the right triangle of combinations at 7:10 which also relates to creating an input counter in circuit maker 2?! Amazing! Now rotate the x and y values so that n is the height-map for each combination and include the xor values for if numbers are only single integers. 0 would be in its own plane because we are only counting 1's here.
This video deserves way more attention, it's great
I feel like the ball problem may have been under specified. If I assume fungibility of the balls (eg. I don't care if I have your $1 bill or my $1 bill it's $1) then when I return to the box I'll either see red or green:
Scenario 1:
Balls:
RG
Remaining:
G
Scenario 2:
RR
Remaining:
R
Options:
RG
Probability:
50% (frequency of R/total possibilities)
The 66% case would kick in if balls were not interchangeable (eg. red ball 1 , red ball 2 and red ball 3 exist) or I have the ability to pick from the other box (eg. I can pick from Box 1 even though my first pick was Box 2).
I may be off here but I feel like it depends on the parameters of the problem
Just because the two balls are the same color doesnt mean they are the same ball. There are three red balls each equally likely to be picked. Picking one of the two in the same box leads to an outcome where the other ball in the box is red.
@@michaelshaffer6717it only asked if the next ball would be red, not if it would be a specific red ball or anything. There are only two outcomes, red or green, and each outcome has a 50% chance of being true.
2/3 would only be true if the balls were all in the same box.
I agree with your statement, the probability THE NEXT BALL IS RED is 50% because it only differs in which box you picked
@@axelarnesson5066 But the odds about which box you picked from aren't 50-50, is the tricky bit. The fact you got a red ball means it's twice as likely you picked from the red-red box, because if you'd chosen from the red-green box, it's half likely you would have gotten a green ball. At first, there's a 50-50 chance you picked from either box, but getting a red ball is more information, which changes the odds. What you're thinking is that it's equally likely you picked from one of two boxes; what you really know is that it's equally likely you've found one of three red balls, and two of the red balls share a box with another red ball.
What you are missing is that if you randomly pick the GR box, you have 50% chance to get the wrong ball, and then you might try from the ither box
Start : Randomly pick option 1 or 2
Option 1 : You randomly pick the RR box (50%), you get a R ball (100%) and the other is R. So 50% to get R
Option 2 : You randomly pick the RG box (50%), then you have 50% to get the R one and see that the other is G (so a total of 25%), but also 50% to get the G one, then you go back to Start
So you have 50% to get RR and the other is R, 25% to get RG and randomly pick the R one and the other one is G, but on the remaining 25% you restart with a non-zero probability to then pick the RR one and get a R. So it has to be more likely to get R as the other ball
Try it yourself, get 2 dices (but you have to know which one you roll) and a coin then foip the coin to pick a dice. For the first dice, whatever you roll, you note 1. For the 2nd dice, you note 6 if and only if you get a 6. Otherwise you do the coin flip again. Now, i’m pretty sure that you see how it’s rigged. It’s exaclty the same thing for the 2 balls problem
This was really mind blowing 🤯 thank you for this video i hope it gets more views
Love your videos, fun and simple - keep it up, excited to see more content from you
It is a shame so little people get to see this part of maths
For a good reason
lmao yes@@chasekwas
True, but its practical applications are little, unless you want to go into obscure branches of statistics.
They do and they frequently make bad decisions because they don't do the math. There are lots of psychological fallacies that lead to everything from bad purchases to burning money for bad investments to voting for the wrong leaders that result because of it.
what, the part that is just trivia and funny scenarios? ew
The periodic numbers one isn't actually true, as it is actually just a semantic argument that relies entirely on the ways we write numbers rather than the actual value itself. Sure, if you add it by continuosly getting a zero and carrying the one, you will get an infinite amount of zeros, but you'll also never finish this calculation. It'd be like if you added 1 to a hundred digits of 9s, but stopped halfway and claimed it equaled 0. If you just finished the calculation, you'd eventually get a 1 in the first digit. Same with the infinite one, just that you can never finish the calculation.
Not only that, but this trick only works because we use a base 10 system, which is an arbitrary choice. If you converted the infinite 9s number to hexadecimal and added 1 to it, you would no longer get an infinite amount of 0s and would instead get something entirely different which obviously does not equal -1. Instead, this trick would work with an infinite amount of Fs in hexadecimal.
I see what you mean but not at all the case, first they are called p-adic not periodic the p in p-adic stands for prime as these are usually used in a base with prime numbers so the number changes based on what base your in ...9999999 in 10-adic is different to ...9999 in hexadecimal. just as 10 in base 10 is different to 10 in hexadecimal. and no you wouldn't always take the result as true but certain parts of math have uses for these p-adic numbers and using these results
The p-adic numbers are just different mathematical objects. Veritasium did a video on them and showed how they provided a shortcut to an answer to a practical problem about the regular old real numbers that would be hard to find otherwise. I've never studied p-adics, but the idea is that you introduce a different metric on p-adics than the one you are used to so that a sequence like 9, 99, 999, 9999, 99999, ... actually has a limit just like the sequence 0.9, 0.99, 0.999, ... has a limit in the real numbers with the Euclidean metric.
@@oLouis53 @JustinBA007, you are both right in a sense. Whilst the actual value does indeed not equal -1 in any base for that matter if you consider it differently (like was done in the video) you can make an argument for it converging on something and it does have an actual use in mathematics. It simply depends on what rules you are using for your allowed operations. It's kind of like Fairey addition in that it isn't mathematically correct but still useful.
@@oLouis53ok, that makes sense, but that is not made clear at all in this video. The way it's presented in this video makes it sound like he is saying that an infinite number of all 9s is actually equal to -1, which is just not true in normal math. Also, not to mention that the chapter of the video is titled "periodic numbers," which only further confuses things.
This principal is actually fundamental to how computers perform arithmetic, namely how they handle subtraction by adding: Negative numbers in a computer are actually written this way, i.e. -1 is 0b11111111 for an 8-bit binary number. This means it's trivial to negate numbers (just flip all the bits and add 1) and you can just treat them like all other numbers. To give a decimal example, we can represent 1000 values with a three digit decimal number, 000 - 999. this effectively defines [501, 999] as [-499, -001] in a way that preserves addition and subtraction. For instance, 209 - 057 becomes 209 + 943, which is 1,152. Since we're stuck with three digit numbers, we drop that leading one and end up with 152 (i.e. 209 - 57)!
For last one it doesnt only matter how much money you are willing to pay but also how many times you are willing to play, looking at numbers best way is to play 1000 times and max money for that is 5,6$
I love this channel so much, I hope you get popular soon!
Bro just speedran every Vsauce2 video 💀
The idea that an infinite string of 9s, written as "...999999999" (with an infinite number of 9s going off to the left), is equal to -1 is a speculative and controversial concept in number theory.
A complete bullshit, to be specific.
wrong.
It's basically the same thing as the Two's Complement format that computers often use to deal with negative numbers.
@@matswessling6600where did you get ...333=-3 from? It would be -1/3. Both by doing ...999 / 3 (-1 / 3) and by doing 0 - ...6667 (0 - 1/3)
p-adic numbers aren't controversial lol
Please keep uploading such videos ❤
I learned way more interesting math than in school! You need more subs! You are so underrated! Please make more videos I really enjoy them! I just subbed to you and turned on all notifications!
Super underrated video. Nice job!
This is incredible. It took my teacher 4 weeks to stumble over this and you nailed it in 8 minutes. I wish professors in universities would teach like you! Just amazing!
As a stats tutor it felt good to get most of these right haha great video !
Very well done, I hope this will be blessed by the algorithm
i literally had the idea of the thumbnail one this morning, i had never even heard of p-adic numbers and now im following a rabbit hole. ty 4 dat
The additive persistence one is not that surprising, as the sum of the digits of a number has an additive persistance of that number -1, so for example the sum of the smallest number with an additive persistance of 3 (199) 's digits is just 19, the smallest number with an additive persistence of 2. This is true for 4 and 5, and i assume it's also true for the others as well.
I'm trying to get my head around the last one.
I set up a simulation in excel where it randomly picks a 0 or a 1 (my head or tail), if it's a 1 then it pays out based on the number of 0s that came beforehand. I then add up the total winnings and divide by the number of games played and it does this a million times (well slightly more: the number of rows permitted in excel). The average win per game seems to settle at around 4-5, but there are spikes where it gets really big. There's every possibility that I'm not calculating this correctly, but also I am aware that each time I'm doing this I'm working out the average win for that particular set of games rather than what happens generally. There are a ridiculous number of ways to put 0s and 1s together when you have a non-fixed number of either of them, so I guess every now and then there might be a combination that averages vastly more than what I'm seeing here.
The general trend is upwards though, but when I plot game# against average win so far I see big spikes (huge wins) followed by decays.
Yeah well basically you are never going to see the massive upside because the average wins per round grows at a logarithmic rate. You really have to look at it analytically:
Expected value is the sum of all possible outcomes each multiplied by their payout. if i bet 2 dollars on a coin flip coming up heads, my expected value is (2*0.5)+(0*0.5), which is 1 dollar.
Each possible outcome in our game, one more tails than the previous one, is a (1/2)^n chance of happening times a 2^n payout .Round one is a 50% chance of making 2 dollars, so that's an expected payout of 1. round two is a 25% chance of making 4 dollars, which is also one. Do you see what is happening? every round has an expected payout of 1 dollar, so when we sum them together we get a result of an infinite expected payout.
Remember, the values he shows in the chart is average wins _per round_. how much money you would make is actually that times the number of rounds played.
vsauce2 has a great video on this: czcams.com/video/RBf1s4TassI/video.html
I've also simulated this in python and for a million games played it only tends towards 6$ if I'm understanding the rules of the game correctly. Not sure what's wrong.
@@copalo343Sounds like you're potentially getting a similar result to me . Some of my outputs tended to around 6, but I got different results each time I ran the simulation because you do get different outcomes as every one of the 1 million coin flips I do was calculated each time I ran it.
As I understand it, the longer it takes before you flip a win, the more money you get, but I'm not sure if we're considering what we ought to pay for each flip, or what we pay to flip the coin until it wins.
It's possible that maybe what you need to do is consider a run of (e.g.) four losses and then a win to be one "game" and do that a million times rather than just do one million coin flips. You end up with a significantly larger dataset with a different number of coin flips each time. I'm not sure how to handle that in excel but I think I could do it in matlab as that can handle changing matrix sizes quite happily, plus I could log multiple results and see what sort of distribution the final outputs have. I have very little experience in python. This might be a task for tomorrow or wednesday.
@@magnusbruce4051 Would for sure like to see what that looks like, could you get back to me when you've done this?
@@copalo343I'll try to remember to do that if I'm successful. I'm a bit too tired to do it tonight (plus I need to install matlab for this computer) and I have a band rehearsal on tuesday night so I might not get chance to do much on it then. Realistically Wednesday is the earliest I'm going to get a chance to start work on it and I will definitely need some time to re-acquaint myself with matlab commands as I've not used it for a few years.
I love your channel. Keep it up
this is one of my favorite math things - facts that are super unintuitive but are still, well, facts. my favorite thing about math is how weird it can be while still being correct!!
The one with the red and green balls is like Betrands Box paradox, where that version has a 3rd box consisting of 2 green balls.
Also wouldve been cool if you did the Multiplicative Persistance as well as Additive
and also kinda like the monty hall paradox, where you always change because monty hall will never reveal the door with the money, so if you change, there is 2/3 chance you win, if you dont, its 1/3, because its 1/3 chance you guessed it right first, and 2/3 you guessed wrong
But 50 per cent is the correct answer.
@@atepomarok9339yeah your right, they did the math wrong, the probability of pulling a red ball after one red ball is 100%… not 200%, multiply, not add.
@@Coolgirl_ig_ it's 2/3, i suggest looking up conditional probability
You're wrong on a couple of these, at least as your presented them. The way your providing the information is not the same as wherever you took these problems from.
Sources plz
@@williamwalsh4743 I typed out a whole explanation already, and then accidentally deleted it. The potatoe one seemed off, but I'm not sure about that one. The red ball green ball one is wrong fs. Without further specification, it makes most sense that you would be randomly choosing a box, and then randomly selecting a ball from within. The way he calculated assumed that you have an equal probability of drawing every ball from either box, which is not true.
@@maniacpwnageking1. The potato one is fine. The dry mass stays the same, so it should always have 1 kg, work from there.
2. The ball one isn’t wrong. He just said you randomly choose a red ball, why would you assume that he first randomly chooses a box? That’s your assumption which you have chosen to make.
Sometimes(p=1/4), you will get green ball and skip that. When you get red(p=3/4) you, obviously, have 1 outcome to get next green and 2 to get next red.
Why would you make implicit reference to the Archimedean property for showing 0.999 repeating is the same as 1 for the reals, and then immediately change to the 999... p-adic example, not even mentioning that the p-adics are a completely different set of numbers that don't have the Archimedean property?
I may be wrong, because I'm not the best at math, but in the first paradox (the one that explains that ...9999999 = -1) the problem here is that you're assuming that this number has an end (because then you can't add 1 using the arithmetic operations, in this case addition), which is impossible because you said it yourself, the numbers are infinite.
Now, EVEN if this was correct (which, like I said, it's mostly not) the moment you get at the first numbers (which are 99999...) when you add the carried 1 to the nine at the most left, you get 100000... . However, like I said, this is probably impossible, because then this number would no longer be infinite, because in the last example I gave, there IS an end.
i think youre right, ...999 does not equal -1
@@shieldgenerator7 Of course, that's what a paradox is supposed to do, something contradictory, but yeah, that's why the paradox doesn't work out
@@ctanimations8488actually after learning about p-adic numbers, it turns out ...999=-1 is TRUE. It's not made clear in the video, but p-adic numbers use an entirely different math system where addition operations like that in the video are valid.
in the math system we're used to, it's not valid, but in the p-adic number system, it is valid.
@@shieldgenerator7 Thanks for the clarification, like I said, I'm not the best at math so this makes it much clearer. Maybe I'll investigate later about p-adic numbers, once again, thanks
Had to stop watching this because I was getting angry.
No views? Lets fix that!
This deserves more views than it have.
Great vid! However, i don't get why there must be something between two numbers for these numbers to count as "different". Is that some axiom i'm not aware of?
Agreed. The point of saying greater or less than means you know the values of the numbers and you compare them. If I had a number line with just the integers then could I say that because I cannot fit a number between 4 and 5 then they are the same number? No. By this logic every number would have the same value.
In the real numbers, if x and y are distinct, then there always exists a third number that is (x + y)/2.
@@martind2520 Thanks for bringing this up.
@andrewfontecchio6257 I just found this en.m.wikipedia.org/wiki/0.999...
An interesting read.
yes there seems to be such an axiom. The trichotomy of the reals. For any pair of real numbers either x=y, xy holds. Assume x
This is a fun video and that's great but I do need to point out ...999 does not infact =-1
For anyone who is interested you cannot simply add one to an infinitely large number (well you actually can using ordinals but not in this case) thus to understand the value of such a number you would need to approximate it using something like an infinite series which you would find diverges to infinity.
Sum of all +ve naturals is -1/12 🤓
buddy they're called adics and yes .999 equals -1 in that number system, even though obviously in our "normal" number system you can't do this
Nope this is just how padic works
Noooooo, reeeeeally? That's why I gave this video a dislike.
@@samueldeandrade8535 nope not rlly
The ball one doesn't make sense though. There are 3 outcomes but 2 of the outcomes are combined as 1 since they're both red balls.
(Nvm I understand now)
I wouldn't say combined but I agree with you because the 3 outcomes which he mentioned (and thus used to conlude a probability of 1/3) are not equiprobable if he chose a random *box*
you may be combining them, but that is actually wrong. Imagine you also number the balls. Clearly, you have a 25% chance for pulling each ball. Keeping the numbers, there are 3 numbers that result in pulling a red (let's say 2,3,4). Each of these is equally likely, so 1/3 each. Now, of the set (2,3,4), how many are from box B?
@@johnbyrnes6621 Ohh wait yea I understand, 1/3 is green and 2/3 is red. So that's 33% and 67%.
Nvm I was wrong. Here is the python code that I wrote, feel free to point out any logical errors
import random
redcount = 0
greencount = 0
remainderoftrials = 0
for i in range(10000):
initialboxchoice = random.randint(1,2)
box1 = ['g','r']
box2 = ['r','r']
if initialboxchoice == 1:
removed = box1.pop(random.randint(0,1))
chosenbox = box1
else:
removed = box2.pop(random.randint(0,1))
chosenbox = box2
if removed != 'g':
if chosenbox == ['r']:
redcount+=1
elif chosenbox == ['g']:
greencount+=1
else:
print('ERROR')
else:
remainderoftrials+=1
print(f"Other ball was green: {greencount}")
print(f"Other ball was red: {redcount}")
print(f"Remainder of trials, i.e. those where the first ball was green: {remainderoftrials}")
print(f"Experimental probability that the second ball in that box was also red: {redcount/(redcount+greencount)}")
P.S. Ik that the code could be written much better, I just wanted to make super sure that it was error free
Conclusion: Probability = 2/3
Hmm...no.
The question ask.. what is the chance of another ball in the box you just chosen is red. It does not ask you to choose a ball from the remaining 3 balls.
amazing channel
Thanks for once again showing that infinity is not a number but rather an idea. In the real (finite) world, one wouldn't have enough time to continually play the "hope I don't flip heads yet" game for anything more than a few dollars.
Actually, the problem with the 2 boxes is misleading. The probability of either box is equal, therefore the probability that the other one is green is 50%. The other box has a 50% chance so there are 2 possibilities of being a red ball at 25%. So to sum up:
Pulling Red ball from A: 50%
Pulling Red ball A from B: 25%
Pulling Red ball B from B: 25%
Still 50/50
The probability of box 1 is 1/3 as it is said a green ball is never drawn first, and there are 3 total red balls. With 3 possible starts there is box 1 (1/3 chance) with ball 2 being green, box 2 (1/3 chance) with ball 2 being red, and box 2 (1/3 chance) with ball 2 being red. 2/3 chance of ball 2 being red, and 1/3 chance of ball 2 being green
The probability of picking either box is equal, but you have been told that a red ball has been selected. You can draw a probability tree to visualise it. There are four possible balls at the start and after you select the first red ball, there are 2 red balls and 1 green ball remaining so the probability of picking a red ball is 2/3 since you still don’t know which box you are in.
Edit: Nevermind, I misunderstood the problem.
Original comment:
I suppose it depends on whether we assume that we have to choose the box first and then the ball or that we go straight to choosing the ball.
If we have to choose the box first:
1. box A (50%) --> red ball (100%) --> 50% × 100% = 50%
2. box B (50%) --> red ball A (50%) --> 50% × 50% = 25%
3. box B (50%) --> red ball B (50%) --> 50% × 50% = 25%
1 results in green ball, so 50% chance of choosing a green ball. 2 and 3 result in red ball, so 25% + 25% = 50% chance of choosing a red ball.
If we choose the ball immediately:
1. red ball from box A (33⅓%) --> 33⅓% for green ball
2. red ball A (33⅓%) --> 33⅓% for red ball
3. red ball B (33⅓%) --> 33⅓% for red ball
1 results in green ball, so 33⅓% chance of choosing a green ball. 2 and 3 result in red ball, so 33⅓% + 33⅓% = 66⅔% chance of choosing a red ball.
(Sorry for any spelling or grammar mistakes I might have made, English is not my first language)
the key to this problem is that the first ball you get is always a red ball. 33% its from box a. 67% chance its from box b.
@@endengineer2441 for the first situation, you said the probability of picking a red ball out of box A is 100%, but it's not; there's still a green ball. it's 50%. so since the probabilities of each of them are equal (50%, 50%, 50%), then it's out of three, not out of four.
Actually if you have 9999... repeating and add one it doesn't give you 0 since there is (at least in some sense) 1. The fact that you cannot reach the end (or like in this case beginning) of it doesn't mean it doesn't exist so its both logical and mathematical error
True for our normal number system we use. In this case we were in a p- adic system . It has a different definition.
you can't reach the end because there isn't one. the sequence is infinite by definition
This is great! Can you share the source code of the video with us?
The 2/3 comes from if the balls were all in the same box and you picked one from that one box. Separating into two boxes creates a separate problem with the constraint that you must pick from the same box.
It’s a given that the first ball is red, so you either have picked from the first box (a 0% chance of the next ball being red), or from the second box (a 100% chance that the next ball is red). Since there are two boxes to pick from, and each box can only give a single outcome (positive or negative respectively), the answer is still 50%.
The question becomes 2/3 if you have them all in the same box, because if you have picked a red, then you could either pick the second red ball, the third red ball, or the green ball. That’s 2 of 3, or a 2/3 chance.
Let's analyse the problem with the rules of conditional probability to find the definite answer. For two random events A and B, the probability of A given B, written as P(A|B) is equal to the probability of A and B occurring together (P(A and B)) divided by the probability of B occurring at all (P(B)).
In this problem we have two boxes, A and B. A contains one red ball and one green ball, B contains two red balls. We pick a box at random, then pick a ball at random from that box and note that its colour is red. We then want to know the probability that the other ball is also red. This probability is precisely the probability that we picked Box B - if we picked Box A, then the other ball is definitely green, whereas if we picked Box B, all balls are red.
For the sake of brevity, we will call "picking Box B at random" B and "Picking a red ball out of a box first" R. We want to know the probability that we picked box B given that we picked a red ball out of a box first. This is written P(B|R). By the law of conditional probability we know that P(B|R) = P(B and R)/P(R). So we want to know P(B and R) and P(R). P(R) is the probability of picking a red ball out of a box first. There are 4 possible scenarios with equal probability: We pick Box A and pull the red ball out first, we pick Box A and pull the green ball out first, we pick Box B and pull red ball 1 out first or we pick Box B and pull red ball 2 out first. As you can see, in 3 of these scenarios we pull a red ball out first, so P(R) = 3/4.
But what is P(B and R)? Well, using the law of conditional probability again, we can note that P(R|B) = P(R and B)/P(B). How does this help? Well, we can rearrange this equation to get P(R and B) = P(R|B) * P(B). We can also note that P(B and R) = P(R and B), as both are the probability of events R and B both occurring. P(B) = 1/2, as we have a 50:50 chance of picking either box at random at the start. P(R|B) is also very simple: if we know that we picked Box B to start, then our chances of pulling a red ball out first are 100%, as all balls in Box B are red. So P(B and R) = P(R and B) = P(R|B) * P(B) = 1 * 1/2 = 1/2.
Putting all this together, we know that P(B|R) = P(B and R)/P(R) = 1/2 / (3/4) = 1/2 * 4/3 = 2/3. This might seem counter intuitive. The important takeaway is that knowing the outcome of the first pull informs you about the probability that you are in a world where you picked Box B earlier. It's sort of similar to the famous Monty Hall problem.
i thought i was a math nerd...... i see now how wrong i was lol. @@LuciferLuckless thanks for the breakdown, easy to understand 😁
The counter-intuitive point is that, if for the case we pull a red ball out, there's only 1/3 chance for the box to be A, and a 2/3 chance for the box to be B
You and 19 likes are wrong. Why not get a box and try it yourself. This is called conditional probability or bayes theorem
Top commenter is right, if there is an equal chance of pulling from each *box*, it doesn't matter how many red ones are in the box on the right. The probability only becomes 2/3 if each *ball*, not each box, has the same probability of being chosen, in which case the separate boxes are just a misdirect and have nothing to do with the problem.
2:54 if you randomly choose a box with 50/50% chances, then the probability is 50%. You're talking about choosing a random ball with some magic, but to actually take a ball you need to choose a box before
for the probability problem, It stated that you're taking a ball out of a random box and that the ball is red. If you took the red ball from box A then the next ball is green. If you took the red ball from box B the next ball from the same box is red so it should be 50-50. You stated that we take a red ball out of a random box. The two options are taking a red ball from box A or from box B since you're randomly choosing the box that you choose. I'm kinda confused tbh.
edit: I now realize where I was wrong
It's conditional probability, doesn't make much sense at first but if you want to learn more, you can check bayes' theorem. For this problem, the question is "what is the probability that the other ball is red if the first is also red" which can be written as P(R2 | R1) with R1 : "the first ball is red" and R2 : "the second ball is red". You can calculate each probability separately but using a tree like shown in the video is way more intuitive and easy for this case (correct me if i'm wrong)
I agree. It should be 50%. He said ”a random box” so it’s 50/50 which box he chooses. If he had said ”a random ball from these two boxes” it’s another story. It’s super important to be clear about these things when handling probability.
@@dominobuilder100It's 66% because of the part where you already took one ball out and know that it's red. The 2 red box is more likely than 1 red 1 green to get you a red ball first, hence getting red first makes it probable that you picked the 2 red box.
No each ball in box b is 25% box b has only red balls so that's 100% which is 50%. Of both boxes. There is only 1 red ball in box a and the chance that you picked it first assuming you got a red ball first and chose box a is 100% and the other ball in there is 100% green so 0% that both balls in box a are red and 50% + 0% is 50% boom.
@@loganblakely3448 2/4 chance that you pick the box with 2 reds and get a red ball, and 1/4 chance that you pick the 1 green 1 red box and get a red ball, and 1/4 chance of getting the green ball.
You got a red ball, so the case where you get a green is now excluded.
You're left with the 2/4 of the 2 reds box, and the 1/4 of the red green box.
That's how we get the 2:1 chance, or 66%
something similar is that the percent of children who are single children is significantly lower than the percent of parents who have a single child (provided the avg children per parents is above 1)
Did they count each individual parent separately?
Intersting. Never heard this b4
Oh i get it why. Even a pair of parents is counted as one.
Imagine an exaggerated village with 2 pair of parent
One parent has 1 children
One parent has 100 children
When computed as ratio, you are one child in a 101 child that is single
Very fun video!
the potato paradox is wrong, if 1% dry mass = 1kg, 2% dry mass isn t 1kg, because 1% of water evaporates, it means it s more dry mass, because if it t less water, the potate is more "dryer"
The term "dry mass" is refering to the mass of the object that is made out of molecules other than H2O. That is not going to change as the water evaporates.
Infinite 9s repeating plus one isn't just 0, there is a carry of one after all infinite nines, so I would describe it as 1 with infinite 0s afterwards, no different than 1.0 repeating multiplied by ten to the power of infinity.
Yeah, pretty stupid math. Obviously not equal to -1.
While I do agree with you, I think the point was to say that since you can never reach the last 9 to make a 1 (maybe calling it the first 9 would be more accurate), it's like putting a bunch of 0's until the end of time and beyond. Therefore, the sum is 0, making the number -1.
To me it's just a matter of how you look at it. Because you can think about it the way you did and make it exactly what you said "1.0 repeating multiplied by ten to the power of infinity"
I personally find it more logical "your" way
This is not actually the case, because you can't actually represent a number that way, because the 1 has no position.
Any digit you ask about is 0. So the result is actually 0.
It forms a completely coherent number system for it to work like this even if it's a little counter intuitive.
@@andrewdevlin8756Look at the title
@@SlackowYes. These 5th graders are finally understanding what math is. Wait until they get to 0th power.
U deserve more subs this is just free knowledge 🔥🔥❤️
This is why I hate when we put an answer down for an equation that can never be completed, 1/3 can't equal 0.3 recurring, because we can't calculate forever and saying it is 0.3 recurring is just an estimation, just like it is true for ...99999. we can't assume that adding 1 to this equals zero as we have an infinitely large 1 waiting to be put on the end of this equation, the true number it equals in 1 ...0000. as this is the best representation for this
The reason for Potato paradox to be even a paradox is because real potato doesn't have 99% of its content as water.
It's hard to find some everyday food item to have 99% water, maybe diet coke.
Yes, things are counterintuitive when you explain the problem in a misleading way
1:08 There is a 1 leading all the zeros
No there isn't, because there is no "start" to the number.
The probability that the second ball is red:
Case 1: you took it from box A
0% chance
Case 2: you took it from box B
100% chance
"What is the probability that the second ball in this box is also red?" 50%
Case 1: you took the only red ball from box A
Case 2: you took one of the red balls from box B
Case 3: you took the other red ball from box B
You can get actual boxes and balls and do the experiment yourself (have someone else randomize them or use your own system to,) and the other ball will be red 2/3rds of the time. I tried it myself.
Before you pick, the chance of picking a box is 50:50, but after you get a red, now with this new information, the chance has changed.
Same like you can teleport randomly anywhere in the world with equal chance, now u teleport, you see an arab, what is the chance that u teleport to an arab country
Me who pays 1 dollar to play the game because it is not specified that we have to find the probability
Now do the part where ...9999999 is also _not_ equal to -1, as you can subtract 1 and get ...9999998, not -2.
...999998 + 2 = 0, therefore ...999998 = -2
But ...99999998 *is* equal to -2.
The ...999 example doesn't convince me at all. If you add 1 to it, you'd have an 1 after the infinite 000s. So it'd be more a 1...000 thing rather than a 0, and if there's no mathematical notation for it, blame mathematicians, not me.
Next time you're gonna tell me that summing all integers gives us -1/12.
I can answer that for you, but still take in count I could be wrong
you may say that that number is infinity
since that number is the sum of infinite non-converge geomtric series.
(9+90+900+9000...)
which makes sense with why you can't just simply add one into it and ask what is the number's representation in decimal base.
from the statement of the video that this number is equall to -1 ( if this statement was true..)
we could construct a proof by contradiction that this number does not exist
by saying that a number is equall to -1, we already asumed that he does exist
we can easily say that 9+90+900 is bigger that 1
we get x>1 and x=-1 , a contradiction
therefore x does not exist.
Also there is a similar thing going on with the sum of all integers gives us -1/12
(if it was true)
we already know that 1+2+3... is bigger than one
which gives us a contradiction
but I have a question is infinity a number? or can we say infinity exists?
@@ronflypotato4242 9 plus any number of positive ints will remain greater than 0, -1 is less than 0, (aka ...999 > 0 > -1), ...999 = -1 contradicts this. This is the same as your proof, right?
Yeah, the argument in the video seems like just semantics imo. Like, sure, if you do the addition manually and carry the one you get an infinite number of 0s, but that's just a consequence of the way we write numbers. That doesn't make the actual value -1.
yeah i used to believe "adding the natural numbers gives us -1/12" but it's more like "adding the natural numbers would diverge and not have a value, but if we extrapolated from other results we'd get -1/12"
The last one is weird. If there is a fixed expected value why would the average win grow with attempts? I mean it was like looking for the average win of a lotto ticket and the same thing happens. OK it's at night and I am super tired. Maybe I look at it again tomorrow.
Because it only goes to infinity when you take into account extremely rare events, 20 flips until heads for example gives 2^20 dollars but also has a probability of 1/2^20, and the more simulations the more likely to get those.
And 5 seconds. You had ONE JOB.
I got the red ball problem right! I think it's the same principle as the Monty Hall problem.
2:12 only if you do not know the box you chose from, because it really does not matter which red you chose, but yeah if you took one from box then if you swap boxes aronud, 66-33 , but if you choose and it is red, choosing next is 50% if you scramble it will act like if it is one box anyway
Probability Problems like box one can be fixed quite easily with Bayesian thinking! Although its not very natural to think of them when sample space is very small... So, it gets very paradoxical.. I think the ball problem is equivalent to monty hall problem?
But dont we add a number to the left of the decimal? unless its a decimal
Also matching problems are just dearrangements? And then the formula of derrangements just converge to 1/e as the sum goes to infinity!
You need more subs my dude
awesome video
BRO How has no one seen this yet, this is so inetersting
You need to include 1+2+3+4+......= -1/12 (and sometimes -1/8).
This video was AMAZING. i need a full series of this. This gave so much vsauce vibes
Mathologer did a good video on this:
czcams.com/video/YuIIjLr6vUA/video.html
Well that sum is not counterintuitive, its just not true. It has to be true and proven in order to be counterintuitive.
@@UltraAryan10 You can make the same argument about n-adics here.
@@RuthvenMurgatroyd P-adics can be treated as a different branch of numbers seperate from real numbers. This sum, people try to put into natural numbers set and thats just not true.
@@UltraAryan10 the sum can be defined mathematically just like ...999999 is my point even though both are divergent and have no _real_ meaning.
the anwser to the probability problem is 50%, the balls from the second box cannot travel into your chosen box via osmosis, so you only have a single ball that is either red or green, not three balls that are red, red, and green
You picked a ball randomly and it was red. Given this, three cases are possible:
• You picked the red ball from box A. (The next ball will not be red)
• You picked ball 1 from box B. (The next ball will be red)
• You picked ball 2 from box B. (The next ball will be red)
2/3 of the cases are valid for our conditions. Probability is 2/3. It is not 50% because the probability of picking a red ball from box B is higher than that of box A.
The chance of the first ball being red is higher if the box you picked is the one with 2 reds. That's why it isn't 50%
Slab amogus
I think if you make that example with people and the color is represented by gender it's easier for you
Just do to experiment yourself and get 2/3, it's not hard to set up
When you said "P, in fact," I thought for a second that you said "being f*cked"
10-adic numbers cant represent 1/2 nor 1/5, which is why its a lot more useful to represent them while p is prime
Yes it can, 0.5 and 0.2.
@@martind2520Decimal numbers doesn’t exists in the P-adic system
@@quentind1924 Yes they do. P-adic numbers have no issue with decimals. The only thing you have to avoid is _infinite_ decimals.
The reason they tend of a avoid decimals is that when p is prime, the decimals are unnecessary, but when p is not prime, you need the decimal part to be able to represent all the numbers.
I don't understand the last game. When does it end? What is the choice I have to make? How much I I bet per coinflip? Or per total game until it ends? Do I have to say how many times I want to flip until I get heads? Because if we play once until heads comes up, I don't think it would be wise to bet infinite money. Or do I have to say at which flip heads comes up and if I'm right I win?
I really don't understand what I bet on and how long it is played aka the winning conditions and what choice I make that I bet on.
the point is that expected value is not always the whole picture
The game ends immediately when you get a head for the first time, if tails , continue playing.
You get the money according to how many times the coin is flipped in a run of a particular game, according to the table, the longer the run the bigger the money.
The problem ask what is the maximum money you SHOULD pay to participate in this game., as an entrance fee to play a round.
Here's a fun demonstration. For a number x of length k, the periodic number 0.(x) = x/99...9 where there are k values of 9. Try it yourself. Therefore 0.(9) = 9/9 = 1.
P.S. A demonstration for that formula is that 10^k * 0.(x) = x.(x) = x + 0.(x)
So (10^k - 1) * 0.(x) = x
0.(x) = x/(10^k -1) = x/99...9 where 9 is repeated k times.
thats a neat fact!
0.(9) isn’t in the P-adic system. ...999 is not the same thing than 0.999... at all
@@quentind1924they werent talking about p-adic numbers at all
@@shieldgenerator7 It’s literally written during the entirety of it
@@quentind1924 i dont understand why you think @masscreationbroadcasts is talking about p-adic numbers
I showed my friends this video as a p-adic primer and it broke they brains. Thanks.
what do you use to make these
The sine problem only has irrational numbers if you're using degrees
Bro really said 99% of gambling addicts quit before they hit big 💀
Full support 🎉🎉🎉
To say that 0.999...=1, you have to be a little more careful. It's easy to show that the largest number that is possibly smaller than 1 is 0.999..., and hence either 0.999...=1 xor there are no numbers between 0.999... and 1 (and 0.999... is smaller than 1). But to show that it must be 1, you still have to use e.g. that the set of real numbers is dense - there cannot be no numbers between any two real numbers. Of course, you can also easily sum the infintie series of 0.9+0.09+0.009+... as it is a geometric series.
After seeing the last expected outcome game, I noticed another way to find 0.999...=1 using probability calculus. Note that { the sum over all positive integers n of [the probability of getting the first head at the nth toss] } must be 1, since in each game [you must get the first head at the nth time for some positive integer n if you do end up getting a head] AND [the chance that you never get a head is limit n goes to infinity of (1/2)^n, which is 0], and summing over the probabilities of all [the possibilities of non-zero probabilities] must yield 1. And this probability sum is also a geometric series 1/2 + (1/2)^2+(1/2)^3+..., so this example tells us that you can make an exact analogy between the probability sum and the geometric series.
For the series 0.9+0.99+0.999+... = the sum over all positive integers n of 9/10^n, let's consider a random number generator that generates all numbers from 0 to 9 with equal probability - let x be an arbitrary number from 0 to 9, then the probability of getting [any number that isn't x] for the first time after using the generator exactly n times would be 9/10^n. As before in the expected outcome game, { the sum over all positive integers n of the probability of getting [any number that isn't x] for the first time after using the generator exactly n times] } must be 1 by probability calculus, and must also be equal to the sum over all positive integers n of 9/10^n, and hence we obtain 0.999... =1.
That potato one isn't surprising; it's just weirdly worded. I wouldn't call reducing the amount of water by a small amount (basically half) dehydration. It's a semantic trick question. What is actually happening is you are taking away 50 kg of water (so 50/99 % of the water, or basically half), and yet leaving the 1 kg of non-water, so of course, the result is half of 100 kg which is 50 kg; as you took away 50 kg of water, but specified this in a deceptive way. Of course this is a deliberate trick question, which is why it is worded that way.
The probability problem's solution makes no sense at first, but if you think about it a lot it starts to make a little sense but still not really
Well, if you use the number line argument, 0.9 repeating could also be 0.9 repeating, and at the end an 8. There is no number between them.
The whole point is that there _isn't_ an end to the 9s.
Let x = 0.99999....
Then 10 x = 9.99999...
Then 10 x = 9 + x
Ergo 9 x = 9, and x = 1
So 0.9999.... = 1
QED
@@stevenvanhulle7242 That one works.
I don’t agree with the first demonstration because I don’t believe 9 repeating is a number, therefore 9 repeating + 1 would not be zero, it would be undefined. Feel free to correct me if I’m wrong on that.
Very fun video!
It’s a different number system. In this "mathematical world", ...999 exists. But if in real situation someone asks you what ...999 is equal to, say that this number doesn’t make any sense unless you are in the P-adic system
0.9 recurring is equal to itself
Yes and it is also equal to 1.
0:09
There is no number between 0.000(repeated)1, so that equals 0?
MY POINT EXACTLY, YOU CAN DO THAT INFINITELY FOR EVERY POSSIBLE CONCIEVABLE NUMBER SO AT SOME POINT IT STOPS MATTERING AND YOU CAN JUST SAY THAT EVERY NUMBER IS BASICALLY THE SAME NUMBER, ITS RIDICULOUS
The problem is that you're adding a one after repeated zeroes, that doesn't make any sense. 0.000... = 0
great video
This video has the perfect length for a snack