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.
@The9thDoctor but it's wrong whether it's ckybyerintuitve or not. There's 365 or 366 days in a year..so there's no way with only 23 ppl that's the percentage of sharing same birthday without added constraints..that problem is clearly wrong..see what I mean..since for example the odds of teo ppls haring the same birthday is (1/365)(1/365) which is way less than 23 percent
@noomade my maths not wrong..of course it's 1/365*1/365 each term represents one of the two ppl..your matht is wrong 1 times 1/365 is just 1/365..did you mistyped something?
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.
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
...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
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.
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
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)!
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...
@@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.
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.
@@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
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 $5 and no one is sitting there playing 10 billion times
For last one it doesnt only matter how much money you are willing to pay but also how many times you are willing to play, looking at numbers best way is to play 1000 times and max money for that is 5,6$
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 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 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.
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.
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%)
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
@@diegonals Well I don't remember what was in my head when I wrote this, but now after rewatching this problem I think it's really obvious why the probability is 2/3. So yeah, I was wrong.
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
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 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.
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?
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.
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: ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-RBf1s4TassI.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.
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.
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 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
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
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.
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 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.
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!
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
@@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.
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%
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.
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
I hope we can come to grasp the concept of infinity better in the future, it is a really prominent one in the universe and thus could be really helpful in understanding it and how it word. Which in turn could help us advance as a civilization further (hopefully leading to human domination of the universe).
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.
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
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.
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!
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
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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.
I don't agree with the boxes ones. You pick a box and you're guaranteed a red ball. So the problem becomes whether you picked the box with 2 red balls or not: 50/50. For the record, I do understand Monty Hall and I don't think there's a connection to it like some comments suggest. In Monty hall you get additional information and the option to modify your choice, here you don't get anything. Your initial 50/50 choice dictates if your second ball (from the same box) is red or green.
No, the boxes one is correct. You're right in that your chance to draw two red balls is 50%. It's just that the additional information the question gives you says that you're in a subset that only happens 75% of the time (the *first* ball you draw is red). So the chance of the second ball being red, given that the first one was red, is 50%/75%, or 2/3 (which I think would also have been a much more intuitive way of visually demonstrating it in the video). I agree that this isn't connected to Monty Hall though. That's an entirely different issue.
@@ffbotha Ah. I think i misinterpreted the problem. My assumption was that the first ball is guaranteed to be red. But yes of course, if the first ball being red is a probability, then it's correct.
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.
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 ...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"
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.
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.
I dont know about this but from what i learned in school ... 999 could be written as the lim of 10^n-1 as n approaches infinity and this lim does in fact reach infinity
The …999 is provably divergent and therefore cannot have a limit of -1. On the other hand .999… is a convergent sequence so we can find its limit (which is 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.
...999999=-1 reminds me of how in 2's-complement binary integers "largest" value (all ones) is also -1. You feel like that behavior only happens with closed set of numbers like binary integers with finite number of bits, but in reality when you have concept of infinity you can do same thing with infinite set, you just add infinite number of digits in front of it!
2:10 Box A and Box B: 50% + 50% Red ball in Box A: 50% Red ball in Box B: 50% + 50% Red ball from Box A: 25% Red ball from Box B: 50% Chance of taking out an red ball: 25% + 50% = 75%
You didnt account for the assumption that the first ball was already taken out and found to be red. This eliminates the possibility of drawing the green ball first, so it goes from 3/4 to 2/3.
I roll a dice, i see the result and i say that it’s smaller or equal to 4 (and i can’t lie. If you don’t trust me then we do it with a dice that makes a very loud sound when it rolls 5 or 6, and you heard nothing. That way, you can be sure that it is smaller or equal to 4). What’s the probability that it’s a 1 ? With the ball puzzle, it’s the same thing : Box A + box B = 50%+50%. Box B red ball A 25% (other one is red), box B red ball B 25% (other one is red), box B red ball 25% (other one is green). And box a green ball 25% but you see that it’s a green ball so it doesn’t work therefore you discard it and it’s 0% Yes, now you only have 75% of anything, but let’s say that on the remaining 25% you reset which leads to all 3 scenarios with equiprobability. So it’s 33% to get box A red ball (other one is green), 33% to get box B red ball 1 (other one is red) and 33% to get box B red ball 2 (other one is red). So ⅔ to have the other one be red
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?
The "periodic numbers" aka second section is wrong. it takes advantage of series notation to give an answer that seems accurate, however isn't. As the proof your basing it on used series notation, something originally meant to be used to determine if a series is infinite. while series notation certainly has other usages, it is ultimately an approximation utilizing an abstraction-rather than a "number"-. the proof basically says proves that .9 repeated k times is equivalent to 10^k-1. which is true, and can be approximated-aka rounded-to be 10^k. furthermore, it is also true, that 10^k can be even be considered equivalent-to .9 repeated k times-for most practical purposes, perhaps even theoretical purposes, however IT IS NOT THE SAME. Ultimately this proof is based on a theoretical number, specially an infinite series of 9s, a number ideal for obscuring a lie due to faults in our base ten number system. which is then increased by finite number, namely 1. and then uses this to make a rather convincing lie... I probably (defiantly) didn't explain this perfectly, if you disagree with me, go learn and research yourself, either you'll learn something yourself, and somehow prove me wrong, or find out WHY I'm right in your own way. Someone agreeing with me, or a more intelligent conversationalist our there somewhere, I look forward to either alternatives eagerly. Closing Note: infinity is weird, and saying 1=.9999...etc is kinda right, as it certainly is for any rounding purposes, however to say that they are the SAME thing feels like an abuse of infinity which hinges on the idea that it cannot simply repeat forever... as for the -1=999 proof, that is almost certainly an abuse of infinity. as infinity is more of a concept than an existing number, so its possible to use it in misleading ways. looping back for a moment, any proof attempting to solve it either way MUST deal with infinity, as there are an infinite amount of numbers between 1 and 2. it seems far more accurate to say that .9999...etc should be the theoretical number closest number to 1 approaching from the negative side,-if it were to exist-. TLDR: mathy no work right when infinity and a finite number added.... go research to become smarter and more interesting. If you want I guess.
@@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.
It's misleading, this is supposed to be an integer (...9999999.0). Still, this is only useful if you're doing periodic math. Usually it's more useful to call this number infinity
For the last one. What matters is not the expected value that you win, but if you have a chance of at least 50% to end with more money, or at most 50% to end with less money. Because of the chance for high returns, let's choose at most 50% to end with less money. So if we only play once, I would say I want to give 2 dollars to play. In reality, no lottery will make games like this, because they want the opposite: games where people want to give more than the expected value to play, not less than the expected value.
So if someone gives you ⅓ chance to win a million dollars but you have to pay 1 dollar to play it you wouldn’t ? I agree that his value isn’t really fitting the real world because nobody would gamble his house to have 50% to double the value +1 dollar (even if mathematically it’s worst trying), but your reasonning also has big flaws
You're right, it depends case by case. 2 dollars is a rather 'save choice'. Was reasoning that expected value is not the only thing that matters, and that that is logically not wrong. Another thing that matters is also how much money you have. For your case, one dollar I would do, and somewhat more too ;), but indeed certainly less than 500000. I think the basic message is that although you could think that our instincts are wrong when you calculate something out, a lot of times our instincts are not wrong, it's just that the calculation perspective is too one-sided and misses something important. @@quentind1924
6:38 theres actually a 51% chance to flip heads and 49% chance to land tails when starting from heads 😳 With this we can say (51/100×1)+(49/100×5)= 2.94 and therefore you would lose money overtime if you bet 3$ every time as long as they start from heads
solve for the characteristic polynomial of (i | x | i | y 12 | z | 0.8262 | z 72.6 | 0.6263 | i×8 | x 9 | y | i | 153) (this is a challenge for people and the answer is in replies)
