3:14 "There's a literally infinite number of variations of this problem, so we don't have time to reveal them all." *UNLESS* you explain each one in half the time you used for the previous one!
Challenge problem: There are infinitely many balls left in the urn at noon. In particular, all balls except the multiples of 5 are left. In fact, on the nth step, ball 5n is removed. This is equivalent to saying that on the nth step, the following balls are in the urn: all balls from 1 to 9n, excluding all multiples of 5 up to 5n. This claim can be proven using induction. The claim holds for n=1, since balls 1 through 9 (=9n) are put in, and 5 (the median) is taken out. Now suppose the claim holds for some n; then after the nth step, we are left with balls 1 through 9n, excluding 5, 10, 15, ..., 5n. Now let's see what happens on the (n+1)th step. Balls 9n+1, 9n+2, ..., 9n+9 are added. Now, in order for the claim to hold, we want to show that 5n+5 is the median of the balls currently in the urn. There are 8n+9 balls in the urn total (9n+9, minus the n which have been removed). Further, there are 4n+4 balls whose numbers are less than 5n+5 (because 4n+4 is 5n+4, minus the n balls which have been removed, each of which had numbers less than 5n+5). Consequently, there are (8n+9)-(4n+4)-1 = 4n+4 balls in the urn whose numbers are greater than 5n+5. Therefore, ball 5n+5=5(n+1) is in the middle, so it gets removed, thus proving the claim for (n+1). Okay, so every multiple of 5 gets removed, and every ball that gets removed is a multiple of 5, leaving infinitely many balls in the urn.
For anyone wondering what the deleted comments were: someone didn't see (n+1) and thought Florence was suggesting putting the balls 9n+1, 9n+2, ..., 9n+9 in the urn on the nth step, and he pointed out that you would need to start at n=0 if you did it that way, or you would never add the first nine balls. Interesting that the commenter pointed out that if you do what Florence said, but on the nth step instead of step n+1, but still started at n=1, you would remove all of the balls that are one less than a multiple of five
Congratulations Florence! You are the winner of this challenge. Please send us an email @ pbsinfiniteseries@gmail.com with your shirt size and shipping address.
I think "adding infinitely many zeroes" is worth slightly more consideration. First, it is very important that we add *countably* many zeroes when we want to conclude a sum. Second, there are two simultaneous limits involved: the limit over the number of steps and the limit over the highest ball number included in the disjunction. Just taking one limit then the other doesn't necessarily mean you get the correct answer, there are cases when you can get any answer you want just by manipulating how fast you let each limit approach infinity with respect to the other.
I am currently studying mathematics, and I am literally in love with it. I was just so confused about continuing to do that and make math my entire career and switching to medical school, but when I see how beautiful is this universal language I push myself more to continue this path since I began it already. You guys have played a huge role in doing that, so please continue the amazing work and make sure that you are a very successful, professional and unique channel. 💓💓
I didn't prove it, I just wrote down a few scratches on a piece of paper, but it seems like only multiples of 5 will get removed from removing the median. So infinite.
You're (possibly) thinking of the median of the added balls. I.e., removing 5 from 1- 10, 15 from 11-20, etc. But that's not what's happening here. You're removing 5(or 6, as one definition for a median of an even number of values is the average of the two middle values, which would be 5.5 in this case)) from 1-10, 11 from 1-20 (minus the 5 that was removed), 16 (or 17) from 1-30 (minus the previous 2 removed), etc. It all still seems silly to me, though. You're still adding infinite sets of (10 - 1) balls, heh.
For the median case she was only adding 9 balls in each step, so you'd add 1-9 and then take out 5, then add 10-18 and take out 10, then add 19-27 and take out 15. It would only be multiples of five that get taken out.
Am I missing something, or is the idea of a time restriction arbitrary here? Saying "At noon" seems to be a roundabout way of letting N tend to infinity, and would have the same result
Vriskanon Probably just a kind of response to a critic of thinking "how things would look like after infinite steps" "Well...you can't get to infinite steps, because this would take infinite time! You will never get there!" "Well, Mr. Smartpants, now I will do each subsequent task in half time and it will take me a finite time to end! Take that!"
Vriskanon You realize that number of balls tend to infinity as N -> infinity, and the limit is also infinity. Thus the results would differ. Unnecessarily rigorous proof: Let a_n be the number of balls at step n. By definiton, if the limit of sequence a_n is infinity, for every R>0 there exists N>0 such that a_n>R for every n>N. Choose N=ceil(R). Now number of balls is a_N=9*ceil(R)>R. Sequence a_n is increasing, thus a_n>a_N>R for n>R. QED. This reasoning doesn't work anymore with supertasks, because now you are in a sence going _beyond_ the limit - you terminate at some point and you have a (more or less) well defined final result that can be analyzed. With limits you never terminate, you just keep going and going, there is no "final step" (technically there is no final step in supertasks also, but I hope you get my point).
+Tetraedri_ The limit of the number of balls in the urn as the number of steps tends to infinity is infinity, but the limit of the set of balls in the urn as the number of steps tends to infinity is the empty set. That's true whether you borrow a trick from Zeno to get there in finite time or not. The question then is whether the limit value for the number of balls should be the limit of the number of balls or should be the number of balls in the limit set. Generally, it's accepted to take the actual limit set and count the balls - which makes the answer zero.
rmsgrey True, didn't think about the limit set. However, if we didn't terminate at some well defined time, there is no meaning in saying "after we're done." If you still was considering only the cardinality of the set, it would tend to infinity, thus at any time there are non-zero amount of balls in the urn, tending to infinity. The limit set has to be reached for the question to be sensible, and without supertasks it can't be reached.
Alejandro Bravo the limit of x*y where x is an expression that goes to infinity, and y is an expression that goes to 0, could go to different values depending on how x and y go to infinity and 0 respectively, but if y is a constant 0, then the limit will be 0, because the expression is a constant 0
Problem is that 0 in this case is not constant 0. It was derived from convergence so it should had been taken into account. I have no idea if it changes the results but when doing calculations with infinity you can't substitute infinity in one place without substituting it in all others and it is what happens here. First infinity is substituted when calculating value of infinite series and then when multiplying the result by infinity.
for any single ball, the probability of it not being picked after infinitely many steps is the number zero. the number zero has no limits in it's definition, it's plain and simple, add zero to itself infinitely many times and you will get zero.
It is: given that number of steps goes to infinity the probability goes to zero. Simple example why you can't split limits: Let's say A = lim x/(2*x^2) so we say given that x goes to infinity the A goes to zero. Then we ask, given infinitely many tries (n) what is probability of event A happening (let's call it event b): B = lim n * A Using naive interpretation of (A is constant 0) A = 0 so B = 0. If instead we substitute the full expression for A as the result isn't constant 0 it is just close to 0 and given that both those limits have the same rate (they do): B = lim(n) lim(x) n * x/x^2 = inf * inf / (2*inf^2) = 1/2
The video has a typo, but the cited article has a correct sequence of (9n)/(9n+1). Basically the power operation should not be there, it's multiplication. But then the term 28/29 should not be there as 28 is not a multiple of 9... argh, this video is seriously messed up! but the material is interesting.
5:29 The terms of that series do not resemble 9^n/(9^n+1). The formula would give the second term as 81/82, which is different than 18/19, and so forth. (Edit: Never mind; other poster points out that probably 9*n/(9*n+1) was what was meant) I assume some of the reasoning was not shown; most likely the series are related by an inequality and 9^n/(9^n+1) is an upper bound on the value that the nth term can take, so the smaller terms in the true series only make the probability of the ball remaining smaller.
For the supertask at 1:16, the amount of remaining balls isn't 0, it's 9∞. The only balls remaining after step ∞ are balls ∞+1 through 9∞. Just because the numbers of those balls are greater than ∞ doesn't mean that they don't exist. No matter how you look at it, you're adding 9 balls each step. If you do the tasks ∞ times, you'll always get 9∞ balls regardless of the subtraction order.
Major twist: In the deterministic version of the paradox, it doesn't matter how quickly you add the balls -- you can go 10 at a time, as described in the video, or double the number you're putting in at every step. It won't change the fact that every ball is eventually removed. In the stochastic version, though, it could. Yes, the infinite product giving (say) the probability that ball 1 is still in the urn at noon converges to zero -- its logarithm is an infinite sum that grows at least as fast as (some scaled version of) the harmonic series. But this fact depends on the fact that the number of balls is the same at every step. If the number of balls added at a given step grows over time, the product could converge to a non-zero probability!
So, when adding all the probabilities together, what you do is taking the limit of the number of balls N->infinity, while summing up the series P_1+P_2+...+P_N, where each P_i also depends on N. It looks like you cannot just interchange the limit and the summation. Each P_i goes to zero as N becomes infinite but does the sum of the P_i go to zero as well? I don't find that obvious.
So, I rarely have an actual grasping of maths, and generally only watch these videos each week to have my mind blown. This week, my mind was blown because I actually understood it all! even more, I have enough confidence to answer the challenge question... The part I found hardest about this challenge was finding the median of the series, having removed specific balls... So, If you add balls 1-9, then take out the median, you will take out ball #5. then, if you add balls 10-18, and take out the median of the remaining balls, you will take out ball #10. repetition #3, you will remove ball #15. this pattern repeats, where repetition #n, you will remove ball #n*5, which means that balls that are not a multiple of 5 will never have the opportunity to be removed. Because there are balls that can never be removed being added with each repetition, after infinite repetition, there will still be infinite balls in the urn. To help find the median of each repetition, I used R, feel free to check it out/point out errors in my code; I have set it to 100 repetitions, which shows the pattern very clearly. i = setdiff(1:9, (median(1:9))) while(max(i) < (900)){ a = (max(i)+1):(max(i)+9) b = c(i, a) i = setdiff(b, (median(b))) } setdiff(1:(max(i)), i) #setdiff makes a series, with the second argument being individual numbers in the series being removed; I use this to remove the median. #c(i, a) is a function that combines the series i (which is the last repetition), and a (the balls being added). #I have to use 3 steps to add the new balls, combine the balls into the pool of *all* the remaining balls, and then to remove the correct ball. #the final line shows all the balls that were removed
As n approaches infinity, 9(1)/[9(1)+1] x 9(2)/[9(2)+1] x ... x 9(n)/[9(n)+1] approaches 0. It does not equal zero. However I can get behind the concept of calculus and say ball 1 has zero chance of remaining after infinite steps. This equation doesn’t apply for every ball however. If we look at the balls as groups of ten this equation only applies to the first group (balls 1-10). For group 2 (balls 11-20) they are not subject to the first step and are guaranteed to be there for the second step where the balls from group 1 only have a 90% chance of being there for step 2. Group 2’s equation should look like 9(1)/[9(1)+1] x 9(2)/[9(2)+1] x ... x 9(n)/[9(n)+1] x [9(1)+1]/9(1) And group X’s equation should look like 9(1)/[9(1)+1] x 9(2)/[9(2)+1] x ... x 9(n)/[9(n)+1] x [9(1)+1]/9(1) x [9(2)+1]/9(2) x ... x [9(x-1)+1]/9(x-1) It’s easy to see that there are terms that cancel out, and the equation can be simplified to, 9(x)/[9(x)+1] x 9(x+1)/[9(x+1)+1] x ... 9(n)/[9(n)+1] as long as X is a finite value there will remain an infinite amount of terms that don’t cancel out and there fore there is still a zero percent chance of any ball in group X remaining after infinite steps. However, as the amount of steps increase so do the amount of groups. In fact since they increase at a 1-1 rate we can say for any step n there is a group n. And it’s equation can be written as, 9(1)/[9(1)+1] x 9(2)/[9(2)+1] x ... x 9(n)/[9(n)+1] x [9(1)+1]/9(1) x [9(2)+1]/9(2) x ... x [9(n-1)]/9(n-1) Cancel all terms and you are left with, 9(n)/[9(n)+1] As n approaches infinity this equation approaches 1. That is to say the odds of any given ball in group n remaining after step n is 100% after infinite steps. Similarly for any group (n-x) the equation gets reduced down to, 9(n-x)/[9(n-x)+1] x 9(n-x+1)/[9(n-x+1)+1] x ... x 9(n-x+x)/[9(n-x+x)+1 And as n approaches infinity, for every finite x, this equation approaches 1. So there are an infinite amount of groups that each have 10 ball all with a 100% probability of remaining after an infinite amount of steps.
For the challenge problem, the urn will contain infinitely many balls at noon. To prove this, let m_n be the number on the ball which is removed during the nth step of the described process. I will prove that for all n >= 1, m_n = 5n. We proceed by induction. For the base case, one can check that m_1 = 5. For the inductive step, we can see that at the nth step, we begin with 8(n-1)+9 balls and remove the (4(n-1)+5)th ball, which will leave us with 8n balls at the end of the nth step. We begin step n+1 by adding 9 balls, for a total of 8n+9 balls, and we will remove the (4n+5)th ball. Our urn now has 4(n-1)+4 = 4n balls with numbers below m_n and 4n+4 balls with numbers below m_(n+1). Thus there are precisely 4 balls with numbers between m_n and m_(n+1). Since we have not removed any balls with numbers above m_n, and since m_n = 5n by our induction hypothesis, we know that m_(n+1)=5(n+1). Therefore, for all n >= 1, m_n = 5n. Finally, since infinitely many natural numbers are not multiples of 5 (exercise left for the reader), infinitely many balls remain in the urn at the end of this process.
~ 6:30 the probability that You would have removed any one ball isn't zero, it simply converges into zero, meaning that the more steps that You perform, the closer You get to zero, meaning that it's infintessimaly small, n/infinity. when You add up this infinite series of infintessimales (not 0 but rather n/infinity) You get (n/infinity)·infinity, not 0·infinity. when You simplify (n/infinity)·infinity, You are left with n, a 1 in 1 chance of any given ball remaining after an infinite number of steps. since the proper term is not literally 0 but rather n/infinity (which many people would incorrectly simplify into a literal 0) the 2 infinities cancel each other out when You keep adding this infintessimaly small probability (multiplying it by infinity).
SORRY that's wrong at 2:11 your logic says that by Nth iteration I have removed total N balls. But the total number of balls you have put inside the urn till then are 10N!
As you drew the comparison to the darts board: Hitting a particular point on the board has chance 0, but hitting the board has chance 1. So here, why can't we have every single ball have probability 0 of remaining in the urn, but hav the probability that any ball stays in the urn to be 1? In fact, adding up infinetely many zeroes in 6:33 seems like an indeterminate form (infinity * 0), especially as those zeroes aren't really zeroes but an infinite product that approaches zero. I think everyone that ever heard something about limits should get really sceptical at that point...
'Physics tells us that space and time are not infinitely divisible' Physicist Wheeler said time is squeezed into a moment, while space is the sum of all the events happening to you. What follows from the big bang is 'Sequential and simultaneity'.
0 - In cases where we remove lowest number, median and random number ball. Because in these cases we perform operation on all balls in the urn. ∞ - In cases where we remove last ball (ball number = step*10) Because in this case we perform operation only on balls introduced at Nth step.
This is my solution to the proposed problem: After some tests, we can see that we are taking out multiples of 5. In the first step, we take out ball number 5. In the second, the ball number 10. Generally, we take out the ball of number 5n in the nth step. Proof: The median of a distribution of k terms (k is an odd number) is given by (k+1)/2 In the problem we have: (9+1)/2 = 5 (the position of the first ball to remove) We now have 8 balls and add 9 more: (17+1)/2 = 9th position And so on. But, we have to add an extra term to find the number of the ball directly. Every time we look for the next median, we pass through a position that we removed in the last step, so we need to add 1 to compensate and get the correct number. For example, in the second step, the median is the 9th position (as seen before) but it doesn't correspond to the number 9. Indeed, it corresponds to the number 10, once we removed the ball number 5 and it's not more on the group. Generally, in the nth step we have: Ball number = (9 + 8(n-1) + 1)/2 + (n-1) (That is, the median position and the extra term to correct the number) Simplifying, we get: Ball number = 5n So, after infinitely many steps, we removed all the multiples of 5 and all the other balls remained in the box. Answer: Infinity balls. (Sorry if my English is not good 😅. I'm from Brazil)
Daniel Fernandes eu acho que fica mais claro se você separar as duas ações, primeiro adicione as bolas de 10 em 10 até o infinito e depois remova as bolas de 1 em 1 até o infinito
Daniel Fernandes - Imagine that instead of adding 10 balls each time, the urn is already full of an infinite amount of balls. If you then take out 1 ball an infinite number of times, then the urn will be empty. Adding balls each step just makes it so that the urn will have an infinite number of balls in it by the time it's noon.
Same. In the first example where she adds ten then removes the smallest, how would there be zero at noon if the net change is 9 balls after each subtask? You add ten, remove one, add ten, remove one, etc. The only way I see it becoming zero is if you stop adding balls. Stating that in the Nth step you remove the Nth ball is valid but it leaves out that in the Nth step you have 9 * Nth balls.
Daniel Fernandes It's based on the assumption that 9*infinity is exactly equal to 1*infinity so it doesn't matter how many balls you add so long as you take 1 away.
Oh Boy! People get very angry when you challange their intuitions about infinity. I wish infinite amounts of patience to be delivered until the end of next week (perhaps you can ration it to have some to spare later)
6:45 0+0+0... Is not necessarily 0. This could be thought of as 0*infinity, which is undefined. The zeroes being summed together aren't exactly zero, they are infinitecimals. They behave like zero in most cases, but not when you are dealing with an infinite ammount of them.
Probability measure is countably additive. This means that we can sum the probabilities of a countably infinite amount of disjoint events. In our case we have precisely that, so the result follows.
This is the problem I still have with seeing infinity as a number. I've been taught that any number is a concept, but that infinity especially should not be treated as a number. That's how we get ridiculous statements like 1 = 0 after subtracting ∞ from the equation ∞ + 1 = ∞. Even saying there are an infinite number of even and odd numbers, and an infinite number of only even numbers (despite there being twice as many total numbers) yields different sizes of infinity. This is baffling to me and infinity as a concept should always be the same. This is why we should fundamentally change the way we think about infinity, and I cannot accept the current model philosophically.
It is quite clear that in all ways you described to remove the balls, the number of balls are increasing at each step. For example, when we insert 10 balls and take out one at each step, there must be ∞ balls left because at nth step, 9n balls are left. Therefore, at ∞th step, ∞ balls must be left. Also there is no paradox as ∞ balls are put and ∞ are taken out. But, ∞-∞ is an indeterminate form. Therefore, the right calculation must be in the way I described above. Am I right in your viewpoint? Please let me know.
I think the paradox can be resolved by booking the balls into that well known hotel containing an infinite number of rooms, as follows:Number the first room 10^-1, the second 10^-2 and so on. Place ball numbered k in the room numbered 10^-k. Define a “room occupancy index” number q as the sum over all occupied rooms of 1 multiplied by the room number. For example q = .1001 would indicate rooms 1 and 4 are occupied, all other rooms are empty.According to the first definition of the problem, balls numbered 1 to 10 will be allocated to rooms numbered 10^-1 to 10^-10, before removing ball number 1. So q(1) = 0.0111111111 is the room occupancy index after this iteration. The number of rooms occupied (ie balls still in the vase) is equal to the total of the nonzero digits in q(1) - ie 9.After each subsequent iteration the room occupancy index will increase by 10 digits, comprising an extra zero at the start and 9 ones at the end.After n iterations, the room occupancy index will be equal to q(n) = 0.000…0001111….111 where there are n zeros and 9n onesAs n tends to infinity, q(n) tends to zero. The paradox seems to arise if this is wrongly interpreted as implying that there are no balls left in the vase. But q(n) contains 9n non zero digits so the vase contains 9n balls. Paradox resolved.According to the second definition of the problem, every tenth ball is removed. In that case we have q(n) = 0.11111111101111111110…… which clearly does not tend to zero - the vase contains 9n balls. No paradox arises.According to the stochastic definition, q(n) will again contain 10n digits, of which 9n digits equal one and n digits equal zero but this time the zeros will be distributed at random. Again the vase clearly contains 9n balls. No paradox.
Perfect proof of the different size of infinities. Removing any ball infinite many times means infinite balls have been removed, even though the actual number of balls goes up by 9 an infinite number of times.
6:40 - the problem with this logic is that 'infinitely small' or 'approaches zero' is replaced with "equals zero". Adding a bunch of 'infinitely smalls' is basically an integral function. you're adding up a bunch of infinitely small values approaching infinity.
No, you have a product from 1 to infinity. That one you can evaluate to be exactly zero. Then, that product (or its value zero) are summed up, and every term of the sum is the same (the product is constant with respect to the summation index). Therefore, we are actually summing over actual zeros and that means the end result is zero.
when you always remove the ball with the median value there will be infinite balls in the urn at noon, right? because the median value constantly gets bigger and the balls with a number below the median value will approach infinity.
Answer to the question: infinite. On each the nth step we got 8(n-1)+9=8n+1 balls. The median is the (4n+1)th ball. We removed n-1 balls so the ball we take out (the (4n+1)th ball) got the number 5n on it. So the only balls we take out are those which got a multiple of 5 on them. Hence, every other ball stays in the urn, and we got infinitely many of them.
On the challenge problem, Add 9 balls and remove the median, then add the next nine balls and remove the new median, repeat. A nice pattern emerges as we see the balls that are multiples of 5 are removed. This can be proven: When we pick the nth ball there are 9n-(n-1) balls in the urn. We can find the median first by adding 1 and dividing by 2. Then we add n-1 to account for the balls that having been taken out. The nth ball removed is equal to (9n-(n-1)+1)/2+(n-1) =(8n+2)/2+n-1=4n+1+n-1=5n. Now since there are 4 balls before each ball taken out and after the previous ball, there will be 4n+4 balls in the urn after n steps, as n goes to infinity we get infinity balls left.
Supertasks should be including derivatives-so the first task is no different outcome than the second because it doesn't 'matter' whether you call it, the lowest numbered ball: it's going to fill up not diminish...
7:01 What I don't get: Each single infinitesimal point on the dart board has probability 0. And yet if you sum them up they result in a dartboard with an area and a larger than 0 probability of getting hit. So far so good. But then why can we sum up the 0 probabilities of the infinitely many balls (6:33) and conclude that their sum is 0?
Dragon Curve Enthusiast Those are two different sizes of infinity: the number of balls is countably infinite. The number of points on a dartboard is UNcountably infinite. That is, you can't assign a unique integer to each point on a dartboard the way you can with balls. Countably many probabilities can be summed. Uncountably many cannot, at least not in the usual way.
Because the probability of a countable union of random events is less than or equal to the sum of the probabilities of those random events (the formula shown at 6:36 ). Since every individual random event has probability of exactly zero their sum is also zero and hence so is the probability of their union (since probabilities can't be negative by definition.)
Thanks for your answers! It would be an overstatement if I said "thanks, that makes sense", but at it's a starting point and definitely helped me. These kinds of infinity-involving problem solutions always seem very arbitrary and opinion-based to me, which they shouldn't be. Like: In the first super task, where I always remove the ball with the lowest number, what keeps me from saying: In each round I add 10 balls and remove one. Thus I add 9 balls each round -> after n rounds I have infinitely many balls. Her explanation (1:56) doesn't really make sense to me. In the round where I remove the nth ball, I must have handled a ball with the number 10*n and put it into the urn. So, is n is not the "largest" number in this infinity either. Thus, why can't I conclude the nth ball is not the last one in the urn, because all the balls from n+1 to n*10 are still in there? Why can she just conclude that I took out the nth ball and thus the urn is empty when there must have been larger balls that are still in there? Could it be that there is also an uncountable (or at least a larger) infinity at play? because there are infinitely many (n) numbers of rounds, but there is a definitely larger number of balls involved.
Dragon Curve Enthusiast Any time you can talk about the nth item, you're dealing with countable infinity. Anything that can be mapped to integers is countable. For example, the set of even integers seems like it should be smaller than the list of all integers. But if you apply the function f(e) = e/2 to the even numbers, the result is just the set of all integers, each even number mapping to a unique integer, and vice versa. So the set of even numbers is equal in cardinality (or "size") to the set of all integers, even though the set of all integers contains all of the even numbers and more. This would be impossible with any finite set. (If this explanation is confusing, look up "Hilbert hotel".) An example of uncountable infinity is the real numbers. You can't map them all on to the integers in a way that includes all of the irrational numbers. (Too much to explain here, but look up "Cantor diagonalization".) The reason it's different from adding 9 balls every time is sort of like the thing with even numbers: some things that make sense with finite numbers don't make sense with infinity. You have to be a lot more careful. In the case where we remove the lowest ball, then for every ball, we can determine exactly which round it is removed from the urn. It doesn't matter how many balls are added. After an infinite number of rounds, you can ask is ball n in the urn? And I can say, "No, that ball was removed in round n." If we just add 9 balls every round, I CAN'T say that. The case for the stochastic version is trickier, but for each ball, I can say "It has Almost Certainly been removed", and it turns out that's enough for countably many balls.
If you remove the median after adding 9 balls each pass, the total balls left when time runs out is infinitely many. The number of balls in our urn each pass is equal to the number of passes, n , times nine less the number of medians removed, (n-1), which equals (9n-(n-1)). This simplifies to 8n+1. Our median's position is therefore described by 4n+1 because the median is the number for an ordered group of numbers which has an equal amount of numbers greater and less than it. Therefore, if our group of numbers has a size 8n+1 the size of the group of numbers on either side of the median should equal (8n/2). The position of our median is then described by 4n+1 because it's the number right after the group of size 4n. I keep referring to it as position and not value because we must keep in mind that we remove the median. This leads to the nice result that the original position (i.e. the number on the ball) of any given median after (n-1) medians have been removed will be given by the expression (4n+1+(n-1)) which simplifies to (5n). Since there are infinitely many numbers that aren't a multiple of five, there will be infinitely many balls left over.
Is it possible to address the probabilities of the stochastic supertasks more precisely using a number system which includes infinitesimals, such as the Hyperreals? I have a sort of in-born aversion to assigning a zero-probability to something which actually does have a chance of occurring, even if we can't describe that chance using our usual mathematical language...
The challenge problem ( +9 rmv med) easy there will be infinitely many balls left because the number which is being removed keeps going up and there will be infinitely many unremoved balls left which are never going to be removed. For ex, step 1: 5 step 2: 10 step 3: 15 step 4: 20 like this only the multiples of 5 will be removed.
"There's a literally infinite number of variations of this problem, so we don't have time to review them all..." I have an idea: We cover the first one a half-minute before noon, then cover the second one a quarter-minute before noon, the third one an eighth-minute minute before noon, and so on. And when we've covered all of them, we can break for lunch.
0 * inf is indeterminate The individual probabilities are so tiny that they're virtually 0, individually. However, none of them ever multiplies by 0, and therefore the possibilities were never truly 0 in the 1st place, but rather very (indefinitely) small. When you sum up those chances, you're not summing up 0's, but rather a lot of (infinitely many) small but non-zero values, making it non-zero. Whether or not it goes to infinity, however, depends on the terms. Also, as Matt Parker explains it here: watch?v=sf5OrthVRPA , that any one ball being left in the urn is amazing for that ball, but it's not amazing that any balls remain. The chances to remain are only small for individual balls, not for the collective of balls. Here's an analogy: single atoms hardly weigh anything to us, but a lot of them can make a grain of sand, and a (seemingly) infinite amount of sand (or sand particles) can make up beaches, or even entire planets tl;dr: A lot of (indefinitely) small possibilities multiplied by an infinite number of balls = a non-zero value, meaning balls WILL remain in that urn.
It's possible to add infinitely many probabilities that are 0 and get a positive number. For example, the probability that a number between 0 and 1 is rational is exactly 0, but (obviously) there exist rational numbers.
Ok, for the first 10 balls probability might be calculated correctly, but as steps go on, ball number 11 has a probability to stay of 18/19 * 27/28 * ... For ball number 21 it starts from 27/28. For an infinite step, the probability starts with infinity divided by infinity.
At 3:58, you said the chance of picking any one ball at step three is 1/29. However, there are actually 28 balls in the urn at that point, as you have thirty balls(the number of balls in the urn without removal) minus two balls(the total number of balls removed by the first and second steps). This equals twenty-eight balls. At step four, you have a one in thirty-seven chance, at step five, a one in forty-six chance, at step six, a one in fifty-five chance, and so on. Just a comment.
The fact that the probability that of a particular ball is left in the urn by noon is zero is not necessarily equivalent to the fact that any ball is left in the urn by noon is zero, since the probability of say ball 2 remaining is not independent of the probability of ball one remaining. The conditional probabilities p(2 remains|1 remains), p(2 remains| ~1 remains), etc, would need to be calculated
There is an error when you compute the probability of the balls to be removed. We always add 10 balls then we remove 1, so the fractions must be: 9/10, 18/19, 27/28, 36/37, ..., 9n/(9n+1)
3:13 I think it will be infinite, because you will be left with groups of 4 balls like (1, 2, 3, 4) , (6, 7, 8, 9) ... and so on, so there isn't a change for all the balls to get pulled out of the urn
Challenge problem: There are infinitely many balls remaining at the end. Reasoning. Suppose ball labelled n was removed. Then ball n+1 will have the same `rank' that ball number n had at the moment ball n was removed. This means ball n+1 will never be removed, because the median `rank' increases on each turn. Thus for every ball removed, there is (at least) one ball that is not removed. Infinitely many balls are removed, so infinitely many balls remain.
I may be misinterpreting this, but as far as I can tell, each paradoxical example has aleph null balls being added and aleph null balls being removed. However, if I put two balls in and take one ball out (without numbering them) an infinite amount of times, I should end up with an infinite number of balls in the urn. What I think this means is that at the end of any of these processes, there are an infinite number of balls still in the urn, but none of them are cardinally numbered.
How do we know that 0 + 0 + 0 ... = 0? If I split a board of length 1 into N pieces, their individual size approaches 0 as N approaches infinity, while their sum is still 1. The probabilities are approaching 0 as you approach infinity, so I fail to see the difference.
You have two different limits, the limit of n -> infinity of factors in the product, and the limit of k -> infinity of the summands in the sum. But every summand is exactly the same product (the factors do not depend on k), so you can take the limit n -> infinity in every summand and see that that limit is zero. Then, you add the summands 0, 0, 0, ... together and get zero as an answer. The difference is in whether you take the same limit or two distinct ones. In your example, the summands 1/N approach zero as N goes to infinity and the sum is over N. As N increases, you add smaller and smaller numbers together, but you always also increase the amount of terms you are adding together. In the sum in the video on the other hand, you add infinite products together that are all independent of the summation index (that I called k because it really doesn't matter what you call it apart from n which is already taken). So you first have to determine what value the product approaches as you increase n without bound, and that value is zero, as can be seen in the PDF linked in the video description. After you have taken that limit, and verified that it is indeed zero, you can rewrite the sum over the product over (9n)/(9n+1) as the sum over zero (actual, real zero, not "a sequence approaching zero"). That sum is indeed zero, since its partial sums (where k ranges from 1 to 1, then from 1 to 2, then from 1 to 3, etc., etc.) are: 0, 0, 0, 0, ... The partial sums are all zero, and therefore the limit of the partial sums is also zero.
Your random balls having 0% probability just shows how haphazard and shortsighted mathematics is on the abstract level. You can't just "jump" to zero. The actual probability of each ball is an infinitesimal (1/inf), not actually a "hard zero". Add up an infinite number of infinitesimals, you would sensibly get 1. 1/n x n = 1 1/inf x inf = 1 So the chance of each ball remaining is zero, but the chances of any balls remaining is 1 (100%). Basically you've just re-done the Devil's Staircase.
Well no point commenting it will probably be re-uploaded over the minor errors in the fractions. :( Does anyone remember when people just added annotations?
"at n-th step you remove the n-th ball which means all of them are removed after infinetly many steps" but i think at that infinite step you would have put in more 10 balls of which none is removed because only the lowest numbered ball is removed. If i'm wrong correct me!!
What if different sized infinities are involved -- I know there are different classes of infinities that can be bigger than others. Those are not the same thing, but it seems to me the conclusion about this more a sign that the infinity is being treated is flawed. Saying (10 - 1) * inf = 9 * inf = inf also works if if infinity is all just infinity, but leads to infinity = 0, which cannot be true, thus the paradox, two conflicting values. Someone else might claim this totally wrong, but that's my reasoning, and too me shows that this concept of "infinity" as one particular thing, as though it was some specific ultimate value, is flawed.
Yugaldeep Singh I somewhat have to agree. if you compare the set of balls you add and set of balls you remove, for each ball you remove you can find 9 balls you added. even though they are both infinite, one must be a larger infinity since they can't match up. isn't this like how 0 to 2 is larger than 0 to 1 even if they are both infinite? IDK I'm not a perfect mathematician
Ryan Lynch but at step 'i' you're removing ball labelled 'i'. If you continue to do this till infinity, there won't be any balls left in the urn. That's the paradox. There isn't a bigger infinity in the picture here. The point was to show that you could perform the exact same experiment and get different outcomes depending on how you perform it.
At 7:20 i dont think that both the scenarios are same. In the second case we certainly dont have any ball remaning as the nth ball is certainly removed at the nth step, while in the first case its just that the probability of any ball remaining is 0. which does not mean the ball cannot remain at noon , its just that the probability is 0.
It seems that as long as all balls are selectable for removal by the "end" of the infinite series (meaning possible in all steps), the total probability will tend to zero. A selection strategy like choosing the last one, or the median ball, leaves the majority of ever-increasing balls ineligible for selection, tending toward infinity.
I am at 4' into the video and my expectation is, that she will soon say: Na, I'm just f* with you. There is no urn that can hold infinitely many balls.
While we want to assume (somewhat correctly) that on the Nth step, we will remove the Nth ball, and there will still be N*9 balls in the urn, the real paradox is that we will be doing the task infinite times. Therefore, to remove the infinith ball would be the end of the task, which would mean that all the balls were ultimately removed. And that just messes with everyone's brain.
Median Ball Urn Challenge: Each iteration we add 9 balls, so after n iterations we have added 9n balls. Since we remove the median each time, after n iterations we have removed n median balls. Leaving the total number of balls in the urn after iteration n to be 9n -n = 8n The median at step n (Before we remove it) is in position (9n - (n-1) +1)/2 = (8n +2)/2 = 4n+1 From this we can see that we never remove a ball which is in position which is a multiple of 4 and the total number of balls in the urn (8n) tends to infinity as n grows large. There will be an infinite number of balls in the urn when the supertask is completed.
For the ern problem: the series 10-1+10-1+10-1+... does not converge, so the "final" number of balls is undefined. 5:32 Does it really converge to zero? 6:36 You add zero infinite times. 0 times infinity is not always 0.
There's a subtlety to what they mean here by "how many balls are there after an infinite number of steps?" They do not take it to mean it to be what the number of balls tends to after an infinite number of steps (this would, as you point out, lead to infinity). Really what they are asking is: "How many balls are never removed after a finite number of steps?" Since the Nth ball is removed at the Nth step, each ball will be removed after a finite number of steps. Why they framed it in the latter sense probably has to do with 1) the history of the problem and 2) the fact that probability is theoretically supported by measure theory where this framing is more natural.
The urn problem isn't a problem of addition, but a set theoretic problem (so you don't try to calculate how many balls but instead you look at the set of balls in the urn and add and remove specific balls from that set and ask whats in the set at the end?)
No what? It looks like this: You start with the empty set, then add the numbers 1 to 10 to your set,t hen you remove the number 1. Then you add the nubmers 11 to 20 and remove 2. There is no number remaining if you do that infinitely often. There are no limits involved
I dont think it is empty too. On the n-th step: You add (10n - 9), (10n - 8), (10n - 7), ... (10n) and you remove n. Then you have the balls (n+1) to (10n) in the urn.
How does the deterministic supertask work? Shouldn't you be looking at both the negative change *and* the positive change? In that case the answer is infinity, after adding up infinite nines.
with infinite sets, the order in which you add and remove elements matters. With finite numbers kardinals (amounts of stuff) and ordinals (orders of stuff) have identical properties, but with infinite amounts they simply don't. There is a fundamental difference between having infinite number of racers and ending up last in that infinite race.
2:21 Uh... lim {n->∞} (20-1)*n = lim {n->∞} 19n Pretty sure it diverges. Put another way: after the Nth step, you've removed N balls and kept 19 times N balls. Is there something I'm missing?
Its very easy to understand the mistake made in the modeling of the problem. Natural numbers (and the math associated with it) represent counting. If you try to count but your reference (origin) is removed (as in the first case, where the smallest ball gets iteratively removed), you simply cannot count in a straight forward way, and infinity(quantity) minus infinity(reference) gets wrongly interpreted as 0. In the second case (where the biggest ball is removed iteratively) the reference is not removed, and a clear mathematical model(way of counting) 10*n - n = 9*n = infinity makes counting possible. The second model can easily be used to interpret the first case (exchanging the indexes of the balls removed), as the indexes of the balls are irrelevant to the task of counting balls. This shows that there is no paradox, there is inadequate modeling of the problem. Despite this video failure, the channel beatifully explained infinity modeling in previous videos.
Why not use half the video to talk about the first scenario, a quarter to talk about a second, and so on until you cover them all?
lol
3:57 Am I being dumb right now!? There are only 28 ball in the urn --- should it not be (1/28) chance?!?
and 5:06 again - should it not be (27/28) chance
Also, at 5:19, the generalized formula uses ekponentiation of n when it should use multiplication of n.
Microwave44 and Underscore Kartoffel
Omg you're all right o.o
at 5:16, it's not (9^n)/(9^n+1).
It's (9*n)/(9*n+1)
Everything I know is a lie
3:14 "There's a literally infinite number of variations of this problem, so we don't have time to reveal them all."
*UNLESS* you explain each one in half the time you used for the previous one!
Why is nobody talking about Matt showing up!!????!
Christopher Bocksell I don't know why, but I find that paradox so cool
Mind blown
and what happened to him? Why is he tiny??
Oh that's just Matt. He transcends time, space, and youtube channels every now and then
Because they're too busy talking about the incorrect formulas.
"What happens when you add randomness to a supertask"
Me: Uh-oh...
Challenge problem: There are infinitely many balls left in the urn at noon. In particular, all balls except the multiples of 5 are left.
In fact, on the nth step, ball 5n is removed. This is equivalent to saying that on the nth step, the following balls are in the urn: all balls from 1 to 9n, excluding all multiples of 5 up to 5n. This claim can be proven using induction.
The claim holds for n=1, since balls 1 through 9 (=9n) are put in, and 5 (the median) is taken out.
Now suppose the claim holds for some n; then after the nth step, we are left with balls 1 through 9n, excluding 5, 10, 15, ..., 5n. Now let's see what happens on the (n+1)th step. Balls 9n+1, 9n+2, ..., 9n+9 are added. Now, in order for the claim to hold, we want to show that 5n+5 is the median of the balls currently in the urn. There are 8n+9 balls in the urn total (9n+9, minus the n which have been removed). Further, there are 4n+4 balls whose numbers are less than 5n+5 (because 4n+4 is 5n+4, minus the n balls which have been removed, each of which had numbers less than 5n+5). Consequently, there are (8n+9)-(4n+4)-1 = 4n+4 balls in the urn whose numbers are greater than 5n+5. Therefore, ball 5n+5=5(n+1) is in the middle, so it gets removed, thus proving the claim for (n+1).
Okay, so every multiple of 5 gets removed, and every ball that gets removed is a multiple of 5, leaving infinitely many balls in the urn.
I did the same reasoning. Thanks for typing the proof
Brian Raynor n is the index of the step. There is no 0th step
Brian Raynor I said that balls 9n+1 through 9n+9 are added on the (n+1)th step, my dude.
For anyone wondering what the deleted comments were: someone didn't see (n+1) and thought Florence was suggesting putting the balls 9n+1, 9n+2, ..., 9n+9 in the urn on the nth step, and he pointed out that you would need to start at n=0 if you did it that way, or you would never add the first nine balls.
Interesting that the commenter pointed out that if you do what Florence said, but on the nth step instead of step n+1, but still started at n=1, you would remove all of the balls that are one less than a multiple of five
Congratulations Florence!
You are the winner of this challenge. Please send us an email @ pbsinfiniteseries@gmail.com with your shirt size and shipping address.
I think "adding infinitely many zeroes" is worth slightly more consideration. First, it is very important that we add *countably* many zeroes when we want to conclude a sum. Second, there are two simultaneous limits involved: the limit over the number of steps and the limit over the highest ball number included in the disjunction. Just taking one limit then the other doesn't necessarily mean you get the correct answer, there are cases when you can get any answer you want just by manipulating how fast you let each limit approach infinity with respect to the other.
I am currently studying mathematics, and I am literally in love with it.
I was just so confused about continuing to do that and make math my entire career and switching to medical school, but when I see how beautiful is this universal language I push myself more to continue this path since I began it already. You guys have played a huge role in doing that, so please continue the amazing work and make sure that you are a very successful, professional and unique channel. 💓💓
Any update
5:04 i think It should be 27/28. :P
I noticed that too
Underscore Kartoffel jajaja when i finished the comment i noticed how fast you are guys! Everybody notices, clever! jajs
They also wrote 9^n/(9n+1) instead of 9n/(9n+1).
We're not the first to notice it.
I didn't prove it, I just wrote down a few scratches on a piece of paper, but it seems like only multiples of 5 will get removed from removing the median. So infinite.
You're (possibly) thinking of the median of the added balls. I.e., removing 5 from 1- 10, 15 from 11-20, etc. But that's not what's happening here. You're removing 5(or 6, as one definition for a median of an even number of values is the average of the two middle values, which would be 5.5 in this case)) from 1-10, 11 from 1-20 (minus the 5 that was removed), 16 (or 17) from 1-30 (minus the previous 2 removed), etc.
It all still seems silly to me, though. You're still adding infinite sets of (10 - 1) balls, heh.
For the median case she was only adding 9 balls in each step, so you'd add 1-9 and then take out 5, then add 10-18 and take out 10, then add 19-27 and take out 15. It would only be multiples of five that get taken out.
Ah, right. I wasn't paying enough attention.
Ralph Strocchia Slow down Fermat
Am I missing something, or is the idea of a time restriction arbitrary here? Saying "At noon" seems to be a roundabout way of letting N tend to infinity, and would have the same result
Vriskanon Probably just a kind of response to a critic of thinking "how things would look like after infinite steps"
"Well...you can't get to infinite steps, because this would take infinite time! You will never get there!"
"Well, Mr. Smartpants, now I will do each subsequent task in half time and it will take me a finite time to end! Take that!"
+Luiz Sarchis Just Zeno's Paradox, rehashed, and alegedly solved.
Vriskanon You realize that number of balls tend to infinity as N -> infinity, and the limit is also infinity. Thus the results would differ.
Unnecessarily rigorous proof: Let a_n be the number of balls at step n. By definiton, if the limit of sequence a_n is infinity, for every R>0 there exists N>0 such that a_n>R for every n>N. Choose N=ceil(R). Now number of balls is a_N=9*ceil(R)>R. Sequence a_n is increasing, thus a_n>a_N>R for n>R. QED.
This reasoning doesn't work anymore with supertasks, because now you are in a sence going _beyond_ the limit - you terminate at some point and you have a (more or less) well defined final result that can be analyzed. With limits you never terminate, you just keep going and going, there is no "final step" (technically there is no final step in supertasks also, but I hope you get my point).
+Tetraedri_
The limit of the number of balls in the urn as the number of steps tends to infinity is infinity, but the limit of the set of balls in the urn as the number of steps tends to infinity is the empty set.
That's true whether you borrow a trick from Zeno to get there in finite time or not.
The question then is whether the limit value for the number of balls should be the limit of the number of balls or should be the number of balls in the limit set. Generally, it's accepted to take the actual limit set and count the balls - which makes the answer zero.
rmsgrey True, didn't think about the limit set. However, if we didn't terminate at some well defined time, there is no meaning in saying "after we're done." If you still was considering only the cardinality of the set, it would tend to infinity, thus at any time there are non-zero amount of balls in the urn, tending to infinity. The limit set has to be reached for the question to be sensible, and without supertasks it can't be reached.
6:39 0+0+0+0+0+... infinite times isn't the same as 0 x infinity which is an indetermination?
Alejandro Bravo the limit of x*y where x is an expression that goes to infinity, and y is an expression that goes to 0, could go to different values depending on how x and y go to infinity and 0 respectively, but if y is a constant 0, then the limit will be 0, because the expression is a constant 0
Didn't know that, thanks!
Problem is that 0 in this case is not constant 0. It was derived from convergence so it should had been taken into account. I have no idea if it changes the results but when doing calculations with infinity you can't substitute infinity in one place without substituting it in all others and it is what happens here.
First infinity is substituted when calculating value of infinite series and then when multiplying the result by infinity.
for any single ball, the probability of it not being picked after infinitely many steps is the number zero.
the number zero has no limits in it's definition, it's plain and simple, add zero to itself infinitely many times and you will get zero.
It is: given that number of steps goes to infinity the probability goes to zero.
Simple example why you can't split limits:
Let's say A = lim x/(2*x^2) so we say given that x goes to infinity the A goes to zero.
Then we ask, given infinitely many tries (n) what is probability of event A happening (let's call it event b):
B = lim n * A
Using naive interpretation of (A is constant 0) A = 0 so B = 0.
If instead we substitute the full expression for A as the result isn't constant 0 it is just close to 0 and given that both those limits have the same rate (they do):
B = lim(n) lim(x) n * x/x^2 = inf * inf / (2*inf^2) = 1/2
5:17 9^2 = 81
9(2) = 18
*ILLUMINATI CONFIRMED!!!!11!1!11*
The video has a typo, but the cited article has a correct sequence of (9n)/(9n+1).
Basically the power operation should not be there, it's multiplication.
But then the term 28/29 should not be there as 28 is not a multiple of 9...
argh, this video is seriously messed up! but the material is interesting.
The typos don't affect the outcome. Both products are zero.
no, the product is immeasurably close to zero but nonzero since none of the terms involved in the multiplication are zero.
+phr00t The product of an infinite multiplication is the limit of the partial products. In this case, that limit is 0.
5:29 The terms of that series do not resemble 9^n/(9^n+1).
The formula would give the second term as 81/82, which is different than 18/19, and so forth.
(Edit: Never mind; other poster points out that probably 9*n/(9*n+1) was what was meant) I assume some of the reasoning was not shown; most likely the series are related by an inequality and 9^n/(9^n+1) is an upper bound on the value that the nth term can take, so the smaller terms in the true series only make the probability of the ball remaining smaller.
I did an Actuarial Science degree and I fucking loved the Stochastic Processes module.
For the supertask at 1:16, the amount of remaining balls isn't 0, it's 9∞. The only balls remaining after step ∞ are balls ∞+1 through 9∞. Just because the numbers of those balls are greater than ∞ doesn't mean that they don't exist. No matter how you look at it, you're adding 9 balls each step. If you do the tasks ∞ times, you'll always get 9∞ balls regardless of the subtraction order.
Major twist: In the deterministic version of the paradox, it doesn't matter how quickly you add the balls -- you can go 10 at a time, as described in the video, or double the number you're putting in at every step. It won't change the fact that every ball is eventually removed. In the stochastic version, though, it could. Yes, the infinite product giving (say) the probability that ball 1 is still in the urn at noon converges to zero -- its logarithm is an infinite sum that grows at least as fast as (some scaled version of) the harmonic series. But this fact depends on the fact that the number of balls is the same at every step. If the number of balls added at a given step grows over time, the product could converge to a non-zero probability!
On the Nth step you add ten balls and then remove one. There are a minimum of nine left after any given turn.
Today I learned that 9^2= 18 and 9^3=28! Cool :)
6.46 You cannot conclude that adding infinitely many zeroes sum up to zero. And especially, in this case it doesn't.
thank you for mentioning vsauce!
For median problem, infinitely many will be left. Only the ones divisible by 5 will be removed
2:11
That's ∞ - ∞, and it's not necessarily zero. N doesn't accept ∞ itself.
What order would you have to add and remove the balls, for there to be negative one-twelfth of a ball in there at noon?
So, when adding all the probabilities together, what you do is taking the limit of the number of balls N->infinity, while summing up the series P_1+P_2+...+P_N, where each P_i also depends on N. It looks like you cannot just interchange the limit and the summation. Each P_i goes to zero as N becomes infinite but does the sum of the P_i go to zero as well? I don't find that obvious.
So, I rarely have an actual grasping of maths, and generally only watch these videos each week to have my mind blown. This week, my mind was blown because I actually understood it all! even more, I have enough confidence to answer the challenge question...
The part I found hardest about this challenge was finding the median of the series, having removed specific balls...
So, If you add balls 1-9, then take out the median, you will take out ball #5.
then, if you add balls 10-18, and take out the median of the remaining balls, you will take out ball #10.
repetition #3, you will remove ball #15. this pattern repeats, where repetition #n, you will remove ball #n*5, which means that balls that are not a multiple of 5 will never have the opportunity to be removed. Because there are balls that can never be removed being added with each repetition, after infinite repetition, there will still be infinite balls in the urn.
To help find the median of each repetition, I used R, feel free to check it out/point out errors in my code; I have set it to 100 repetitions, which shows the pattern very clearly.
i = setdiff(1:9, (median(1:9)))
while(max(i) < (900)){
a = (max(i)+1):(max(i)+9)
b = c(i, a)
i = setdiff(b, (median(b)))
}
setdiff(1:(max(i)), i)
#setdiff makes a series, with the second argument being individual numbers in the series being removed; I use this to remove the median.
#c(i, a) is a function that combines the series i (which is the last repetition), and a (the balls being added).
#I have to use 3 steps to add the new balls, combine the balls into the pool of *all* the remaining balls, and then to remove the correct ball.
#the final line shows all the balls that were removed
As n approaches infinity,
9(1)/[9(1)+1] x 9(2)/[9(2)+1] x ... x 9(n)/[9(n)+1]
approaches 0. It does not equal zero. However I can get behind the concept of calculus and say ball 1 has zero chance of remaining after infinite steps.
This equation doesn’t apply for every ball however. If we look at the balls as groups of ten this equation only applies to the first group (balls 1-10). For group 2 (balls 11-20) they are not subject to the first step and are guaranteed to be there for the second step where the balls from group 1 only have a 90% chance of being there for step 2. Group 2’s equation should look like
9(1)/[9(1)+1] x 9(2)/[9(2)+1] x ... x 9(n)/[9(n)+1] x [9(1)+1]/9(1)
And group X’s equation should look like
9(1)/[9(1)+1] x 9(2)/[9(2)+1] x ... x 9(n)/[9(n)+1] x [9(1)+1]/9(1) x [9(2)+1]/9(2) x ... x [9(x-1)+1]/9(x-1)
It’s easy to see that there are terms that cancel out, and the equation can be simplified to,
9(x)/[9(x)+1] x 9(x+1)/[9(x+1)+1] x ... 9(n)/[9(n)+1]
as long as X is a finite value there will remain an infinite amount of terms that don’t cancel out and there fore there is still a zero percent chance of any ball in group X remaining after infinite steps.
However, as the amount of steps increase so do the amount of groups. In fact since they increase at a 1-1 rate we can say for any step n there is a group n. And it’s equation can be written as,
9(1)/[9(1)+1] x 9(2)/[9(2)+1] x ... x 9(n)/[9(n)+1] x [9(1)+1]/9(1) x [9(2)+1]/9(2) x ... x [9(n-1)]/9(n-1)
Cancel all terms and you are left with,
9(n)/[9(n)+1]
As n approaches infinity this equation approaches 1. That is to say the odds of any given ball in group n remaining after step n is 100% after infinite steps.
Similarly for any group (n-x) the equation gets reduced down to,
9(n-x)/[9(n-x)+1] x 9(n-x+1)/[9(n-x+1)+1] x ... x 9(n-x+x)/[9(n-x+x)+1
And as n approaches infinity, for every finite x, this equation approaches 1. So there are an infinite amount of groups that each have 10 ball all with a 100% probability of remaining after an infinite amount of steps.
For the challenge problem, the urn will contain infinitely many balls at noon. To prove this, let m_n be the number on the ball which is removed during the nth step of the described process. I will prove that for all n >= 1, m_n = 5n. We proceed by induction. For the base case, one can check that m_1 = 5. For the inductive step, we can see that at the nth step, we begin with 8(n-1)+9 balls and remove the (4(n-1)+5)th ball, which will leave us with 8n balls at the end of the nth step. We begin step n+1 by adding 9 balls, for a total of 8n+9 balls, and we will remove the (4n+5)th ball. Our urn now has 4(n-1)+4 = 4n balls with numbers below m_n and 4n+4 balls with numbers below m_(n+1). Thus there are precisely 4 balls with numbers between m_n and m_(n+1). Since we have not removed any balls with numbers above m_n, and since m_n = 5n by our induction hypothesis, we know that m_(n+1)=5(n+1). Therefore, for all n >= 1, m_n = 5n. Finally, since infinitely many natural numbers are not multiples of 5 (exercise left for the reader), infinitely many balls remain in the urn at the end of this process.
Surely you meant 9n/(9n+1) - instead of 9ⁿ/(9ⁿ+1). After all the number of balls in the urn grows linearly, not exponentially.
~ 6:30 the probability that You would have removed any one ball isn't zero, it simply converges into zero, meaning that the more steps that You perform, the closer You get to zero, meaning that it's infintessimaly small, n/infinity. when You add up this infinite series of infintessimales (not 0 but rather n/infinity) You get (n/infinity)·infinity, not 0·infinity. when You simplify (n/infinity)·infinity, You are left with n, a 1 in 1 chance of any given ball remaining after an infinite number of steps. since the proper term is not literally 0 but rather n/infinity (which many people would incorrectly simplify into a literal 0) the 2 infinities cancel each other out when You keep adding this infintessimaly small probability (multiplying it by infinity).
6:32
"0+0+0+0... ∞ =0"
Tell that to integrals.
Sorry for the Necro, but this is countable infinity, while integrals are an uncountable infinity.
SORRY that's wrong at 2:11
your logic says that by Nth iteration I have removed total N balls. But the total number of balls you have put inside the urn till then are 10N!
As you drew the comparison to the darts board: Hitting a particular point on the board has chance 0, but hitting the board has chance 1. So here, why can't we have every single ball have probability 0 of remaining in the urn, but hav the probability that any ball stays in the urn to be 1? In fact, adding up infinetely many zeroes in 6:33 seems like an indeterminate form (infinity * 0), especially as those zeroes aren't really zeroes but an infinite product that approaches zero. I think everyone that ever heard something about limits should get really sceptical at that point...
'Physics tells us that space and time are not infinitely divisible' Physicist Wheeler said time is squeezed into a moment, while space is the sum of all the events happening to you. What follows from the big bang is 'Sequential and simultaneity'.
0 - In cases where we remove lowest number, median and random number ball.
Because in these cases we perform operation on all balls in the urn.
∞ - In cases where we remove last ball (ball number = step*10)
Because in this case we perform operation only on balls introduced at Nth step.
This is my solution to the proposed problem:
After some tests, we can see that we are taking out multiples of 5. In the first step, we take out ball number 5. In the second, the ball number 10. Generally, we take out the ball of number 5n in the nth step.
Proof:
The median of a distribution of k terms (k is an odd number) is given by (k+1)/2
In the problem we have:
(9+1)/2 = 5 (the position of the first ball to remove)
We now have 8 balls and add 9 more:
(17+1)/2 = 9th position
And so on.
But, we have to add an extra term to find the number of the ball directly. Every time we look for the next median, we pass through a position that we removed in the last step, so we need to add 1 to compensate and get the correct number.
For example, in the second step, the median is the 9th position (as seen before) but it doesn't correspond to the number 9. Indeed, it corresponds to the number 10, once we removed the ball number 5 and it's not more on the group.
Generally, in the nth step we have:
Ball number = (9 + 8(n-1) + 1)/2 + (n-1)
(That is, the median position and the extra term to correct the number)
Simplifying, we get:
Ball number = 5n
So, after infinitely many steps, we removed all the multiples of 5 and all the other balls remained in the box.
Answer: Infinity balls.
(Sorry if my English is not good 😅. I'm from Brazil)
In the median case, the urn ends up full, because the median keeps going up, so am increasing number of balls can no longer be removed.
I still don't get how you can add 10, take 1 and end up with 0 after infinite iterations
Daniel Fernandes eu acho que fica mais claro se você separar as duas ações, primeiro adicione as bolas de 10 em 10 até o infinito e depois remova as bolas de 1 em 1 até o infinito
Daniel Fernandes - Imagine that instead of adding 10 balls each time, the urn is already full of an infinite amount of balls. If you then take out 1 ball an infinite number of times, then the urn will be empty. Adding balls each step just makes it so that the urn will have an infinite number of balls in it by the time it's noon.
Same.
In the first example where she adds ten then removes the smallest, how would there be zero at noon if the net change is 9 balls after each subtask? You add ten, remove one, add ten, remove one, etc. The only way I see it becoming zero is if you stop adding balls.
Stating that in the Nth step you remove the Nth ball is valid but it leaves out that in the Nth step you have 9 * Nth balls.
By having infinite balls by noon and then removing infinite balls by noon, you end up with 0 balls
Daniel Fernandes It's based on the assumption that 9*infinity is exactly equal to 1*infinity so it doesn't matter how many balls you add so long as you take 1 away.
Oh Boy! People get very angry when you challange their intuitions about infinity. I wish infinite amounts of patience to be delivered until the end of next week (perhaps you can ration it to have some to spare later)
6:45
0+0+0... Is not necessarily 0.
This could be thought of as 0*infinity, which is undefined.
The zeroes being summed together aren't exactly zero, they are infinitecimals. They behave like zero in most cases, but not when you are dealing with an infinite ammount of them.
Probability measure is countably additive. This means that we can sum the probabilities of a countably infinite amount of disjoint events.
In our case we have precisely that, so the result follows.
Sum of elements where limit for each item is zero is not zero.
There are infinite examples on calculus.
Am I overlooking something or is the formula at 5:05 wrong? It should be 9/10 x 18/19 x 27/28, right?
This is the problem I still have with seeing infinity as a number. I've been taught that any number is a concept, but that infinity especially should not be treated as a number. That's how we get ridiculous statements like 1 = 0 after subtracting ∞ from the equation ∞ + 1 = ∞. Even saying there are an infinite number of even and odd numbers, and an infinite number of only even numbers (despite there being twice as many total numbers) yields different sizes of infinity. This is baffling to me and infinity as a concept should always be the same. This is why we should fundamentally change the way we think about infinity, and I cannot accept the current model philosophically.
It is quite clear that in all ways you described to remove the balls, the number of balls are increasing at each step. For example, when we insert 10 balls and take out one at each step, there must be ∞ balls left because at nth step, 9n balls are left. Therefore, at ∞th step, ∞ balls must be left. Also there is no paradox as ∞ balls are put and ∞ are taken out. But, ∞-∞ is an indeterminate form. Therefore, the right calculation must be in the way I described above. Am I right in your viewpoint? Please let me know.
I think the paradox can be resolved by booking the balls into that well known hotel containing an infinite number of rooms, as follows:Number the first room 10^-1, the second 10^-2 and so on.
Place ball numbered k in the room numbered 10^-k. Define a “room occupancy index” number q as the sum over all occupied rooms of 1 multiplied by the room number. For example q = .1001 would indicate rooms 1 and 4 are occupied, all other rooms are empty.According to the first definition of the problem, balls numbered 1 to 10 will be allocated to rooms numbered 10^-1 to 10^-10, before removing ball number 1. So q(1) = 0.0111111111 is the room occupancy index after this iteration. The number of rooms occupied (ie balls still in the vase) is equal to the total of the nonzero digits in q(1) - ie 9.After each subsequent iteration the room occupancy index will increase by 10 digits, comprising an extra zero at the start and 9 ones at the end.After n iterations, the room occupancy index will be equal to q(n) = 0.000…0001111….111 where there are n zeros and 9n onesAs n tends to infinity, q(n) tends to zero. The paradox seems to arise if this is wrongly interpreted as implying that there are no balls left in the vase. But q(n) contains 9n non zero digits so the vase contains 9n balls. Paradox resolved.According to the second definition of the problem, every tenth ball is removed. In that case we have q(n) = 0.11111111101111111110…… which clearly does not tend to zero - the vase contains 9n balls. No paradox arises.According to the stochastic definition, q(n) will again contain 10n digits, of which 9n digits equal one and n digits equal zero but this time the zeros will be distributed at random. Again the vase clearly contains 9n balls. No paradox.
Perfect proof of the different size of infinities. Removing any ball infinite many times means infinite balls have been removed, even though the actual number of balls goes up by 9 an infinite number of times.
3:18 but you just mentioned a way you can
Mankind's Supertask:
Avoid extinction before 2020
PlayTheMind
We’re kind of failing that right now
PlayTheMind Thanks for pointing out you don't know the definition of what is being talked about.
ouch
It's okay, we can do infinite work from 2019 to 2020, we will just double our speed each step of the way!
Snatch n Grab think he made that comment on purpose, but maybe he can be trolling, or you could be trolling, or I could be too.
6:40 - the problem with this logic is that 'infinitely small' or 'approaches zero' is replaced with "equals zero". Adding a bunch of 'infinitely smalls' is basically an integral function.
you're adding up a bunch of infinitely small values approaching infinity.
No, you have a product from 1 to infinity. That one you can evaluate to be exactly zero. Then, that product (or its value zero) are summed up, and every term of the sum is the same (the product is constant with respect to the summation index). Therefore, we are actually summing over actual zeros and that means the end result is zero.
when you always remove the ball with the median value there will be infinite balls in the urn at noon, right? because the median value constantly gets bigger and the balls with a number below the median value will approach infinity.
Answer to the question: infinite.
On each the nth step we got 8(n-1)+9=8n+1 balls. The median is the (4n+1)th ball. We removed n-1 balls so the ball we take out (the (4n+1)th ball) got the number 5n on it. So the only balls we take out are those which got a multiple of 5 on them. Hence, every other ball stays in the urn, and we got infinitely many of them.
at 5:04 - Probability for step 3 should be 9/10*18/19*27/28 and at 5:24 for step 4 probability product of ( 9*m/9*m+1) ( for all n from 1 to n )
On the challenge problem,
Add 9 balls and remove the median, then add the next nine balls and remove the new median, repeat.
A nice pattern emerges as we see the balls that are multiples of 5 are removed.
This can be proven: When we pick the nth ball there are 9n-(n-1) balls in the urn. We can find the median first by adding 1 and dividing by 2. Then we add n-1 to account for the balls that having been taken out. The nth ball removed is equal to (9n-(n-1)+1)/2+(n-1) =(8n+2)/2+n-1=4n+1+n-1=5n. Now since there are 4 balls before each ball taken out and after the previous ball, there will be 4n+4 balls in the urn after n steps, as n goes to infinity we get infinity balls left.
I'll be glad to become a patron. Thanks for these videos!
Thank you Nestor! That is awesome to hear. -Rusty
Shouldn't it be a 1 in 28 chance, and shouldn't it be 9n/(9n+1) and not 9^n/(9^n+1)
Supertasks should be including derivatives-so the first task is no different outcome than the second because it doesn't 'matter' whether you call it, the lowest numbered ball: it's going to fill up not diminish...
P.S. derivatives are not the final solution, as Collatz Conjecture shows, but fractal derivatives are still a ways off...
7:01 What I don't get:
Each single infinitesimal point on the dart board has probability 0. And yet if you sum them up they result in a dartboard with an area and a larger than 0 probability of getting hit. So far so good. But then why can we sum up the 0 probabilities of the infinitely many balls (6:33) and conclude that their sum is 0?
Bulls eye (pun intended)
Dragon Curve Enthusiast Those are two different sizes of infinity: the number of balls is countably infinite. The number of points on a dartboard is UNcountably infinite. That is, you can't assign a unique integer to each point on a dartboard the way you can with balls.
Countably many probabilities can be summed. Uncountably many cannot, at least not in the usual way.
Because the probability of a countable union of random events is less than or equal to the sum of the probabilities of those random events (the formula shown at 6:36 ). Since every individual random event has probability of exactly zero their sum is also zero and hence so is the probability of their union (since probabilities can't be negative by definition.)
Thanks for your answers!
It would be an overstatement if I said "thanks, that makes sense", but at it's a starting point and definitely helped me. These kinds of infinity-involving problem solutions always seem very arbitrary and opinion-based to me, which they shouldn't be.
Like: In the first super task, where I always remove the ball with the lowest number, what keeps me from saying: In each round I add 10 balls and remove one. Thus I add 9 balls each round -> after n rounds I have infinitely many balls.
Her explanation (1:56) doesn't really make sense to me. In the round where I remove the nth ball, I must have handled a ball with the number 10*n and put it into the urn. So, is n is not the "largest" number in this infinity either. Thus, why can't I conclude the nth ball is not the last one in the urn, because all the balls from n+1 to n*10 are still in there? Why can she just conclude that I took out the nth ball and thus the urn is empty when there must have been larger balls that are still in there?
Could it be that there is also an uncountable (or at least a larger) infinity at play? because there are infinitely many (n) numbers of rounds, but there is a definitely larger number of balls involved.
Dragon Curve Enthusiast Any time you can talk about the nth item, you're dealing with countable infinity. Anything that can be mapped to integers is countable. For example, the set of even integers seems like it should be smaller than the list of all integers. But if you apply the function f(e) = e/2 to the even numbers, the result is just the set of all integers, each even number mapping to a unique integer, and vice versa. So the set of even numbers is equal in cardinality (or "size") to the set of all integers, even though the set of all integers contains all of the even numbers and more. This would be impossible with any finite set. (If this explanation is confusing, look up "Hilbert hotel".)
An example of uncountable infinity is the real numbers. You can't map them all on to the integers in a way that includes all of the irrational numbers. (Too much to explain here, but look up "Cantor diagonalization".)
The reason it's different from adding 9 balls every time is sort of like the thing with even numbers: some things that make sense with finite numbers don't make sense with infinity. You have to be a lot more careful. In the case where we remove the lowest ball, then for every ball, we can determine exactly which round it is removed from the urn. It doesn't matter how many balls are added. After an infinite number of rounds, you can ask is ball n in the urn? And I can say, "No, that ball was removed in round n." If we just add 9 balls every round, I CAN'T say that.
The case for the stochastic version is trickier, but for each ball, I can say "It has Almost Certainly been removed", and it turns out that's enough for countably many balls.
Me an actual engineer: fascinating
Also Me: heheheh she said balls
(I wonder how many times she said balls)
If you remove the median after adding 9 balls each pass, the total balls left when time runs out is infinitely many. The number of balls in our urn each pass is equal to the number of passes, n , times nine less the number of medians removed, (n-1), which equals (9n-(n-1)). This simplifies to 8n+1. Our median's position is therefore described by 4n+1 because the median is the number for an ordered group of numbers which has an equal amount of numbers greater and less than it. Therefore, if our group of numbers has a size 8n+1 the size of the group of numbers on either side of the median should equal (8n/2). The position of our median is then described by 4n+1 because it's the number right after the group of size 4n. I keep referring to it as position and not value because we must keep in mind that we remove the median. This leads to the nice result that the original position (i.e. the number on the ball) of any given median after (n-1) medians have been removed will be given by the expression (4n+1+(n-1)) which simplifies to (5n). Since there are infinitely many numbers that aren't a multiple of five, there will be infinitely many balls left over.
Is it possible to address the probabilities of the stochastic supertasks more precisely using a number system which includes infinitesimals, such as the Hyperreals? I have a sort of in-born aversion to assigning a zero-probability to something which actually does have a chance of occurring, even if we can't describe that chance using our usual mathematical language...
The challenge problem ( +9 rmv med)
easy
there will be infinitely many balls left because the number which is being removed keeps going up and there will be infinitely many unremoved balls left which are never going to be removed.
For ex, step 1: 5
step 2: 10
step 3: 15
step 4: 20
like this only the multiples of 5 will be removed.
*talking about structures of 2D bees*
"There's a POINT this"
HAHA! I see what you did there
That's straight up kooky dukes.
You never gave a reasonable explanation that the union < sum is still valid for infinite sums.
Umbrall That's Boole's theorem, it's not specifically mentioned in the video but it is explicitly mentioned in their proof linked in the description.
"There's a literally infinite number of variations of this problem, so we don't have time to review them all..."
I have an idea: We cover the first one a half-minute before noon, then cover the second one a quarter-minute before noon, the third one an eighth-minute minute before noon, and so on. And when we've covered all of them, we can break for lunch.
0 * inf is indeterminate
The individual probabilities are so tiny that they're virtually 0, individually. However, none of them ever multiplies by 0, and therefore the possibilities were never truly 0 in the 1st place, but rather very (indefinitely) small. When you sum up those chances, you're not summing up 0's, but rather a lot of (infinitely many) small but non-zero values, making it non-zero. Whether or not it goes to infinity, however, depends on the terms.
Also, as Matt Parker explains it here: watch?v=sf5OrthVRPA , that any one ball being left in the urn is amazing for that ball, but it's not amazing that any balls remain. The chances to remain are only small for individual balls, not for the collective of balls.
Here's an analogy: single atoms hardly weigh anything to us, but a lot of them can make a grain of sand, and a (seemingly) infinite amount of sand (or sand particles) can make up beaches, or even entire planets
tl;dr: A lot of (indefinitely) small possibilities multiplied by an infinite number of balls = a non-zero value, meaning balls WILL remain in that urn.
At 5:05 it should be 27/28, not 28/29.
It's possible to add infinitely many probabilities that are 0 and get a positive number.
For example, the probability that a number between 0 and 1 is rational is exactly 0, but (obviously) there exist rational numbers.
Ok, for the first 10 balls probability might be calculated correctly, but as steps go on, ball number 11 has a probability to stay of 18/19 * 27/28 * ... For ball number 21 it starts from 27/28. For an infinite step, the probability starts with infinity divided by infinity.
At 3:58, you said the chance of picking any one ball at step three is 1/29. However, there are actually 28 balls in the urn at that point, as you have thirty balls(the number of balls in the urn without removal) minus two balls(the total number of balls removed by the first and second steps). This equals twenty-eight balls. At step four, you have a one in thirty-seven chance, at step five, a one in forty-six chance, at step six, a one in fifty-five chance, and so on. Just a comment.
The fact that the probability that of a particular ball is left in the urn by noon is zero is not necessarily equivalent to the fact that any ball is left in the urn by noon is zero, since the probability of say ball 2 remaining is not independent of the probability of ball one remaining. The conditional probabilities p(2 remains|1 remains), p(2 remains| ~1 remains), etc, would need to be calculated
the fact that the probability of any ball*
The conditional probabilities don't need to be calculated, you missed the
3:56 Seriously?
5:18 Denominator should be 9n + 1, not (9^n) + 1.
After the third step isn't it a 1/28 chance of being removed?
There is an error when you compute the probability of the balls to be removed. We always add 10 balls then we remove 1, so the fractions must be: 9/10, 18/19, 27/28, 36/37, ..., 9n/(9n+1)
3:13 I think it will be infinite, because you will be left with groups of 4 balls like (1, 2, 3, 4) , (6, 7, 8, 9) ... and so on, so there isn't a change for all the balls to get pulled out of the urn
Also in the probability that ball 1 remains in urn after n times, the 3rd term should be 27/28, given that it is 9^n/[(9^n)+1]
Challenge problem: There are infinitely many balls remaining at the end.
Reasoning. Suppose ball labelled n was removed. Then ball n+1 will have the same `rank' that ball number n had at the moment ball n was removed. This means ball n+1 will never be removed, because the median `rank' increases on each turn. Thus for every ball removed, there is (at least) one ball that is not removed. Infinitely many balls are removed, so infinitely many balls remain.
3:58, you would have a 1/28 chance, not 1/29.
I may be misinterpreting this, but as far as I can tell, each paradoxical example has aleph null balls being added and aleph null balls being removed. However, if I put two balls in and take one ball out (without numbering them) an infinite amount of times, I should end up with an infinite number of balls in the urn. What I think this means is that at the end of any of these processes, there are an infinite number of balls still in the urn, but none of them are cardinally numbered.
Ryan Taglang That seems a nice solution. "There are no balls in the urn ... which have a natural number index associated with them."
How do we know that 0 + 0 + 0 ... = 0? If I split a board of length 1 into N pieces, their individual size approaches 0 as N approaches infinity, while their sum is still 1. The probabilities are approaching 0 as you approach infinity, so I fail to see the difference.
You have two different limits, the limit of n -> infinity of factors in the product, and the limit of k -> infinity of the summands in the sum. But every summand is exactly the same product (the factors do not depend on k), so you can take the limit n -> infinity in every summand and see that that limit is zero. Then, you add the summands 0, 0, 0, ... together and get zero as an answer.
The difference is in whether you take the same limit or two distinct ones. In your example, the summands 1/N approach zero as N goes to infinity and the sum is over N. As N increases, you add smaller and smaller numbers together, but you always also increase the amount of terms you are adding together.
In the sum in the video on the other hand, you add infinite products together that are all independent of the summation index (that I called k because it really doesn't matter what you call it apart from n which is already taken). So you first have to determine what value the product approaches as you increase n without bound, and that value is zero, as can be seen in the PDF linked in the video description. After you have taken that limit, and verified that it is indeed zero, you can rewrite the sum over the product over (9n)/(9n+1) as the sum over zero (actual, real zero, not "a sequence approaching zero"). That sum is indeed zero, since its partial sums (where k ranges from 1 to 1, then from 1 to 2, then from 1 to 3, etc., etc.) are:
0, 0, 0, 0, ...
The partial sums are all zero, and therefore the limit of the partial sums is also zero.
Your random balls having 0% probability just shows how haphazard and shortsighted mathematics is on the abstract level.
You can't just "jump" to zero. The actual probability of each ball is an infinitesimal (1/inf), not actually a "hard zero".
Add up an infinite number of infinitesimals, you would sensibly get 1.
1/n x n = 1
1/inf x inf = 1
So the chance of each ball remaining is zero, but the chances of any balls remaining is 1 (100%).
Basically you've just re-done the Devil's Staircase.
Well no point commenting it will probably be re-uploaded over the minor errors in the fractions.
:(
Does anyone remember when people just added annotations?
I do Abram. I remember it very fondly. -Rusty
Pepperidge Farms remembers.
Sounds like a good plan to approach thesis writing!
Damn, really need to become enlightened to understand Infinity now.
"at n-th step you remove the n-th ball which means all of them are removed after infinetly many steps" but i think at that infinite step you would have put in more 10 balls of which none is removed because only the lowest numbered ball is removed. If i'm wrong correct me!!
There is no "infiniteth" step because there will be another step after it. The steps never end. Try to comprehend that. It sure took me a long time.
so i think the balls should not become zero.
What if different sized infinities are involved -- I know there are different classes of infinities that can be bigger than others. Those are not the same thing, but it seems to me the conclusion about this more a sign that the infinity is being treated is flawed. Saying (10 - 1) * inf = 9 * inf = inf also works if if infinity is all just infinity, but leads to infinity = 0, which cannot be true, thus the paradox, two conflicting values. Someone else might claim this totally wrong, but that's my reasoning, and too me shows that this concept of "infinity" as one particular thing, as though it was some specific ultimate value, is flawed.
Yugaldeep Singh I somewhat have to agree. if you compare the set of balls you add and set of balls you remove, for each ball you remove you can find 9 balls you added. even though they are both infinite, one must be a larger infinity since they can't match up. isn't this like how 0 to 2 is larger than 0 to 1 even if they are both infinite? IDK I'm not a perfect mathematician
Ryan Lynch but at step 'i' you're removing ball labelled 'i'. If you continue to do this till infinity, there won't be any balls left in the urn. That's the paradox. There isn't a bigger infinity in the picture here. The point was to show that you could perform the exact same experiment and get different outcomes depending on how you perform it.
At 7:20 i dont think that both the scenarios are same.
In the second case we certainly dont have any ball remaning as the nth ball is certainly removed at the nth step, while in the first case its just that the probability of any ball remaining is 0. which does not mean the ball cannot remain at noon , its just that the probability is 0.
It seems that as long as all balls are selectable for removal by the "end" of the infinite series (meaning possible in all steps), the total probability will tend to zero. A selection strategy like choosing the last one, or the median ball, leaves the majority of ever-increasing balls ineligible for selection, tending toward infinity.
I am at 4' into the video and my expectation is, that she will soon say: Na, I'm just f* with you. There is no urn that can hold infinitely many balls.
7:47 smol matt! so cute! x3
I need Matt as my physics conscience
While we want to assume (somewhat correctly) that on the Nth step, we will remove the Nth ball, and there will still be N*9 balls in the urn, the real paradox is that we will be doing the task infinite times. Therefore, to remove the infinith ball would be the end of the task, which would mean that all the balls were ultimately removed. And that just messes with everyone's brain.
Median Ball Urn Challenge:
Each iteration we add 9 balls, so after n iterations we have added 9n balls.
Since we remove the median each time, after n iterations we have removed n median balls.
Leaving the total number of balls in the urn after iteration n to be 9n -n = 8n
The median at step n (Before we remove it) is in position (9n - (n-1) +1)/2 = (8n +2)/2 = 4n+1
From this we can see that we never remove a ball which is in position which is a multiple of 4 and the total number of balls in the urn (8n) tends to infinity as n grows large.
There will be an infinite number of balls in the urn when the supertask is completed.
So our existence in the universe is of probability zero, but it did happened. God somehow don't roll a die.
Correction: the equation 9^n/(9^n+1) should actually be 9*n/(9*n+1) ... and like others have said the step 3 fraction should be 27/28, not 28/29
For the ern problem: the series 10-1+10-1+10-1+... does not converge, so the "final" number of balls is undefined.
5:32 Does it really converge to zero?
6:36 You add zero infinite times. 0 times infinity is not always 0.
There's a subtlety to what they mean here by "how many balls are there after an infinite number of steps?" They do not take it to mean it to be what the number of balls tends to after an infinite number of steps (this would, as you point out, lead to infinity). Really what they are asking is: "How many balls are never removed after a finite number of steps?" Since the Nth ball is removed at the Nth step, each ball will be removed after a finite number of steps. Why they framed it in the latter sense probably has to do with 1) the history of the problem and 2) the fact that probability is theoretically supported by measure theory where this framing is more natural.
The urn problem isn't a problem of addition, but a set theoretic problem (so you don't try to calculate how many balls but instead you look at the set of balls in the urn and add and remove specific balls from that set and ask whats in the set at the end?)
If you think the urn as a set, the limit diverges.
No what? It looks like this: You start with the empty set, then add the numbers 1 to 10 to your set,t hen you remove the number 1. Then you add the nubmers 11 to 20 and remove 2. There is no number remaining if you do that infinitely often. There are no limits involved
I dont think it is empty too. On the n-th step: You add (10n - 9), (10n - 8), (10n - 7), ... (10n) and you remove n. Then you have the balls (n+1) to (10n) in the urn.
How does the deterministic supertask work? Shouldn't you be looking at both the negative change *and* the positive change? In that case the answer is infinity, after adding up infinite nines.
with infinite sets, the order in which you add and remove elements matters. With finite numbers kardinals (amounts of stuff) and ordinals (orders of stuff) have identical properties, but with infinite amounts they simply don't. There is a fundamental difference between having infinite number of racers and ending up last in that infinite race.
The probability at 3:58 is wrong (it should be 1/28), and your generalized statement at 5:19 isn't right (it should be (9n+9)/(9n+10))
2:21
Uh...
lim {n->∞} (20-1)*n = lim {n->∞} 19n
Pretty sure it diverges.
Put another way: after the Nth step, you've removed N balls and kept 19 times N balls. Is there something I'm missing?
Its very easy to understand the mistake made in the modeling of the problem. Natural numbers (and the math associated with it) represent counting. If you try to count but your reference (origin) is removed (as in the first case, where the smallest ball gets iteratively removed), you simply cannot count in a straight forward way, and infinity(quantity) minus infinity(reference) gets wrongly interpreted as 0.
In the second case (where the biggest ball is removed iteratively) the reference is not removed, and a clear mathematical model(way of counting) 10*n - n = 9*n = infinity makes counting possible.
The second model can easily be used to interpret the first case (exchanging the indexes of the balls removed), as the indexes of the balls are irrelevant to the task of counting balls. This shows that there is no paradox, there is inadequate modeling of the problem.
Despite this video failure, the channel beatifully explained infinity modeling in previous videos.