Could you say that since every point has a uniform distribution and if the coin flip does not bring the point to the end it is simply picking another initial point thus the outcome is solely depended on the coin?
@midou6104
4 күн бұрын
7:38 You should prove that f(x) increase in a construction speed Right! Right!?
@elunedssong8909
5 күн бұрын
Before watching video: Notice that from the perspective of the mid point, going to the right is identical probabalistically to going to the left. Therefore: The inverse probability of any statement is true of the opposite side. Next, notice at 1/2 of the line, the odds of going to 1 (which we will now ONLY talk about these odds, as the inverse is the same thing) is a 50/50 AND it must happen on a single coinflip. Next, let's ask about the point 1/2+1/4 This point is at 3/4ths, and initially the circle is 1/4th wide. But if it goes left, we return to the middle point, who is then a 50/50 (and will conclude the experiment) So we do a 50/50, into another 50/50 to NOT go into the 1. Therefore: at 3/4ths, the probability of going to 1 is a 3/4. Repeat ad neasuem. Now the only question that is left is how does this work when you are not perfectly afixed to a point who is in the series 1/2 + 1/4 + 1/8... or 1/2 - 1/4 - 1/8 ...... Let's study the charachter of this movement from .6 .6-> .2->.4->.8->.6 We find a repeating cycle from the initial position, around back eventually to the initial position. The length of time to leave the side you started on is: what bracket of 1/2+1/4+.... you were in when you started. Therefore: We know for a fact that the 1/2+1/4+1/8.... probabilities define "probability buckets" for their neighbors, which I will now arbitrarily declare to be relatively linear, and therefore: The odds to go to one if you are at .6 is .6 and the odds to go to one if you start at .99 is .99, and vice versa. The odds to go to one at .01 is .01. Maybe I was too quick to finish the problem, but I think i'm satified with my solution. After video: My solution is right? Lets go! The final tidbit at the end, if i had to guess, it is following the bucket sizes. Where the largest relative maximum happens exactly at 1/2+1/4+1/8.... and 1/2 - 1/4 - 1/8 - 1/16th.... Why this is? It seems intuitive, and so its kind of hard to explain, but when you are drifting towards one or the other, being on one of these bucket values makes you have less total coin flips before hitting the end. For instance, 1/2 is a 50/50 to hit into 1 IMMIDIATELY, but so is slightly less than 3/4ths. I would guess this is why the graph is no longer linear, and all bumpy. Edit: oh also there is a special charachteristic of points like 1/2+1/8th. if they fail to go to one, its - (1/2 - (1/2 + 1/8th)) for the next point location = 1/8th. But 1/8th is in the series that's perfectly collapsing. So there is another something special about these points, which directly lead into collapsing coinflips (determenisticly ending, cannot endlessly loop)
@AYMAN-tz6jb
6 күн бұрын
A bit late, but I have found two answer! first : [1, 2, 2, 3, 3, 4, 4, 5] [1, 3, 5, 5, 7, 7, 9, 11] second: [1, 2, 3, 3, 4, 4, 5, 6] [1, 2, 5, 5, 6, 6, 9, 10 ] edit: found third one: [1, 2, 2, 3, 5, 6, 6, 7 ] [1, 3, 3, 5, 5, 7, 7, 9] great video :)
@joaoguerreiro9403
6 күн бұрын
Math and computer science channels are definitely my favourite! Please keep it up ❤️❤️
@Jiglias
8 күн бұрын
my guess without thinking about it too hard: I'd say the probability of it reaching 1 is equal to x. The probability of reaching one if x = 0 is 0%, if x = 0.5 it is 50%, and if x = 1 it is 100%. if we than take x=0.75 than there's a 50% chance of immediately reaching one and then a 50% chance of going to 0.5, which you than have a 50% chance of reaching one, i.e P(reaching one if x = 0.75) = (1+0.5)/2 = 75%. It's not to hard to generalise this, lets take two values that we already know the probability for, call them a and b and let x be their mid point i.e: Let x=(a+b)/2 The probability of reaching a is 50%, and than once there you have P(a) chance of finding one, this is also true with b, so: P(x) = (P(a) + P(b)) / 2 and since we know that in at least two cases p(n)=n we can say P(x) = (a+b)/2 P(x) = x I don't really know how to lay out proofs though, but this made sense to me
@Aleixus
10 күн бұрын
I didn’t really understand the problem… on average, 1/2 will end up with 1 and the other 1/2 will end up with 0…. That’s it I think
@DeclanPeterson-s8n
10 күн бұрын
I have this interesting idea for the Riemann Zeta function. It uses derivatives of the permutation function.
@AalapShah12297
10 күн бұрын
It's very easy to fall into the trap of assuming f(x) = x at 6:24 and call the problem solved. While the answer is still correct, the fact that this simple function becomes a complex fractal for any P≠0.5 is a good reminder that the kind of rigorous analysis you did at the end was indeed necessary for the proof.
@KaiTyrusNelson
11 күн бұрын
Could you prove continuity and then just form a sequence on the dyadic rationals for any real number between 0 and 1?
@user-td4ii9px4s
12 күн бұрын
Great content.
@ccolombe
12 күн бұрын
I have been looking for a channel with some interesting puzzles! Lets go
@ShafinHossain-lt7zf
13 күн бұрын
Loved your video and want more, subscribed!
@PurpleMindCS
12 күн бұрын
Thanks!
@ellielikesmath
13 күн бұрын
i have no idea how fast actual math geniuses must process what you're saying, but i can tell you this content is way, way too fast for me to follow or be all that interested in.
@ShivanshSharma
14 күн бұрын
Collaboration and Clear Clarification.
@viks3864
16 күн бұрын
Great video! It was really well explained and is something which I haven't seen anyone else really talk about. Looking forward to the next part :D
@human7767
16 күн бұрын
Thank you!
@ckq
16 күн бұрын
Bro these types of videos trigger me cause it's so painfully obvious to me that E[x] = x so obviously the probability it's 1 is x and 0 is 1-x. Its basics of random walks which have been studied forever
@ckq
16 күн бұрын
Like why are we wasting 10 minutes? OK the last part was actually kinda interesting since I also independently discovered that by considering a random sequences of 1s and 0s where the probability the next term is 1 is (1/3 a[n-1] +2/9 a[n-2]..... + 0.7)/3 so the maximum probability is 0.9 when all previous values are 1.
@chenyongyan6664
16 күн бұрын
this community is amazing, I'm just too dumb for it😭
@NAFTA2012
16 күн бұрын
Very interesting challenge, I stumbled uppon it just recently. So far I've noticed that there is quite obvious constraint regarding 2 faces of each dice. This simplified brute focing D4 and D6 which gave me possible pattern for first dice. This pattern helped me brute forcing D8 and that lead me to pattern for second dice. So far I'm unable to prove how many alternative dice pairs exist but I'm able to generate alternative dice pair for any even sided pair of dice. I also just noticed requirement for symetric distribution but I'm not certain that this is only possibility.
@nightchicken3517
16 күн бұрын
Pi thousand subscribers. Nice
@dariusgoh5314
16 күн бұрын
Great vid pls blowup
@Chikhi_Salah
16 күн бұрын
That is such a nice problem
@PurpleMindCS
16 күн бұрын
Thanks!
@Hamster_Tool
16 күн бұрын
Can you share video code
@CallmeAnythingXD
17 күн бұрын
lol I love how youtube algorithm recommended me this video. Keep going!
@PurpleMindCS
16 күн бұрын
Thank you!
@madhavgullapalli505
17 күн бұрын
I solved it 30 seconds without proof ... average CS major.
@Mulakulu
17 күн бұрын
I wonder if there is a way to flatten the curve of rolls between 2 and 12 where each is equally likely, or at least as equal as possible
Method #2는 그저 함수가 단조증가함을 나타낼 뿐이며, 6:50의 그래프와 같이 점 사이의 파형이 어떻게 나타나는 지를 이야기 하지 않습니다. 함수는 각 점에서 선형적으로 증가하고 각 점 사이에서 단조증가하지만, 여전히 각 점 사이에 파형이 존재할 수 있습니다.
@TheEGod.
17 күн бұрын
0:30 in an im just gonna guess they are equal, so 1/2. since like if you start with a number less then 1/2, all the probabilitys for it should be equal to the corresponding number greater then 1/2, just opposite, so yeah. im most likely wrong though causw that solution seemed too easy
@berlinisvictorious
17 күн бұрын
Hey can you someday make tutorial on how to make these types of animations, we need way more channels like this!
@PurpleMindCS
17 күн бұрын
I use a library called Manim, created by 3Blue1Brown. The community edition is open-source and pretty well documented: www.manim.community/
@mosesgklein
17 күн бұрын
Here's my non-rigorous attempt at insight into the weird shape of the graph for the biased coin. (It can't be fully rigorous because we haven't rigorously defined the weird shape.) It's based on the argument from @B-nu8ss for the fair coin, based on the expected value of the position being x at every step (also how I solved the original problem). That isn't true for a biased coin -- but it isn't true in a particular way. The more steps, the less likely the point is to defy the bias. Let's suppose the coin is biased toward moving left -- by symmetry you can adapt the argument to the other case. With every step, the expected value becomes smaller. So how many steps does it take to get to an endpoint? That's a random variable, call it Y. For most starting numbers x, Y has geometric distribution: P(Y=y)=1/2^y. But if x is a dyadic rational, say with denominator 2^n, then the distribution of Y is truncated: wherever the geometric distribution would give Y>n, we have Y=n instead. That means for dyadic rationals, especially with small denominator, the probability of getting to 1 (defying the bias in the coin) is slightly higher than a smooth pattern would predict. Those are the points where the graph seems to have a sharp corner, steeper to the left and flatter to the right. That's for P(right)<1/2 -- if P(right)>1/2, then the graph is flatter to the left and steeper to the right of those points.
@HWABAG6
18 күн бұрын
1+1=11 btw
@rohakdebnath8985
18 күн бұрын
binary fraction proof is the simplest and coolest
@IlIlllIlll
18 күн бұрын
Before watching: 💩 After watching: 🗿
@MorallyGrayRabbit
18 күн бұрын
the fair coin graph is a fractal too, its just degenerate
@kylecow1930
18 күн бұрын
oh i remember looking at this! i managed to solve it assuming P was differentiable at zero, this was needed because i got that it should be fractally repeating as you get closer and closer and needed it to look like a straight line somewhere, i didnt spend that long on it so i wonder if this assumption of differentiability is even true in the weighted case, it looks like it could be but far less clearly so
@columbus8myhw
18 күн бұрын
My solution for this problem: . . . After each coin flip, the expected location of the coin does not change. Since the process almost surely terminates, the final expected location will remain x. This implies that the probability it ends at 1 is x and the prob. it ends at 0 is 1-x.
@vamer423
18 күн бұрын
yo wth? only 160 comments. you deserve a heck of a lot more, keep going.
@PurpleMindCS
18 күн бұрын
Thanks so much!
@vamer423
18 күн бұрын
@@PurpleMindCS no need to thank me brother, your content speaks for itself
@lagomoof
18 күн бұрын
The fractal looks like a relative of (if not an instance of) the so-called blancmange curve which, unsurprisingly, is defined in terms of powers of two. Wikipedia currently has an article on it, as does MathWorld.
@IsaacDickinson-tf8sf
18 күн бұрын
A puzzle I’ve tried to solve is the probability that when picking k dominos from a triangular set from double blank to double l, that they form at least one chain that uses every domino picked, Mexican train style. There are countless ways to extend this such as the probability that a certain domino is the beginning of the chain. For k=2, I got (4l-4)/(l^2 +5l +6) and I’ve also figured out k=3, too.
@HL1_EP1
18 күн бұрын
Isn't just adding a constant value to each of the sides of each of the die going to result in the same probability calculations? Although I guess it wouldn't be in the spirt of the solution-finding we should discover.
@dawidolszewski-bi2wr
18 күн бұрын
Great videos, I am looking for more
@PurpleMindCS
18 күн бұрын
Thanks so much!
@alejandromorera3241
19 күн бұрын
I believe there is this book by Barnsley which treats these kinds of self-referential functions. I would highly recommend any viewer which is curious about these kinds of functions to read “Fractals everywhere”.
@maths.visualization
19 күн бұрын
Can you share video code??
@GroundThing
19 күн бұрын
Came here from the follow up, but working through it before watching that, and here's my rough sketch of a proof of the closed form of the probabilities P_0 and P_1, the probabilities that the process converges to 0 or 1 respectively: Starting at x_0, after 1 trial, you have a 50% chance to go to 0 or 1 (which one it winds up at being the floor of 2*x_0, assuming x_0 ≠ 1, which is a trivial case as you will remain at 1 with a 100% probability, and x_0 ≠ 0.5, in which case it should be 0, coupled with the following probability) and a 50% chance to wind up at x_1, which is 2*x_0 if x< 0.5 and 2*x_0-1 if x >= 0.5. This is equivalent to taking the fractional part of 2*x_0. From there, we can observe that P_0(x_0) = 0.5*(1-floor(2*x_0))+0.5*P_0(x_1), and P_1(x_0) = 0.5*floor(2*x_0)+0.5*P_1(x_1). Using recursion, we can continue this process to x_2, to see that P_0(x_0) = 0.5*(1-floor(2*x_0))+0.25*(1-floor(2*x_1))+0.25*P_0(x_2), and P_1(x_0) = 0.5*floor(2*x_0)+0.25*floor(2*x_1)+0.25*P_1(x_2). Continuing this, further, and recognizing that x_1 is the fractional part of 2*x_0, x_2 is the fractional part of 2*x_1, which is also the fractional part of 4*x_0, and in general x_n is the fractional part of 2^n*x_0, after n total trials we get P_0(x_0) = sum as i goes from 1 to infinity of 1/2^i*(1-floor(2*fract(2^(i-1)*x_0))), and P_1(x_0) = sum as i goes from 1 to infinity of 1/2^i*floor(2*fract(2^(i-1)*x_0)). This doesn't look that much cleaner of a way to describe the probability, however the number of "2^i"s gives a key insight to the process: if we consider the binary representation of x_0, floor(2*fract(2^(i-1)*x_0)) gives a 1 if the "i"th decimal (binimal?) digit is a 1, and a 0 if the "i"th decimal(/binimal) digit is a 0. From there, you multiply that number by 1/2^i and add it to the sum, but 1/2^i is simply the number whose "i"th decimal(/binimal) digit is a 1, and 0 otherwise, so P_1(x_0) is the sum of the digits in their respective place values, in binary, of x_0, which is to say x_0, and likewise P_0(x_0)=1-x_0.
@besusbb
19 күн бұрын
Oh hey, this is really interesting, I couldn't help but notice that the problem can be thought of as a Markov reward process with the position of the point being a state with the Markov property, and thought that it could be solved approximately with a program iteratively without any random sampling, that is for example having an array storing probabilities of ending at 1 of various states, let's say V(s) and have initial values with terminal states V(0) = 0, V(1) = 1 and any arbirtrary initial value for 0 < s < 1, and then you continuously update V(s) ← 0.5V(s + r) + 0.5V(s - r) in any order and as long as you do it for all non terminal states it should converge to an approximate solution for a discrete version of the problem; but this isn't why I'm bringing it up but rather that recently when reading through Richard Sutton and Andrew Barto's book about reinforcement learning there was this exercise where you are given a starting amount of money from $1 to $99 and you can place bets on coinflips with at least $1 and at most all the money you have to either lose all of your bet or double your bet, and your job is to find the optimal betting strategy (deciding how much to bet depending on how much you have) to get to $100 with the highest probability. This can be solved by an iterative method much of which is similar to what I described above, but the point is, if you plot the value function (in this case value would be the probability for winning when using the optimal strategy vs how much you currently have) with any winning probability on individual bets <0.5, it looks exactly like the graph showed at 10:21 with that fractal pattern and all which I guess makes sense given the similarity of the 2 problems but also really caught me by surprise. So thanks for the video, I really didn't expect this connection, very cool problem.
@Lespati-wy9dy
19 күн бұрын
Today, I came up with a puzzle: you have a function (c(x) = sin(x)) and a circle r=1. What is the highest y coordinate of the circle where it still touches the curve at each x position? The height is then given by the function h(x). Things I found out: h is oscillating from 1 to 2. Create the function f(x)=c(x)+sqrt(1-(x-x0)^2), x0 being the x-position of the circle, the highest point of f equals h(x)
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