I'm actually more impressed by the ability to look at two random numbers and know if they are coprime within a few seconds.

Me: Can you calculate pi? Matt: 3.04 Me: I think it's a bit higher. Matt: 3.05?

From that day forward it was forever known that Parker pi is equal to 3.052338478336799.

When you're doing a time lapse like this in the future, can you please put an analog wall clock in the frame behind you so we can see it whirl around when you're working?

a suggestion for next year: 1. at a constant rate, dig a tunnel through the center of the earth 2. at the same rate, walk along a path formed by a great circle around the earth 3. divide the walking time by the digging time you can expect some error but it would make for great video

"I probably made some mistakes... I'm kinda hopping the mistakes have averaged out." It's good coming back to older videos and seeing that the spirit of Matt hasn't changed one bit in essence.

There are 120^2=14400 possible pairs of numbers between 1 and 120. 8771out of them are coprime pairs. This gives a value of pi~sqrt(6*14400/8771)~3.13857.

As Pierre Roux suggested, I have used the digits of Pi as a source of random numbers from which to calculate pi! Taking the first 10^6 digits of pi, and splitting them into 10^5 integers of length 10, I got 5*10^4 pairs of random integers (assuming the digits of pi are truly random - this is unproven). Comparison of these integers yields the estimate 3.14099105107796, which is accurate to a 0.00019% error :) PI-Ception!

the montage starts at 3:14, coincidence? i think not

how long until the Parker square jokes when he doesn't perfectly calculate pi

I know all PI digits, not in the correct order though

I'm probably not the first person to mention this, but in Excel you can have it just count a column in its entirety. You don't need to define a set range, you can just put COUNT(C:C) and then you can add more rows at will without having to edit that function.

Euler strikes again. great video!

For those following along at home, these are the methods of calculation he has used in order of decreasing error: calculating the first terms of an infinite series (off by 3.1%), rolling two dice 500 times (off by 2.8%), timing a pendulum of known length (off by 0.433%), weighing a cardboard circle (off by 0.31%), and measuring a circle using pies (off by 0.1035%).

12:20 "63 has a 3 in it, it is 31*3" XD, You need a nap Matt

so if you wanted to do this perfectly, you would need 2 infinitely sided die, which are spheres... PI strikes again

Next year throw darts at a circle

If your happened to get really lucky and select exactly 304 dice, it would give an approximation for pi about: 3.1414043... which is astronomically close :).

3:47 Yes, 169 is a great darts score. It's also impossible to score 169 in just one visit.

"Impossibly the most convoluted way i ever done" - 3blue1brown appears with calculating Pi by number of collisions.

10X10X5 is a Parker cube

Another simpler way: - Take random pairs of numbers in [0, n] and interpret them as x and y coordinates within an n*n square - For each pair, calculate the distance to the origin. - Count the number of pairs for which that distance is less than or equal to n Effectively, you've just determined whether or not that point is inside of a quarter-circle centered at the origin, with a radius n. So, the probability that a pair will be in that group is: p = ((area of circle) / 4) / (area of square) = ((πn^2) / 4) / n^2 = π / 4

Absolutely love you dude, makes maths so much more interesting

Thank you Matt, I was looking forward to getting home and seeing this one. Cheers sir.

Right now i'm gonna write a code so that the range of random number generated is much larger, thanks matt this will be a good test since i'm only a begginer in coding

Actually, the random function gives you a number in the interval [0;1). So the random number is between zero and one and can take on the value 0, but not the value 1. When you multiply by 120, you get a number in [0;120). *The correct way to proceed* at this point is to *round down and add one*. This gives a number in the range [0;119] plus 1, which is in the interval [1;120]. If on the other hand, you round up, you mess with the probabilities, because in case the random number is 0, you get roundup(0·120)=roundup(0)=0, which is outside of your desired range. Also, drawing 120 is slightly less likely because the event of the "direct hit" (where no rounding is required) is missing (which is kind of a sloppy explanation, but I hope y'all get what I mean).

So much love for what you do! Great to see.

The proof for the sum of 1/k^2 is such a beautiful proof, it's absolutely amazing

Tip: If press F9 excel will recalc you table and update randoms.

correction 12:28 63 is 3*21 not 3*31

Thank you Matt. This is going to make the terrific activity when my Year 7 class studies Probability at the end of the year - it links together the work we will have done on prime numbers, and on pi - and we'll have an Excel exploration at the same time.

Waaaay more entertaining than this should have been. Glad you pulled in the Excel how-to. Cheers!

Stand-up maths 2030: calculating points by baking 100 pies of random diameters

2 years and almost 2,400 comments later I'm sure someone has already suggested this, but you can use =RANDBETWEEN(1,120) for a much easier way to get a random integer between 1 and 120. You can also simplify your coprime formula to =(GCD(A1,B1)=1)*1 -Excelphile

I have actually a nostalgia feeling to this video... one of many and many I watched in one stage of my life, but it took me straight back

I'm so glad you show the proofs.

but this was uploaded on the 13'th matt you had one job

"Not a very rigorous proof..." A Parker proof? ;)

Wow how exciting!!!! This exactly what I wanted to watch for more than 2 minutes on my Tuesday afternoon

I was so happy when you were at the science fair I was like isn't that the guy from standupmaths? Thank you you really made my day :)

Hey Matt, I know I'm months behind on seeing your awesome video, but I have a question; Could one potentially generate more results with fewer rolls by, say, rolling 3 'random' numbers (a, b, and c) and comparing 'ab', 'ac', and 'bc' for cofactors and coprimes? Or, would that introduce some kind of error that defeats the purpose of such a method?

Brown paper is too mainstream; it's all about that brown table, dawg!

I ran it on a D120 2500 times and got within 0.6% on the first try. You're the best mathematician of our time Matt! I love watching your videos!

Please continue, I love your videos!

The moment that 6/pi^2 formula came up, I was laughing my ass off, because it occurred to me why you were rolling giant dice, and how you would approach the problem. Pi is used in a lot of really weird stuff, lol.

The probability of 2 random numbers being coprime is only 6/π² if you consider _ALL_ numbers, like, from 1 to infinity. If you only consider numbers within a specific range (such as from 1-120 or 42-63360 or whatever), the probability will be slightly higher than 6/π², and you'll get an experimental approximation that likely undershoots pi. Not sure if that was mentioned or not. Edit: yes it was 8:41

I wish you'd talk more about the sneaky Pi proof. You explanations make things much easier for me to understand!

Thanks Matt. That was a fun little afternoon project.

"that may have been my largest understatement" literal lol

Nice one. I guess the pi estimation would improve with larger sample size. Now try again with 5000 dices.

Happy pi day! I was already planning an excel sim as I was watching - going to get that bad-boy going as soon as I get to work!

Happy PI-Day. This video was awesome! I actually replicated it with slightly bigger max. rnd.numbers (100Mio.) and some more trials (5k) - estimation error is around 0.00-something. Loving it! Thx :)

PI DAY OMG I FORGOT

18:18 "I am all about the underestimates" and posted a Pi day video a day early. Nice.

As soon as you said 6/π², I thought of the Basel problem, but I couldn't prove it before you got to that point in the explanation. But once again, we see that familiar numbers do indeed indicate a connection.

Brilliant! I'm going to do it with C-Sharp. Thanks for another lovely pi-day video.

So, I just had to pull out a cpp file and do it myself. Using numbers ranging from 1 to 2^16 and 1000000 samples, I was able to get PI with 3 digit precision. Kinda impressive considering you can't even notice the time it takes to execute.

When I got here there were 1459 comments. Did you know if you take the sum of the cubes of the digits of 1459, and then take the sum of the cubes of the digits of the result, you'll get back to 1459? Try it!

Matt, I think you've really excelled yourself!

Maths is so beautiful, and it never ceases to surprise and amaze me.

So can we use Parker pi to calculate the circumference of a Parker circle?

x is equal to 6 over the parker square of pi

21:25 pro-tip: use C1:C to specify entire C column starting with C1; or C1:1 to specify entire 1 row starting with C1

This is, what I can say, a mind blowing proof. I am actually from Electronics and Communiation background and Fourier series and transform is our base of Job 😒. I know using Fourier we can get that pi^2/6 =1/1^2 +1/2^2 +.... and we use it many times in electronics. When you first showed the probability it remembered that result which I accustomed of and stopped your video and tried to solve it using that. I failed and then went through your proof. It seemed mind blowing to me.

I wrote a python script for the same which did 100k iterations and the range of random number was (0 to 1M),got 3.1416 as the result,it was amazing to see how the result started to converge towards 3.14 with the increase in number of iterations ☺

Can anyone explain why those dice have only III and VIII?

Now I'm curious to how the variance in the estimate varies with the number of tries. Thank you for including the Excel model

EVERYONE WRITE TWO RANDOM INTEGERS AND WE CALCULATE PI

Greatest common *divisor*, not denominator.

you can also put points in a grid in a circle/square and then do the ratio between points where x^2+y^2 is smaller/bigger than 1. funnily enough this actually corresponds to the integral of the volume of a tube. (unit circle 1 unit above the ground and then volume underneath that)

Hey Matt, I got a question for you. I know you're busy calculating answers to math and probably don't have the time to give an answer to this question, but I'm wondering what thoughts you have regarding the philosophy of math in general. You're always calculating pi, looking at primes etc... and I just wonder how you couch all that with regards to the greater/abstract implications of it all. Or maybe you don't think much about that stuff, that's fine too. Maybe you've waxed on this stuff before in previous content and I've missed it or forgot it, I'll double check. Anyways I really enjoy your content, even if I'm fairly poor with math I appreciate your intellect, wit, and ability to convey your work easily - have a good day.

He would have had to count less if he counted just the dice in the cofactor box.

"To prove..." - we believe you!!!

the idea that most numbers are big just tripped me out, thanks.

WOW! Escher _AND_ origami in the background? Truly a man of taste!

What I'm taking away from this is that trying to calculate pi makes you crazy.

If somebody wants to try it in python: #pi from gcd(m,n) import math as m import random as rand max = 10 ** 6 #maximum random numbers n = 10 ** 3 #number of tries count = 0 for i in range(n): if m.gcd(rand.randint(1, max), rand.randint(1, max)) == 1: count = count + 1 pi = (6 /(count/n)) ** (1/2) error = (pi - m.pi) / m.pi * 100 print('my PI = ' + str(pi)) print('error = ' + str(error) + '%')

I had this quite fun assignment for maths class on my uni where they asked us to approximate pi anyway you wish as long as you could back up your methods. I decided to go by the circumference of a circle being pi*d or in this case we'll use pi*2r and then continued to use the circumference of polygons of "radius" 1 with increasingly more edges. So if you make it easy on yourself you'd have a triangle for example you can divide it up into 3 pieces by drawing a line from the middle to each corner. Now every one of these sections would have an angle of 360/3 degrees. For the circumference, you can name the two "radii" of 1 a and b and slap the cosine rule against it. So in this case sqrt(1^2+1^2-2cos(120)) =~ 1.73. Now triple this for the number of edges and you got yourself your first-order approximation of pi*2r = 2pi = 5.19 -> pi = 2.595. Not that great. But for a triangle, not all that bad either. Now I found that for any given polygon with N edges you can write up this formula for its circumference as a function of N. So the 360/3 would logically become 360/N for an Nth polygon and at the end instead of multiplying it by 3 you'd multiply it by N instead. This leaves us with this beauty of an equation: N*sqrt(2-2cos(360/N)). This allowed me to quickly drag and drop a column in excel showing the progress of polygons approximating pi for each order of N. To have a nice little taste of the result we'll plug in N = 100 and see where we get: 100*sqrt(2+2cos(3.6)) = 6.282.... -> pi = 6.28215/2 = 3.141075. A hundred terms further we suddenly have pi accurate to the third decimal. Now obviously that's very slow progress but considering I had a very simple algorithm to follow I could easily bump up the value of N and reach quite a decent accuracy in this way. For example the 1000th term we get 3.141587 which rounds to two more decimals already. As we allow N -> infinity we of course find the exact value of pi eventually. Nothing too groundbreaking there, just thought it was a fun assignment and I was pretty stoked about my method actually working out :p

Nice explanation of p hacking at the end!

"exactly the same thing yourself..." me: lol, too late I did it in c him: "...in excel"

To recalculate or re-run formulas in excel, simply press F9. At least it works in windows which has a Function Key row.

Love your videos! Just wanted to go e you a heads up when you changed the number to one million you just also needed to change cell E4 to a million as well And you would have gotten results closer to pie! Again love your videos keep them coming!!!

Very cool video! Here's a suggestion for next Pi Day: Run a Monte Carlo simulation of pairs (x, y) of random real numbers from -1 to 1 (or 0 to 1 if you prefer) and count how many are inside the unit circle, i.e., with x^2 + y^2 < 1. The fraction should be π/4, so multiply the final result by 4 to obtain an estimate of π.

Oh God, it hurts so much seeing someone not closing the marker pens they're not using.

Marginal return on effort seems to be limited here. By that I mean, increasing the max number only gets you a few fractions of a percent closer to the real value. Which compares with the infinite sum from last year, where adding a lot of terms adds only a minor and diminishing correction each time. For next year's Pi day, divide 32 by 10, then come to America and visit the Indiana legislature and try and convince them that it should be changed to 3.2 so you'll be correct. Again.

Mathologer did a very good video about the Basel-Problem with the use of Euler's prime formula to derive, where 6/pi2 comes from.

Nice video man! A lot of patience!!!! Can I ask you where did you get that amazing mug?

Matt Parker to be the next Doctor!!

Wow just take a look at the side of the two boxes! Look at how many dices with 3 are facing us! What are the odds of that?

I think this is the first time I've ever heard someone say "cuboid" instead of the bland "box". Definitely an underused and underrated word.

You could use floor(rand()/rand()) to get a random positive integer that goes up as high as the precision of rand allows. I'm not certain if the probability is uniform, though.

Err... I do know how to program but I don't know how to use Excel lol :D

pst... Use RANDBETWEEN in Excel to generate a random integer between two numbers.

I have the same cup! I didnt knew that, nice vid!✌️

I played around on Excel. My spreadsheet now does around ten million trials (maxing out from row 2 to the very bottom, ten times) with the numbers varying between 1 and (2^53)-1, which is the maximum value that the GCD function can deal with. My result for Pi right now is 3.141652, but Excel just died on me leaving a 450 MB file which will probably overheat my computer if I try to open it. After vastly improving the calculation I'm still only getting a few significant figures correct. I didn't have chance to repeat measurements before Excel died on me to see how much variation I'm getting in my results. Really interesting video, by the way - thanks!

What if you alternated rolling dice so each roll was used in 2 calculations. i.e gcd(a,b) , gcd(b,c) , then gcd(c,d) etc. Would that change the probability?

This is a little bit of a Parker pi, isn't it?

You get a gold star. Fascinating.

So much unjustifiable rage at Matt mixing up denominator and divisor

I just realised I know avery digit of pi, just not in the right order.