Generating π from 1,000 random numbers 

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

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

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?

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

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

Euler strikes again. 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.

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

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!

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.

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

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

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

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

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

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

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 :).

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).

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

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

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

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 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 :)

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

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

Absolutely love you dude, makes maths so much more interesting

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.

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

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

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

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.

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

x is equal to 6 over the parker square of pi

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

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

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

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

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

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

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?

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

Please continue, I love your videos!

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!

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!

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.

I'm so glad you show the proofs.

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

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 ☺

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

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

Matt, I think you've really excelled yourself!

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

Matt Parker to be the next Doctor!!

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 :)

Thanks Matt. That was a fun little afternoon project.

"that may have been my largest understatement" literal lol

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

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!

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

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

12:24 "63, which has a 3 in it, Its 3 by 31..." I don't know if I trust his math.

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

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

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

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

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)

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.

PI DAY OMG I FORGOT

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.

u know, people like you and ideas like this simply make the world beautiful.

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.

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?

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

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.

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

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

Greatest common *divisor*, not denominator.

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

Excel also has a RANDBETWEEN(min,max) function.

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.

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

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

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?

if he keeps doing this every year and averages his results they will get there eventually

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

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

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

Nice explanation of p hacking at the end!

You could roll three red die and three blue die and use the multiplying effect (like a d100 pair of dice do) to simulate a much larger sided die.

Something I can't wrap my mind around: how can you describe fairly and randomly picking a number from the unbounded set of all integers?

it's got to be πday in some time zones, but in the UK it's still 3.13-day...

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

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!!!

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) + '%')