This is one of the best videos on your channel. At first the problem seems very random but the solution is so elegant you can't help but be left in awe.
I have a coursework in this and it took me a day to calculate this in matlab. I wish I had found this video earlier. Because it would have saved me a day lol
Get good scrubs, I proved string theory whilst finishing all of mind your decisions videos questions without seeing the solution while taking a shat in the airport at Uganda
i have to watch the second half of the video several times before i can understand the barier 's proof. my English isn't really good. now i know an interesting method for this problem. thanks
the explanation was perfect and clear, thanks a lot. There's also another way that uses both of the approaches you mentioned: using the second approach to show that the intended probability is equal to the expected value then using the second approach to find the Expectation of the mentioned random variable
The graph at 3:08 represent the space of the two variables d and theta. When theta = pi/2 sin(theta) = 1/2 but d can be larger that 1/2 because in general 0
Take this problem one step further and calculate the same probability for a needle of length 2 units. The answer would be 2/pi. However, for a needle of length 3 units, the answer would not be 3/pi.
The solution i thought of is basically : P = averageHeight/2 the height of the needle is sin(theta) in order to find the average height i integrated the sin(theta) : 90 integral (sin(x)dx) 0 by that i basically found the area created from all the different heights the area is equal to 1 radian = 57.295... then i divided the area by 90 to get the average height (just like you divide the area of a rectangle by its width to get its height) 1 radian = 180/PI average height = 180/PI / 90 = 2/PI probability = averageHeight/2 = 2/PI : 2 = 1/PI
what if in the second method we have circle of dia 2r and distance between the parallel lines as t. and 2r < t. In this case, E(circle) is not equal to 2
I calculated probability of intersection with one needle end in distance h from a line, and probability for that (excluding zero ): ½ - arcsin(h) / π this is then integrated ½ h - (sqrt(1 - h^2) + h arcsin(h)) / π + C and not surprisingly, if h 0->1 then area equals 1 / π
The way I see it that when you break it into those two parts they do not overlap. If we begin by imagining as he did a straight line split into two segments x and y we know that these two don't overlap and thus the probability of one hitting plus the probability of the other hitting must equal the probability of the whole line hitting because they fit together without any overlap that could result in double counting. Once this has been said I don't see an issue with joining them in a bent way or even separating them provided there is no overlap. Therefore I see no issues with the remainder of the proof.
Imagine it true and think of its various implications, it may start to make more sense. The first thing I did was assume a needle of length 4 compared to a square of side length 1. Otherwise the proof gets highly technical
Ugh, it's so annoying when people say "OMG, pi just showed up out of NOWHERE!" Clearly it didn't, there are circles hidden in this problem (see 2:04). Pi does not just "spontaneously" pop out of nowhere!
Starting parameters are missing. Position of each needle before it falls. Interference among needles during fall and after grounding. Surface properties, angle of reflection and its dispersion.Probability of being stuck, height, air pressure, turbulence, and so on.
But since the events are essentially random, we can assume that each occurs with equal probability and will not affect the outcome. Also, overcomplicating problems like these obscure the beauty of the solution. :P, just my two cents.
Actually, it is not quite correct, from perspective of "theory of probability". Events are not really random, there are waves =), it's complicated, you know...
+rospotreb pozor idk about you, but I like a simple 1/pi answer more than a blarghhhh something around 0.3 number lmao. and have fun computing that with all your parameters
Edit: the fact that I got within 0.04% of the answer is kinda nuts. Especially considering that I was using euclidean constructions If the angle of the needle is random between 0 and 360 degrees, then it's statistically most likely to drop at a 45° angle (give or take 90 or 180 degrees). This gives it a vertical height of about 0.7 units. If we drop it in a 2 unit tall section with a line going through the center, and measure from the center of the needle, then move the needle from the bottom to the top... (I'm using a simulated ruler/compass and shortcuts for common constructions) I suppose the vertical height (which I've rounded to 0.7) of the needle rotated at 45° can fit in a gap of length 2 units 3+(1/7) times. Or about 3.14 times. So instead of 1 in 3, it's 1 in 3.14, which is about a 0.318% chance. I think.
I do not understand the part when you said when the needle is bent, the probability equation still holds, because when the needle is bent, it can have 2 intersections with a line (like a triangle).
(I think) he is not saying that having bent needle is the same as straight. He is saying that each piece can be accounted for separately as if it was a smaller straight needle. The expected value of one needle with two pieces is the same as the sum of two expected values of single piece.
The video does not say how thick are the lines and what is the width of the needle. Also should we count cases when it is just touching the lines and not crossing? So many unknowns...
They're both infinitely thin, and it doesn't matter whether or not we count the cases of touching and not crossing, since it turns out it doesn't change the probability.
The second proof is brilliant but the case of the needle falling exactly horizontally on the line is a bit worrisome. So it's valid to say that the event contributes 0 to the expected value since it occurs on a set of measure 0, even though it would make the number of crossings uncountably infinite? I guess I can stomach that. Makes me want to take a probability course with real analysis.
Think of it in terms of limits, I have a needle of a given length and it is directly perpendicular to a set of lines (maximizes the number of intersections for this particular needle). As I rotate it, the number of intersections decreases (at some point reaching 1 as long as the length is non-infinite). Why then should it not intersect 0 times as this angle reaches parallelity?
I'll bet that if you have the lines half the distance of the needle, make them wires, run a current circuit from the bottom line around to the top line, and measure the current. Based on the number of needles I'll bet you could do some simple math with the current strength to calculate pi. It's been too long since I've done that kind of calculation to be sure, but you probably could do it.
what confused me was wrong graph in 1 proof..Max value of 1/2*sin(angle) is obviously 1/2 (height of grey zone) but in a picture it looked like it reaches around 0,8. but otherwise great video!
imposter syndrome is strong with this one. good luck to all of my fellow under 130iq people taking probability theory. thank you for this video, this is amazing.
Sorry if this is a really stupid mistake, but I noticed that the integral used radians, as it integrated from 0 to pi/2. I tried integrating in degrees, with the same function but from 0 to 90. However i got 90/pi, or about 28.64, not 1/2. I'm in precalc so please tell me what went wrong so I can learn.
Harnoor Lal when integrating by degrees, the integral of sin(x) isnt -cos(x) +c ; but is -(pi/90)*cos(x) +c. in other words, just dont integrate by degrees
@@tosca1883 And this is why we usually leave the names as they are even if they're not the original authors. Finding the true original author is difficult and causes controversy.
That is the only simple equation that gets more complicated, if you were to do more trig with sine waves or radians because significantly easier with tau
+Minecraftster148790 Puts it into the same form as 0.5mv^2, etc. The half tells us that this comes from integration. Or conversely, d(0.5tau*r^2)/dx = tau*r
I have only done GCSE maths (age 15-16 if you are American) and aren't taught about radians and gradients and stuff yet, but I do know a bit about it as I am a giant maths nerd. I don't know how awkward it is to use pi in sine waves and things yet, but I think I can understand that example that person just gave (on tablet so I can't read their name)
Because we are only looking at the values from 0 to pi/2. Anything outside this range is already represented withtin the bound we set. We also know d is at most 1.
Pi appears because you use radians. What would happen if you used degrees? If radians is the only way to reach conclusions such as this, why use degrees at all?
This problem is always presented without a clear statement of "randomness." Which variable is random according to what distribution? You cannot put a uniform distribution on an infinite measure set.
This is one of the best videos on your channel. At first the problem seems very random but the solution is so elegant you can't help but be left in awe.
I definitely like the calculus one more. It's a pain to learn, but once you do, it is just so freaking useful.
I agree, it's really important.
Same here
I have a coursework in this and it took me a day to calculate this in matlab. I wish I had found this video earlier. Because it would have saved me a day lol
I understood your explanation so much better than numberphiles explanation. thank you
really liked the second one, I would have nerver thought of that !
This is so ridiculously easy, I learned this in 6th grade along with being able to solve Chaos theory while also curing cancer.
Abca209 step up your game. I solved world hunger and poverty while on the toilet in kindergarten.
Get good scrubs, I proved string theory whilst finishing all of mind your decisions videos questions without seeing the solution while taking a shat in the airport at Uganda
Barbier's proof is just brilliant. I liked it a lot. Thanks for showing it
Ah the memories this video wakes in me. Man, I am old. Btw. Great video!
I can't decide which solution I like more. Great video!
i have to watch the second half of the video several times before i can understand the barier 's proof. my English isn't really good. now i know an interesting method for this problem. thanks
This was one of my assignments in a first year IT subject. With a large enough sample it's spot on every time :D
I prefer the calculus proof. It's simpler
one of the most interesting videos I've seen in a while!
the explanation was perfect and clear, thanks a lot.
There's also another way that uses both of the approaches you mentioned:
using the second approach to show that the intended probability is equal to the expected value
then using the second approach to find the Expectation of the mentioned random variable
There’s also a method using double integration. Would be great if that were included as well.
Second method is way more beautiful, however I can't develop the intuitive for linearity of expectation when events are dependent.
for the graph at 3:08, when theta = 1/2 why is d clearly above 1/2? since sin(1/2)=1 and d= 1/2 sin(theta) = 1/2?
The graph at 3:08 represent the space of the two variables d and theta. When theta = pi/2 sin(theta) = 1/2 but d can be larger that 1/2 because in general 0
@@christianfunintuscany1147 yes, but all that means is that d includes both the white and the grey. The gray is where d
Wow you an ilectureonline both came up with the same video idea, on the same day!!! What an incredible coincidence!
Take this problem one step further and calculate the same probability for a needle of length 2 units. The answer would be 2/pi. However, for a needle of length 3 units, the answer would not be 3/pi.
Interesting
The trigonometric method is elegant and can be explained easily.
I learnt this very example in Computer Science II at University in 1975 (!!!!) It was an example of what was called "Monte Carlo analysis".
One of your best videos
3:06 is scale off? 1/2 sin Theta shouldn't never be more than 1/2..that looks higher...
I'd say it looks about right (if the angle is 90( that would equal 1/2) if not its way to big)
The first solution made much more sense.
Really fond of the second proof. I feel like I don't appreciate continuity enough.
The solution i thought of is basically : P = averageHeight/2
the height of the needle is sin(theta)
in order to find the average height i integrated the sin(theta) :
90
integral (sin(x)dx)
0
by that i basically found the area created from all the different heights
the area is equal to 1 radian = 57.295...
then i divided the area by 90 to get the average height (just like you divide the area of a rectangle by its width to get its height)
1 radian = 180/PI
average height = 180/PI / 90 = 2/PI
probability = averageHeight/2 = 2/PI : 2 = 1/PI
3:43 the width is actually pi/2, not one half. If it were one half, the area of the rectangle would be 1/2 and not pi/2...
what if in the second method we have circle of dia 2r and distance between the parallel lines as t. and 2r < t. In this case, E(circle) is not equal to 2
I found an answer with the first method. I saw the second method and I only say wow.
I calculated probability of intersection with one needle end in distance h from a line, and probability for that (excluding zero ):
½ - arcsin(h) / π
this is then integrated
½ h - (sqrt(1 - h^2) + h arcsin(h)) / π + C
and not surprisingly, if h 0->1 then area equals 1 / π
thanks for this. I'm reading Evan's introduction to stochastic differential equations and his explanation of this same problem is extremely lacking.
Happy Pi Day! (At least, it was here when this video was uploaded.)
I like the second proof, but how/why can you assume that E(x+y) = E(x) + E(y) ? (at 6:52)
The way I see it that when you break it into those two parts they do not overlap. If we begin by imagining as he did a straight line split into two segments x and y we know that these two don't overlap and thus the probability of one hitting plus the probability of the other hitting must equal the probability of the whole line hitting because they fit together without any overlap that could result in double counting. Once this has been said I don't see an issue with joining them in a bent way or even separating them provided there is no overlap. Therefore I see no issues with the remainder of the proof.
Imagine it true and think of its various implications, it may start to make more sense. The first thing I did was assume a needle of length 4 compared to a square of side length 1. Otherwise the proof gets highly technical
Really cool video!
So clever man
Ugh, it's so annoying when people say "OMG, pi just showed up out of NOWHERE!" Clearly it didn't, there are circles hidden in this problem (see 2:04). Pi does not just "spontaneously" pop out of nowhere!
I guessed 1/3 at the beginning and can't believe I was so close :D
So would the probability of crossings be the same if u throw a flexible rope of length one? Does that follow from this?
No, but the expected number of intersections will be the same.
A rope may have multiple crossings while a needle won't.
2:01
shouldn't theta be the non-obtuse angle instead of the acute angle?
Or is any right angle acute as well? :/
Starting parameters are missing. Position of each needle before it falls. Interference among needles during fall and after grounding. Surface properties, angle of reflection and its dispersion.Probability of being stuck, height, air pressure, turbulence, and so on.
+rospotreb pozor I don't see why this is relevant though.
not obligatory. though, often imaginary tasks are oversimplified. it leads to rough mistakes in mental composition.
But since the events are essentially random, we can assume that each occurs with equal probability and will not affect the outcome. Also, overcomplicating problems like these obscure the beauty of the solution. :P, just my two cents.
Actually, it is not quite correct, from perspective of "theory of probability".
Events are not really random, there are waves =), it's complicated, you know...
+rospotreb pozor idk about you, but I like a simple 1/pi answer more than a blarghhhh something around 0.3 number lmao. and have fun computing that with all your parameters
LOL it's been 10 years since I last studied this and I got it right
I never felt so confused in math as I am right now
Beautiful!!!
Edit: the fact that I got within 0.04% of the answer is kinda nuts. Especially considering that I was using euclidean constructions
If the angle of the needle is random between 0 and 360 degrees, then it's statistically most likely to drop at a 45° angle (give or take 90 or 180 degrees). This gives it a vertical height of about 0.7 units.
If we drop it in a 2 unit tall section with a line going through the center, and measure from the center of the needle, then move the needle from the bottom to the top... (I'm using a simulated ruler/compass and shortcuts for common constructions)
I suppose the vertical height (which I've rounded to 0.7) of the needle rotated at 45° can fit in a gap of length 2 units 3+(1/7) times. Or about 3.14 times.
So instead of 1 in 3, it's 1 in 3.14, which is about a 0.318% chance. I think.
Found it much easier to follow proof one
At 9:01 when you write the equation E(Pn)
E(x) = x E(1)
That's what was used
I do not understand the part when you said when the needle is bent, the probability equation still holds, because when the needle is bent, it can have 2 intersections with a line (like a triangle).
(I think) he is not saying that having bent needle is the same as straight. He is saying that each piece can be accounted for separately as if it was a smaller straight needle. The expected value of one needle with two pieces is the same as the sum of two expected values of single piece.
Cool !
What if the line spacing and the length of the needle are same?? How can we relate d & l then??
That's definitely a clever proof(Barbier's proof)
The video does not say how thick are the lines and what is the width of the needle. Also should we count cases when it is just touching the lines and not crossing? So many unknowns...
They're both infinitely thin, and it doesn't matter whether or not we count the cases of touching and not crossing, since it turns out it doesn't change the probability.
The second proof is brilliant but the case of the needle falling exactly horizontally on the line is a bit worrisome. So it's valid to say that the event contributes 0 to the expected value since it occurs on a set of measure 0, even though it would make the number of crossings uncountably infinite? I guess I can stomach that. Makes me want to take a probability course with real analysis.
Think of it in terms of limits, I have a needle of a given length and it is directly perpendicular to a set of lines (maximizes the number of intersections for this particular needle). As I rotate it, the number of intersections decreases (at some point reaching 1 as long as the length is non-infinite). Why then should it not intersect 0 times as this angle reaches parallelity?
Not surprising that pi appears here.
Well, after learning Euler's Identity, one can expect e and pi to appear anywhere and everywhere.
I'll bet that if you have the lines half the distance of the needle, make them wires, run a current circuit from the bottom line around to the top line, and measure the current. Based on the number of needles I'll bet you could do some simple math with the current strength to calculate pi. It's been too long since I've done that kind of calculation to be sure, but you probably could do it.
Awesome
You defined it so that touching is a form of intersecting by saying d
Phoenix Fire see what answer you get it you consider your case
The difference that makes in the answer is exactly 0.
good
what confused me was wrong graph in 1 proof..Max value of 1/2*sin(angle) is obviously 1/2 (height of grey zone) but in a picture it looked like it reaches around 0,8. but otherwise great video!
imposter syndrome is strong with this one. good luck to all of my fellow under 130iq people taking probability theory. thank you for this video, this is amazing.
Sorry if this is a really stupid mistake, but I noticed that the integral used radians, as it integrated from 0 to pi/2. I tried integrating in degrees, with the same function but from 0 to 90. However i got 90/pi, or about 28.64, not 1/2. I'm in precalc so please tell me what went wrong so I can learn.
Harnoor Lal when integrating by degrees, the integral of sin(x) isnt -cos(x) +c ; but is -(pi/90)*cos(x) +c. in other words, just dont integrate by degrees
Thank you for the help, I realize my mistake now.
Shouldn't the integral be equal to zero? Sintheta dtheta
More like De Gea needle problem...
Thamer AL Sadoun casillas needle problem
or even neuer needle problem
@@tosca1883 And this is why we usually leave the names as they are even if they're not the original authors. Finding the true original author is difficult and causes controversy.
Looks like the second method might be more easily understood if everyone wasn't so attached to Pi, and maybe were taught to use Tau.
What about pi R squared? Tau over 2 would be a pain and pi would be much better
That is the only simple equation that gets more complicated, if you were to do more trig with sine waves or radians because significantly easier with tau
+Minecraftster148790 Puts it into the same form as 0.5mv^2, etc. The half tells us that this comes from integration. Or conversely, d(0.5tau*r^2)/dx = tau*r
I have only done GCSE maths (age 15-16 if you are American) and aren't taught about radians and gradients and stuff yet, but I do know a bit about it as I am a giant maths nerd. I don't know how awkward it is to use pi in sine waves and things yet, but I think I can understand that example that person just gave (on tablet so I can't read their name)
Yep canceling the 2s was by far the hardest part of that proof.
How the graph of d and theta is created? and why?
and shouldn't sin(pi/2)=1?
how is area of rectangle is pi/2 ? explain please.
Because we are only looking at the values from 0 to pi/2. Anything outside this range is already represented withtin the bound we set. We also know d is at most 1.
I once saw a bbc documentary that claimed there's was no logic behind pi being in the equation for this. Lols
Polygonal needle? You should have stopped at 4:37.
Pi Day should be 22nd of July. 22/7 is more accurate that 3.14
No, people might think pi is rational
Ooh my... The second proof is way more beautiful than my imaginary girl friend
Very interesting video! Also, do you have a degree in mathematics?
My favorite part was to stay with it.
Wait why is the rectangle's area pi/2. Isnt 1*1/2... 1/2 😭
Pi appears because you use radians.
What would happen if you used degrees?
If radians is the only way to reach conclusions such as this, why use degrees at all?
Use degrees, same thing
Degrees are just radians multiplied by 180/PI
Could "Pi" has another value in another universe?
Only if the universe existed in non-euclidean space
no
You're a day early!
See the numberphile video th-cam.com/video/sJVivjuMfWA/w-d-xo.html for an actual demonstration of this (using matches instead of needles).
Calculus proof is much better
I don't see Tau once,
Dislike for disappointment,
Don't like haikus? (crap)
tau is chancer
The most underrated comment of 2 years ago
If they are bent they can intersect in more than 2 points. Imagine wiggly shape
I didn't understand anything about Barbier's Proof method.Maybe its because my English or my Maths.
This problem is always presented without a clear statement of "randomness." Which variable is random according to what distribution? You cannot put a uniform distribution on an infinite measure set.
happy pi day
Annoying mid roll ads
Nice content, but seriously Presh could you put a bit more effort editing your videos please ?
what do you have in mind ? Everything seems fine to me
+Zozeux You can notice that he's forgetting what to read next pretty often in this video.
+Zacharie Etienne I agree. Please do that
+Christoph Michelbach oh yeah Indeed. Yes it could be enjoyable to have a constant flow of talking.
+Zozeux he's a stuttering mess several times. He should have edited that stuff out or re-recorded it
ze dumb way:
50% because it will fall either crossing ze line or not crossing😂😂😂
Please get a Turkish translation in all your videos🙏🏻🙏🏻🙏🏻