Surprising π from probability! Buffon's needle problem

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  • čas přidán 3. 06. 2024
  • A needle of length 1 is randomly dropped on a floor with horizontal lines 2 units apart. What is the probability the needle intersects one of the horizontal lines? The answer might surprise you--it involves the constant π!
    This is known as Buffon's Needle Problem. In this video I present the common proof that uses trigonometry and calculus. I also show a clever proof from Barbier's (I read about this in "Proofs from THE BOOK" which is a collection of elegant proofs).
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Komentáře • 136

  • @user-kh5tv9rb6y
    @user-kh5tv9rb6y Před 8 lety +101

    I definitely like the calculus one more. It's a pain to learn, but once you do, it is just so freaking useful.

  • @rc210397
    @rc210397 Před 4 lety +18

    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.

  • @davidbowl9650
    @davidbowl9650 Před 5 měsíci +2

    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

  • @replicaacliper
    @replicaacliper Před 7 lety +20

    I understood your explanation so much better than numberphiles explanation. thank you

  • @AngryArmadillo
    @AngryArmadillo Před 8 lety +3

    I can't decide which solution I like more. Great video!

  • @guilemaigre14
    @guilemaigre14 Před 8 lety +9

    really liked the second one, I would have nerver thought of that !

  • @ProfDragonVale
    @ProfDragonVale Před 8 lety +33

    I prefer the calculus proof. It's simpler

  • @armanozcan7983
    @armanozcan7983 Před 4 lety +9

    Second method is way more beautiful, however I can't develop the intuitive for linearity of expectation when events are dependent.

  • @bozidard8859
    @bozidard8859 Před 8 lety

    Barbier's proof is just brilliant. I liked it a lot. Thanks for showing it

  • @hosseinheli7155
    @hosseinheli7155 Před 4 lety

    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

  • @donnyofdoom1873
    @donnyofdoom1873 Před 8 lety +2

    This was one of my assignments in a first year IT subject. With a large enough sample it's spot on every time :D

  • @wolfnederpel
    @wolfnederpel Před 8 lety +1

    one of the most interesting videos I've seen in a while!

  • @erikziak1249
    @erikziak1249 Před 8 lety +3

    Ah the memories this video wakes in me. Man, I am old. Btw. Great video!

  • @Abca209
    @Abca209 Před 7 lety +43

    This is so ridiculously easy, I learned this in 6th grade along with being able to solve Chaos theory while also curing cancer.

    • @gamedungeon3088
      @gamedungeon3088 Před 5 lety +7

      Abca209 step up your game. I solved world hunger and poverty while on the toilet in kindergarten.

    • @tamashamas6193
      @tamashamas6193 Před 5 lety +1

      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

  • @trungnguyenquang9737
    @trungnguyenquang9737 Před 8 lety

    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

  • @py8554
    @py8554 Před rokem +3

    There’s also a method using double integration. Would be great if that were included as well.

  • @100nacional100
    @100nacional100 Před 5 lety

    One of your best videos

  • @iaexo
    @iaexo Před 3 lety

    Wow you an ilectureonline both came up with the same video idea, on the same day!!! What an incredible coincidence!

  • @Vivenk88
    @Vivenk88 Před 5 lety

    The trigonometric method is elegant and can be explained easily.

  • @Shawkster6
    @Shawkster6 Před 8 lety

    Really cool video!

  • @patbateman69420
    @patbateman69420 Před 2 lety

    thanks for this. I'm reading Evan's introduction to stochastic differential equations and his explanation of this same problem is extremely lacking.

  • @teodoranasz3531
    @teodoranasz3531 Před 4 lety +3

    Really fond of the second proof. I feel like I don't appreciate continuity enough.

  • @starpawsy
    @starpawsy Před 3 lety

    I learnt this very example in Computer Science II at University in 1975 (!!!!) It was an example of what was called "Monte Carlo analysis".

  • @jordia.2970
    @jordia.2970 Před 3 lety

    So clever man

  • @outti
    @outti Před 8 lety

    Beautiful!!!

  • @UnstoppableImmovable
    @UnstoppableImmovable Před 5 lety +4

    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.

    •  Před 11 hodinami

      Interesting

  • @anasali5202
    @anasali5202 Před 3 lety +1

    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

  • @MrTomBoz
    @MrTomBoz Před 8 lety +1

    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

  • @mr0661
    @mr0661 Před 7 lety +1

    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

  • @Supermario0727
    @Supermario0727 Před 8 lety

    The first solution made much more sense.

  • @codebeard
    @codebeard Před 8 lety +3

    Happy Pi Day! (At least, it was here when this video was uploaded.)

  • @user-wi8fd8fh7m
    @user-wi8fd8fh7m Před rokem

    I found an answer with the first method. I saw the second method and I only say wow.

  • @GoodiTwoShuz
    @GoodiTwoShuz Před 7 lety +10

    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?

    • @christianfunintuscany1147
      @christianfunintuscany1147 Před 4 lety +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

    • @hershyfishman2929
      @hershyfishman2929 Před 2 lety

      @@christianfunintuscany1147 yes, but all that means is that d includes both the white and the grey. The gray is where d

  • @Freakschwimmer
    @Freakschwimmer Před 8 lety

    2:01
    shouldn't theta be the non-obtuse angle instead of the acute angle?
    Or is any right angle acute as well? :/

  • @jurgisjuronis646
    @jurgisjuronis646 Před 9 měsíci

    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!

  • @danilocorreia5104
    @danilocorreia5104 Před 5 lety +2

    I never felt so confused in math as I am right now

  • @Diammmond
    @Diammmond Před 8 lety

    Found it much easier to follow proof one

  • @christianfunintuscany1147

    Cool !

  • @98forestfire
    @98forestfire Před 8 lety

    Awesome

  • @astralchan
    @astralchan Před 7 lety +1

    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...

  • @BerraLilltroll
    @BerraLilltroll Před 7 lety +2

    So would the probability of crossings be the same if u throw a flexible rope of length one? Does that follow from this?

    • @nivpearlman6514
      @nivpearlman6514 Před 6 lety

      No, but the expected number of intersections will be the same.
      A rope may have multiple crossings while a needle won't.

  • @esaedromicroflora1247
    @esaedromicroflora1247 Před 2 lety

    I guessed 1/3 at the beginning and can't believe I was so close :D

  • @rohitmahanandi9299
    @rohitmahanandi9299 Před 3 měsíci

    good

  • @Sam_on_YouTube
    @Sam_on_YouTube Před 8 lety +2

    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.

  • @nicholasw996
    @nicholasw996 Před 8 lety

    Not surprising that pi appears here.
    Well, after learning Euler's Identity, one can expect e and pi to appear anywhere and everywhere.

  • @rospotrebpozor3873
    @rospotrebpozor3873 Před 8 lety +1

    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.

    • @codemurp3244
      @codemurp3244 Před 8 lety

      +rospotreb pozor I don't see why this is relevant though.

    • @rospotrebpozor3873
      @rospotrebpozor3873 Před 8 lety

      not obligatory. though, often imaginary tasks are oversimplified. it leads to rough mistakes in mental composition.

    • @codemurp3244
      @codemurp3244 Před 8 lety

      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.

    • @rospotrebpozor3873
      @rospotrebpozor3873 Před 8 lety

      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...

    • @codemurp3244
      @codemurp3244 Před 8 lety

      +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

  • @rohitpharande5607
    @rohitpharande5607 Před 6 lety +1

    That's definitely a clever proof(Barbier's proof)

  • @SlipperyTeeth
    @SlipperyTeeth Před 8 lety

    You defined it so that touching is a form of intersecting by saying d

    • @JackRule16
      @JackRule16 Před 6 lety

      Phoenix Fire see what answer you get it you consider your case

  • @martinepstein9826
    @martinepstein9826 Před 7 lety +1

    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.

    • @JackRule16
      @JackRule16 Před 6 lety +1

      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?

  • @nomekop777
    @nomekop777 Před rokem

    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.

  • @stewieyan
    @stewieyan Před 4 lety +1

    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).

    • @brachypelmasmith
      @brachypelmasmith Před 2 měsíci

      (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.

  • @prateeksarangi4560
    @prateeksarangi4560 Před 3 lety

    What if the line spacing and the length of the needle are same?? How can we relate d & l then??

  • @GiovaniMoreiraG
    @GiovaniMoreiraG Před 3 lety

    LOL it's been 10 years since I last studied this and I got it right

  • @FTWTitanium
    @FTWTitanium Před 8 lety +3

    Very interesting video! Also, do you have a degree in mathematics?

  • @omfgmouse
    @omfgmouse Před 8 lety +2

    I like the second proof, but how/why can you assume that E(x+y) = E(x) + E(y) ? (at 6:52)

    • @hedgechasing
      @hedgechasing Před 8 lety +2

      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.

    • @JackRule16
      @JackRule16 Před 6 lety

      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

  • @atulmeena9472
    @atulmeena9472 Před 7 lety

    At 9:01 when you write the equation E(Pn)

  • @NetAndyCz
    @NetAndyCz Před 7 lety

    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...

    • @ArticulateDuck
      @ArticulateDuck Před 7 lety

      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.

  • @Germaniumtingz
    @Germaniumtingz Před 8 lety +10

    More like De Gea needle problem...

    • @Dragon30ficationXD
      @Dragon30ficationXD Před 7 lety +1

      Thamer AL Sadoun casillas needle problem

    • @tosca1883
      @tosca1883 Před 7 lety +1

      or even neuer needle problem

    • @GRBtutorials
      @GRBtutorials Před 4 lety

      @@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.

  • @TripleSuccotash1
    @TripleSuccotash1 Před 8 lety +1

    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.

    • @Minecraftster148790
      @Minecraftster148790 Před 8 lety +1

      What about pi R squared? Tau over 2 would be a pain and pi would be much better

    • @TripleSuccotash1
      @TripleSuccotash1 Před 8 lety

      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

    • @echaen1707
      @echaen1707 Před 8 lety

      +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

    • @Minecraftster148790
      @Minecraftster148790 Před 8 lety

      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)

    • @potato-hj9nm
      @potato-hj9nm Před 8 lety

      Yep canceling the 2s was by far the hardest part of that proof.

  • @SuntzuDragon
    @SuntzuDragon Před 7 lety

    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.

    • @giladzxc17
      @giladzxc17 Před 7 lety +4

      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

    • @SuntzuDragon
      @SuntzuDragon Před 7 lety

      Thank you for the help, I realize my mistake now.

  • @nafnist
    @nafnist Před 3 lety

    I once saw a bbc documentary that claimed there's was no logic behind pi being in the equation for this. Lols

  • @stabgan
    @stabgan Před 4 lety

    If they are bent they can intersect in more than 2 points. Imagine wiggly shape

  • @Titurel
    @Titurel Před 8 lety +8

    3:06 is scale off? 1/2 sin Theta shouldn't never be more than 1/2..that looks higher...

    • @leackedimp7169
      @leackedimp7169 Před 8 lety

      I'd say it looks about right (if the angle is 90( that would equal 1/2) if not its way to big)

  • @sjoerdwiesmeijer7231
    @sjoerdwiesmeijer7231 Před 4 lety

    Shouldn't the integral be equal to zero? Sintheta dtheta

  • @199NickYT
    @199NickYT Před 8 lety +2

    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!

  • @charlottegreen7930
    @charlottegreen7930 Před rokem

    How the graph of d and theta is created? and why?

  • @sikemperveraksy7741
    @sikemperveraksy7741 Před 3 lety

    Please get a Turkish translation in all your videos🙏🏻🙏🏻🙏🏻

  • @dopamine6650
    @dopamine6650 Před 3 lety

    I misread the title as piano probability

  • @ProfDragonVale
    @ProfDragonVale Před 8 lety

    You're a day early!

  • @Qqwerty42
    @Qqwerty42 Před 4 lety +1

    See the numberphile video czcams.com/video/sJVivjuMfWA/video.html for an actual demonstration of this (using matches instead of needles).

  • @marcovargasglobant7923

    Could "Pi" has another value in another universe?

    • @ShaneClough
      @ShaneClough Před 5 lety

      Only if the universe existed in non-euclidean space

  • @AmberArchive
    @AmberArchive Před 3 lety +1

    no

  • @mr.peoples901
    @mr.peoples901 Před rokem

    My favorite part was to stay with it.

  • @robl4836
    @robl4836 Před 8 lety +1

    Pi Day should be 22nd of July. 22/7 is more accurate that 3.14

  • @asterrovels6821
    @asterrovels6821 Před 7 lety

    how is area of rectangle is pi/2 ? explain please.

    • @hyrekandragon2665
      @hyrekandragon2665 Před 7 lety

      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.

  • @sk4lman
    @sk4lman Před 6 lety

    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?

  • @kaanuysal9205
    @kaanuysal9205 Před 8 lety

    I didn't understand anything about Barbier's Proof method.Maybe its because my English or my Maths.

  • @sunritroykarmakar4406
    @sunritroykarmakar4406 Před 2 lety

    Calculus proof is much better

  • @tixanthrope
    @tixanthrope Před rokem +1

    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.

  • @markgraham2312
    @markgraham2312 Před 3 lety

    Polygonal needle? You should have stopped at 4:37.

  • @P3achTree
    @P3achTree Před 8 lety

    happy pi day

  • @pinkandblack9429
    @pinkandblack9429 Před 8 lety +3

    I don't see Tau once,
    Dislike for disappointment,
    Don't like haikus? (crap)

  • @zacharieetienne5510
    @zacharieetienne5510 Před 8 lety +14

    Nice content, but seriously Presh could you put a bit more effort editing your videos please ?

    • @frenz2000
      @frenz2000 Před 8 lety +2

      what do you have in mind ? Everything seems fine to me

    • @HansPeter-qg2vc
      @HansPeter-qg2vc Před 8 lety +10

      +Zozeux You can notice that he's forgetting what to read next pretty often in this video.

    • @komolunanole8697
      @komolunanole8697 Před 8 lety +1

      +Zacharie Etienne I agree. Please do that

    • @frenz2000
      @frenz2000 Před 8 lety +3

      +Christoph Michelbach oh yeah Indeed. Yes it could be enjoyable to have a constant flow of talking.

    • @ophello
      @ophello Před 7 lety

      +Zozeux he's a stuttering mess several times. He should have edited that stuff out or re-recorded it

  • @JamesWylde
    @JamesWylde Před 5 lety

    Annoying mid roll ads

  • @plerikmetselaar1152
    @plerikmetselaar1152 Před 8 lety

    ze dumb way:
    50% because it will fall either crossing ze line or not crossing😂😂😂