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Probability and Geometry

  1. Mar 15, 2005 #1
    What is the probability that a dart, hitting a square board at random, lands nearer the center than the edge?

    This was a problem at a past NJUMC test, and I'm not sure how to go about it. It's obviously not a circle, because the equal distance between the center and the edge horizontally and vertically is simply (side/4), while the equal distance between the center and the edge on the diagonals is (side)/(2sqrt(2)).

    So I'm very confused and frustrated :-/

    Thanks for the help.
     
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  3. Mar 16, 2005 #2

    HallsofIvy

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    For a problem like that "probability" is relative area. The point (x,y) (x,y both positive) is "equi-distant" from center and edge if [itex]\sqrt{x^2+ y^2}= min(s-x,s-y)[/itex].
    If, for example s-x is smaller than s-y, you get [itex]x^2+ y^2= x^2- 2sx+ s^2[/itex] or [itex]y^2= -2sx- s^2[/itex], a parabola. It looks to me like the area "nearer the center than the edge" is the region bounded by the parabolas [itex]y= \frac{1}{2}(s^2- x^2)[/itex], [itex]x= \frac{1}{2}(s^2- y^2)[/itex], [itex]y= -\frac{1}{2}(s^2-x^2)[/itex], and [itex]x= -\frac{1}{2}(s^2- y^2)[/itex].
     
  4. Mar 16, 2005 #3
    Thank you ver much for your help, but I'm still having problems. How exactly would you evaluate what you have set up?

    I took a similar approach to yours after reading your suggestions, and ended up getting 3/4. That doesn't sound right to me :)

    Thank you again for your help. I was thinking of exactly what type of region would enclose this, and your saying parabola really helped me out.

    ::EDIT::

    Haha! I may still be wrong, but I love going back after thinking I got a problem wrong, doing it over, and thinking I got it right this time. 29/66.

    I'd still like to hear what you got :)
     
    Last edited: Mar 16, 2005
  5. Mar 17, 2005 #4

    Curious3141

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    This was the most irritating problem I've done in a while because it's fairly tedious and for some reason, I kept slipping up in my calculations. I hope my current answer is correct. :

    [tex]p = 8 - \frac{16}{3}\sqrt{2}[/tex], which is around 0.458

    Definitely not a "neat" rational answer.
     
  6. Mar 17, 2005 #5

    Curious3141

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    Quick, somebody throw darts at a square board a thousand times and confirm my answer experimentally. :biggrin:
     
  7. Mar 17, 2005 #6
    dont you think that is a little big?
     
  8. Mar 17, 2005 #7
    Yeah, that's what I thought too. If you draw it out it looks like it should only be around 1/3 or so.
     
  9. Mar 17, 2005 #8
    I get [tex]\frac{4\sqrt{2}-5}{3}\thickapprox .21895[/tex]
     
  10. Mar 17, 2005 #9

    Curious3141

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    That looks much more correct. I must have made a mistake somewhere, because the answer should be bounded by :

    [tex]3 - 2\sqrt{2} < p < \frac{1}{4}[/tex] and your answer certainly fits. I'll go back to the drawing board and see where I slipped up. It's almost certainly some stupid error.
     
  11. Mar 17, 2005 #10
    i used matlab to run 10,000 simulated dart throwings.
    i got p=.2193. which is close to the theoretical value i got.
     

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    Last edited: Mar 17, 2005
  12. Mar 17, 2005 #11

    Curious3141

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    Zone Ranger, your answer is exactly correct. I made a mistake in taking the wrong element of the area during the integration. Other than that, I think we have the same method. :smile:
     
  13. Mar 18, 2005 #12
    Could you please show how you did it? Thank you!
     
  14. Mar 18, 2005 #13
    look at the first quadrant. assume x<y then set [tex]\sqrt{x^2+y^2}=s-y[/tex]. solving for y we get
    [tex]y=\frac{s^2-x^2}{2s}[/tex]

    find out what x ranges over (find out when x=y). so take [tex]y=\frac{s^2-x^2}{2s}[/tex] and replace y by x to get
    [tex]x=\frac{s^2-x^2}{2s}[/tex]. solve for x you get
    [tex](\sqrt{2}-1)s[/tex] so then [tex]{\int_{x=0}^{(\sqrt{2}-1)s}}\frac{s^2-x^2}{2s}dx=(\frac{2\sqrt{2}}{3}-\frac{5}{6})s^2[/tex] now you have integrated too much so we subtract the triangle that is below the line y=x. ei subtract [tex]\frac{1}{2}[(\sqrt{2}-1)s]^2[/tex]

    but this is only 1/8th of the picture so we multiply the above by 8 and divide by the total area of the square which is [tex]4s^2[/tex]. then you get the result i got above

    ie

    [tex]\frac{[(\frac{2\sqrt{2}}{3}-\frac{5}{6})s^2-\frac{1}{2}[(\sqrt{2}-1)s]^2]8}{4s^2}=\frac{4\sqrt{2}-5}{3}[/tex]
     
  15. Mar 18, 2005 #14

    Curious3141

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    What I did was to find the locus of points that were equidistant from the center and the sides. I used the diagonals to divide the square into 4 quadrants. The equation of the parabola in each quadrant is different, but we only have to consider one of the quadrants because of the symmetry.

    Lower left hand corner = origin, side of square = 2A. Vertices are at (0,0);(2A,0);(0,2A) and (2A,2A). The bottom most quadrant is the one I considered. The equidistant parabola in the bottom quadrant is defined by :

    [tex](x - A)^2 + (y - A)^2 = y^2[/tex]

    Basically, solve for the minimum of that parabola (it's at [tex](A, \frac{A}{2}) [/tex]) and the point at which the parabola intersects the diagonals of the square (those occur at [tex](2 - \sqrt{2}, 2 - \sqrt{2}) and (2 + \sqrt{2}, 2 - \sqrt{2})[/tex]. Then find the area of the parabola bounded by the lines [tex]y = \frac{A}{2}[/tex] and [tex]y = 2 - \sqrt{2}[/tex]. Add that to the area of the triangle in the bottom quadrant that is enclosed within the two diagonals and lies above that area. This area is given by [tex](\sqrt{2} - 1)^2 = 3 - 2\sqrt{2}[/tex]. Add the two areas together, and this is the area for one quadrant that is closer to the center than the sides.

    You need to multiply by 4 and divide by the area of the square, which is [tex]4A^2[/tex]. That gives the answer.
     
  16. Mar 18, 2005 #15

    cronxeh

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    shouldnt it be a double integral..
     
  17. Mar 18, 2005 #16

    Curious3141

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    No. Zone Ranger and I had similar methods, but I'll quote mine because I'm familiar with it.

    The element of area I integrated (bounded by the y-bounds I stated) was :

    [tex]2(x - A)dy[/tex]

    and that gives :

    [tex]2\int_{\frac{A}{2}}^{A(2 - \sqrt{2})}(x - A)dy[/tex]

    where [tex]x = A + \sqrt{(A)(2y - A)}[/tex]

    Then that area needs to be added to the area of the triangle above it in the bottom quadrant, then multiplied by 4 and divided by [tex]4A^2[/tex].
     
    Last edited: Mar 18, 2005
  18. Mar 19, 2005 #17

    cronxeh

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    Well I thought if you have average dart-hit density on that surface, then you would go about calculating the average distance of each hit from origin circle to that point using a double integral..
     
  19. Mar 19, 2005 #18

    Curious3141

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    But that's not what we're doing. We're finding the locus of points that are equidistant from the center and from one of the sides. There is a different parabola in each quadrant (as divided up by the diagonals), and these parabolas together constitute the locus. The required probability is the ratio of the area enclosed by the locus and the total area of the dartboard.

    To find the area enclosed by the locus in one quadrant you only need a single integral.
     
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