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Stick breaking game

  1. Jan 14, 2016 #1
    You are to break a stick into 3 pieces, randomly, which also means the length of each portion is completely arbitrary. What is the probability of those 3 portions successfully forming a triangle of any kind?
     
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  3. Jan 14, 2016 #2

    Samy_A

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    Using the triangle inequality as only rule, I get 1/8. (Not yet sure of that result though.)
     
  4. Jan 14, 2016 #3
    It should be 1/4
     
  5. Jan 14, 2016 #4

    mathman

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    There is a restriction on arbitrary, since the three pieces must add up to the original length. To restate the problem, what is the probability that the longest piece is at least half the total.
     
  6. Jan 15, 2016 #5

    Samy_A

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    Indeed.
     
  7. Jan 15, 2016 #6

    Erland

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    In exactly what sense is the breaking of the stick random? Exactly how is the breaking administered?
     
  8. Jan 15, 2016 #7
    But for an equilateral triangle (or a triangle even nearly equilateral), the longest piece (~33% long) is quite smaller than half the total (50% long).

    edit: Perhaps you meant "the probability that the longest piece is less than half the total"? Because I would agree with that.
     
    Last edited: Jan 15, 2016
  9. Jan 15, 2016 #8
    The way I've interpreted the question is:

    -Given a stick of length 1, make 2 breaks. Each break is at a location that is a uniformly distributed random number between 0 and 1.
    -Piece A is of length: 0 to smaller break location
    -Piece B is of length: smaller break location to larger break location
    -Piece C is of length: larger break location to 1
    -What is the probability that the 3 pieces can be arranged to make any triangle?
     
  10. Jan 15, 2016 #9

    mathman

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    What I meant was - if the largest piece is at least half the total a triangle cannot be formed.
     
  11. Jan 15, 2016 #10

    mfb

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    That is a critical point, but most reasonable choices I see lead to the same answer.

    - represent the stick as [0,1] and make two cuts at random points in (0,1) with independent uniform distributions: 1/4 probability that a triangle can be formed.
    - imagine the stick "bent" as unit circle, make three cuts at random points with independent uniform distributions, take the fractions formed as lengths: still 1/4.
    - imagine the range of possible lengths as the area given by A+B+C=1, make a uniform distribution over this area using one of the usual metrics. Still 1/4.
    - start by finding the length of piece A with a uniform distribution over (0,1), then start finding the length of B with a uniform distribution over (0,1-length(A)). Now we just have 1/6 probability that a triangle can be formed.
     
  12. Jan 19, 2016 #11
    Sorry for the ambiguity in my question.
     
  13. Jan 21, 2016 #12
    I thought I would share the logic I used to solve this problem with some pictures.

    edit: By the way, this is by no means intended to be a proof, or particularly rigorous, just an illustration of the thought process I used.

    First I will show why the longest piece of the 3 segments needs to be less than half the total length in order for a triangle to be possible. Imagine the circles attached at each end of the segments (in figure 1 below) can act as points of rotation. I think it quickly becomes clear from looking at the diagram that one cannot arrange the 3 segments below into a triangle due to one of the segments being greater than half the total length.

    Figure 1.
    1.png

    Secondly, I wanted to show the logic for why the chance of being able to make a triangle from breaking a stick in 2 (uniformly randomly picked) spots would be 1/4.

    Consider your first break. I contend it will be located in the first half of the stick. The reason is that if it lands in the second half, you just flip the stick. Now consider the stick divided into 4 quadrants. Half the time the first break will land in the first quadrant, and half the time in the second quadrant. First let us consider the case in which the first break lands in the first quadrant (figure 2):

    Figure 2.
    2.png
    Now let us consider the case in which the first break lands in the second quadrant (figure 3):

    Figure 3.
    3.png
     
    Last edited: Jan 21, 2016
  14. Jan 21, 2016 #13

    mfb

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    There is no need to split the analysis in two cases, you can make the same argument just using halves instead of quarters.
     
  15. Jan 21, 2016 #14
    Oops! Indeed.
     
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