Probability that a stick randomly broken in three places can form a triangle

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I like questions about geometric probability, and two of my favourite questions here on math.SE are
Probability that a stick randomly broken in two places can form a triangle
and
Probability that a stick randomly broken in five places can form a tetrahedron.



I wondered about the probability that a stick randomly broken in three places can form a triangle. More generally, we can ask for the probability distribution of the number of triangles that can be formed from the four pieces. Since each triple of the pieces has probability $frac14$ of forming a triangle, the expected number of triangles we can form is $frac14cdot4=1$, which is already a rather nice result.



I tried various ways of applying inclusion–exclusion, with or without first ordering the segments by size, but it all seemed too complicated and unenlightening, and I ended up writing a program to output all the inequalities in order to let qhull compute the volumes of the polytopes they define.



The result (confirmed by simulations) is:



beginarrayc
&textprobability&textprobability\
#triangle&text(reduced)&text(unreduced)\hline
0&frac37&frac45105\
1&frac1135&frac33105\
2&frac16105&frac16105\
3&frac4105&frac4105\
4&frac115&frac7105
endarray



You can check that the expected number of triangles comes out as $1$.



While the overall distribution is somewhat complicated, the probabilities that we can form
all triangles ($frac115$), any triangle ($frac47$) and no triangles ($frac37$) come out as nice low fractions, so I thought that maybe there's hope to find an elegant way to compute (one of) them after all. Do you see one? (The three places where the stick is broken are independently uniformly chosen along its length.)




probability alternative-proof inclusion-exclusion geometric-probability




share|cite|improve this question























    up vote
    2
    down vote

    favorite
    1












    I like questions about geometric probability, and two of my favourite questions here on math.SE are
    Probability that a stick randomly broken in two places can form a triangle
    and
    Probability that a stick randomly broken in five places can form a tetrahedron.



    I wondered about the probability that a stick randomly broken in three places can form a triangle. More generally, we can ask for the probability distribution of the number of triangles that can be formed from the four pieces. Since each triple of the pieces has probability $frac14$ of forming a triangle, the expected number of triangles we can form is $frac14cdot4=1$, which is already a rather nice result.



    I tried various ways of applying inclusion–exclusion, with or without first ordering the segments by size, but it all seemed too complicated and unenlightening, and I ended up writing a program to output all the inequalities in order to let qhull compute the volumes of the polytopes they define.



    The result (confirmed by simulations) is:



    beginarrayc
    &textprobability&textprobability\
    #triangle&text(reduced)&text(unreduced)\hline
    0&frac37&frac45105\
    1&frac1135&frac33105\
    2&frac16105&frac16105\
    3&frac4105&frac4105\
    4&frac115&frac7105
    endarray



    You can check that the expected number of triangles comes out as $1$.



    While the overall distribution is somewhat complicated, the probabilities that we can form
    all triangles ($frac115$), any triangle ($frac47$) and no triangles ($frac37$) come out as nice low fractions, so I thought that maybe there's hope to find an elegant way to compute (one of) them after all. Do you see one? (The three places where the stick is broken are independently uniformly chosen along its length.)




    probability alternative-proof inclusion-exclusion geometric-probability




    share|cite|improve this question





















      up vote
      2
      down vote

      favorite
      1









      up vote
      2
      down vote

      favorite
      1






      1





      I like questions about geometric probability, and two of my favourite questions here on math.SE are
      Probability that a stick randomly broken in two places can form a triangle
      and
      Probability that a stick randomly broken in five places can form a tetrahedron.



      I wondered about the probability that a stick randomly broken in three places can form a triangle. More generally, we can ask for the probability distribution of the number of triangles that can be formed from the four pieces. Since each triple of the pieces has probability $frac14$ of forming a triangle, the expected number of triangles we can form is $frac14cdot4=1$, which is already a rather nice result.



      I tried various ways of applying inclusion–exclusion, with or without first ordering the segments by size, but it all seemed too complicated and unenlightening, and I ended up writing a program to output all the inequalities in order to let qhull compute the volumes of the polytopes they define.



      The result (confirmed by simulations) is:



      beginarrayc
      &textprobability&textprobability\
      #triangle&text(reduced)&text(unreduced)\hline
      0&frac37&frac45105\
      1&frac1135&frac33105\
      2&frac16105&frac16105\
      3&frac4105&frac4105\
      4&frac115&frac7105
      endarray



      You can check that the expected number of triangles comes out as $1$.



      While the overall distribution is somewhat complicated, the probabilities that we can form
      all triangles ($frac115$), any triangle ($frac47$) and no triangles ($frac37$) come out as nice low fractions, so I thought that maybe there's hope to find an elegant way to compute (one of) them after all. Do you see one? (The three places where the stick is broken are independently uniformly chosen along its length.)




      probability alternative-proof inclusion-exclusion geometric-probability




      share|cite|improve this question











      I like questions about geometric probability, and two of my favourite questions here on math.SE are
      Probability that a stick randomly broken in two places can form a triangle
      and
      Probability that a stick randomly broken in five places can form a tetrahedron.



      I wondered about the probability that a stick randomly broken in three places can form a triangle. More generally, we can ask for the probability distribution of the number of triangles that can be formed from the four pieces. Since each triple of the pieces has probability $frac14$ of forming a triangle, the expected number of triangles we can form is $frac14cdot4=1$, which is already a rather nice result.



      I tried various ways of applying inclusion–exclusion, with or without first ordering the segments by size, but it all seemed too complicated and unenlightening, and I ended up writing a program to output all the inequalities in order to let qhull compute the volumes of the polytopes they define.



      The result (confirmed by simulations) is:



      beginarrayc
      &textprobability&textprobability\
      #triangle&text(reduced)&text(unreduced)\hline
      0&frac37&frac45105\
      1&frac1135&frac33105\
      2&frac16105&frac16105\
      3&frac4105&frac4105\
      4&frac115&frac7105
      endarray



      You can check that the expected number of triangles comes out as $1$.



      While the overall distribution is somewhat complicated, the probabilities that we can form
      all triangles ($frac115$), any triangle ($frac47$) and no triangles ($frac37$) come out as nice low fractions, so I thought that maybe there's hope to find an elegant way to compute (one of) them after all. Do you see one? (The three places where the stick is broken are independently uniformly chosen along its length.)





      probability alternative-proof inclusion-exclusion geometric-probability





      share|cite|improve this question










      share|cite|improve this question




      share|cite|improve this question









      asked Jul 25 at 16:33









      joriki

      164k10180328




      164k10180328




















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          up vote
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          The lengths of the broken stick pieces $(Y_1, Y_2, Y_3, Y_4)$ are spacings of ordered uniform. The joint distribution is equivalent to
          $$ left(frac X_1 sum_i=1^4 X_i,
          frac X_2 sum_i=1^4 X_i,
          frac X_3 sum_i=1^4 X_i,
          frac X_4 sum_i=1^4 X_iright)$$
          where $X_i$ are iid exponential random variables. If we ordered the above spacings, the distribution of $(Y_(1), Y_(2), Y_(3), Y_(4))$ is equivalent to



          $$ left(frac X_1/4 sum_i=1^4 X_i,
          frac X_1/4 + X_2/3 sum_i=1^4 X_i,
          frac X_1/4 + X_2/3 + X_3/2 sum_i=1^4 X_i,
          frac X_1/4 + X_2/3 + X_3/2 + X_4 sum_i=1^4 X_iright)$$



          Any $3$ of the $4$ pieces cannot form a triangle if and only if
          $$ begincases
          Y_(1) + Y_(2) < Y_(3) \
          Y_(1) + Y_(2) < Y_(4) \
          Y_(1) + Y_(3) < Y_(4) \
          Y_(2) + Y_(3) < Y_(4)
          endcases$$
          Note that the second inequality is implied by the first, and the third inequality is implied by the fourth. So the probability of no triangle being formed is
          $$ beginalign
          &Pr Y_(1) + Y_(2) < Y_(3), Y_(2) + Y_(3) < Y_(4) \
          =& PrBiggfrac X_1 4 + frac X_1 4 + frac X_2 3 <
          frac X_1 4 + frac X_2 3 + frac X_3 2, \
          & frac X_1 4 + frac X_2 3 + frac X_1 4 + frac X_2 3
          + frac X_3 2 <
          frac X_1 4 + frac X_2 3 + frac X_3 2 + X_4Bigg \
          =& PrX_1 < 2X_3, 3X_1 + 4X_2 < 12X_4 \
          =& int_0^infty Pr2X_3 > xPr12X_4 - 4X_2 > 3xe^-xdx
          endalign$$



          Note that
          $$ beginalign
          &Pr12X_4 - 4X_2 > 3x \
          =& int_0^inftyPr12X_4 - 4u > 3xe^-udu \
          =& int_0^infty e^-(3x+4u)/12 e^-udu \
          =& e^-x/4int_0^infty e^-4u/3du \
          =& frac 3 4e^-x/4
          endalign
          $$



          So the integral become
          $$ int_0^infty e^-x/2frac 3 4 e^-x/4e^-xdx
          = frac 3 4 int_0^infty e^-7x/4 dx = frac 3 4 times frac 4 7 = frac 3 7 $$



          This is not a very elegant way but at least it is doable. Looking forward to someone to post a better solution.






          share|cite|improve this answer





















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            up vote
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            The lengths of the broken stick pieces $(Y_1, Y_2, Y_3, Y_4)$ are spacings of ordered uniform. The joint distribution is equivalent to
            $$ left(frac X_1 sum_i=1^4 X_i,
            frac X_2 sum_i=1^4 X_i,
            frac X_3 sum_i=1^4 X_i,
            frac X_4 sum_i=1^4 X_iright)$$
            where $X_i$ are iid exponential random variables. If we ordered the above spacings, the distribution of $(Y_(1), Y_(2), Y_(3), Y_(4))$ is equivalent to



            $$ left(frac X_1/4 sum_i=1^4 X_i,
            frac X_1/4 + X_2/3 sum_i=1^4 X_i,
            frac X_1/4 + X_2/3 + X_3/2 sum_i=1^4 X_i,
            frac X_1/4 + X_2/3 + X_3/2 + X_4 sum_i=1^4 X_iright)$$



            Any $3$ of the $4$ pieces cannot form a triangle if and only if
            $$ begincases
            Y_(1) + Y_(2) < Y_(3) \
            Y_(1) + Y_(2) < Y_(4) \
            Y_(1) + Y_(3) < Y_(4) \
            Y_(2) + Y_(3) < Y_(4)
            endcases$$
            Note that the second inequality is implied by the first, and the third inequality is implied by the fourth. So the probability of no triangle being formed is
            $$ beginalign
            &Pr Y_(1) + Y_(2) < Y_(3), Y_(2) + Y_(3) < Y_(4) \
            =& PrBiggfrac X_1 4 + frac X_1 4 + frac X_2 3 <
            frac X_1 4 + frac X_2 3 + frac X_3 2, \
            & frac X_1 4 + frac X_2 3 + frac X_1 4 + frac X_2 3
            + frac X_3 2 <
            frac X_1 4 + frac X_2 3 + frac X_3 2 + X_4Bigg \
            =& PrX_1 < 2X_3, 3X_1 + 4X_2 < 12X_4 \
            =& int_0^infty Pr2X_3 > xPr12X_4 - 4X_2 > 3xe^-xdx
            endalign$$



            Note that
            $$ beginalign
            &Pr12X_4 - 4X_2 > 3x \
            =& int_0^inftyPr12X_4 - 4u > 3xe^-udu \
            =& int_0^infty e^-(3x+4u)/12 e^-udu \
            =& e^-x/4int_0^infty e^-4u/3du \
            =& frac 3 4e^-x/4
            endalign
            $$



            So the integral become
            $$ int_0^infty e^-x/2frac 3 4 e^-x/4e^-xdx
            = frac 3 4 int_0^infty e^-7x/4 dx = frac 3 4 times frac 4 7 = frac 3 7 $$



            This is not a very elegant way but at least it is doable. Looking forward to someone to post a better solution.






            share|cite|improve this answer

























              up vote
              7
              down vote













              The lengths of the broken stick pieces $(Y_1, Y_2, Y_3, Y_4)$ are spacings of ordered uniform. The joint distribution is equivalent to
              $$ left(frac X_1 sum_i=1^4 X_i,
              frac X_2 sum_i=1^4 X_i,
              frac X_3 sum_i=1^4 X_i,
              frac X_4 sum_i=1^4 X_iright)$$
              where $X_i$ are iid exponential random variables. If we ordered the above spacings, the distribution of $(Y_(1), Y_(2), Y_(3), Y_(4))$ is equivalent to



              $$ left(frac X_1/4 sum_i=1^4 X_i,
              frac X_1/4 + X_2/3 sum_i=1^4 X_i,
              frac X_1/4 + X_2/3 + X_3/2 sum_i=1^4 X_i,
              frac X_1/4 + X_2/3 + X_3/2 + X_4 sum_i=1^4 X_iright)$$



              Any $3$ of the $4$ pieces cannot form a triangle if and only if
              $$ begincases
              Y_(1) + Y_(2) < Y_(3) \
              Y_(1) + Y_(2) < Y_(4) \
              Y_(1) + Y_(3) < Y_(4) \
              Y_(2) + Y_(3) < Y_(4)
              endcases$$
              Note that the second inequality is implied by the first, and the third inequality is implied by the fourth. So the probability of no triangle being formed is
              $$ beginalign
              &Pr Y_(1) + Y_(2) < Y_(3), Y_(2) + Y_(3) < Y_(4) \
              =& PrBiggfrac X_1 4 + frac X_1 4 + frac X_2 3 <
              frac X_1 4 + frac X_2 3 + frac X_3 2, \
              & frac X_1 4 + frac X_2 3 + frac X_1 4 + frac X_2 3
              + frac X_3 2 <
              frac X_1 4 + frac X_2 3 + frac X_3 2 + X_4Bigg \
              =& PrX_1 < 2X_3, 3X_1 + 4X_2 < 12X_4 \
              =& int_0^infty Pr2X_3 > xPr12X_4 - 4X_2 > 3xe^-xdx
              endalign$$



              Note that
              $$ beginalign
              &Pr12X_4 - 4X_2 > 3x \
              =& int_0^inftyPr12X_4 - 4u > 3xe^-udu \
              =& int_0^infty e^-(3x+4u)/12 e^-udu \
              =& e^-x/4int_0^infty e^-4u/3du \
              =& frac 3 4e^-x/4
              endalign
              $$



              So the integral become
              $$ int_0^infty e^-x/2frac 3 4 e^-x/4e^-xdx
              = frac 3 4 int_0^infty e^-7x/4 dx = frac 3 4 times frac 4 7 = frac 3 7 $$



              This is not a very elegant way but at least it is doable. Looking forward to someone to post a better solution.






              share|cite|improve this answer























                up vote
                7
                down vote










                up vote
                7
                down vote









                The lengths of the broken stick pieces $(Y_1, Y_2, Y_3, Y_4)$ are spacings of ordered uniform. The joint distribution is equivalent to
                $$ left(frac X_1 sum_i=1^4 X_i,
                frac X_2 sum_i=1^4 X_i,
                frac X_3 sum_i=1^4 X_i,
                frac X_4 sum_i=1^4 X_iright)$$
                where $X_i$ are iid exponential random variables. If we ordered the above spacings, the distribution of $(Y_(1), Y_(2), Y_(3), Y_(4))$ is equivalent to



                $$ left(frac X_1/4 sum_i=1^4 X_i,
                frac X_1/4 + X_2/3 sum_i=1^4 X_i,
                frac X_1/4 + X_2/3 + X_3/2 sum_i=1^4 X_i,
                frac X_1/4 + X_2/3 + X_3/2 + X_4 sum_i=1^4 X_iright)$$



                Any $3$ of the $4$ pieces cannot form a triangle if and only if
                $$ begincases
                Y_(1) + Y_(2) < Y_(3) \
                Y_(1) + Y_(2) < Y_(4) \
                Y_(1) + Y_(3) < Y_(4) \
                Y_(2) + Y_(3) < Y_(4)
                endcases$$
                Note that the second inequality is implied by the first, and the third inequality is implied by the fourth. So the probability of no triangle being formed is
                $$ beginalign
                &Pr Y_(1) + Y_(2) < Y_(3), Y_(2) + Y_(3) < Y_(4) \
                =& PrBiggfrac X_1 4 + frac X_1 4 + frac X_2 3 <
                frac X_1 4 + frac X_2 3 + frac X_3 2, \
                & frac X_1 4 + frac X_2 3 + frac X_1 4 + frac X_2 3
                + frac X_3 2 <
                frac X_1 4 + frac X_2 3 + frac X_3 2 + X_4Bigg \
                =& PrX_1 < 2X_3, 3X_1 + 4X_2 < 12X_4 \
                =& int_0^infty Pr2X_3 > xPr12X_4 - 4X_2 > 3xe^-xdx
                endalign$$



                Note that
                $$ beginalign
                &Pr12X_4 - 4X_2 > 3x \
                =& int_0^inftyPr12X_4 - 4u > 3xe^-udu \
                =& int_0^infty e^-(3x+4u)/12 e^-udu \
                =& e^-x/4int_0^infty e^-4u/3du \
                =& frac 3 4e^-x/4
                endalign
                $$



                So the integral become
                $$ int_0^infty e^-x/2frac 3 4 e^-x/4e^-xdx
                = frac 3 4 int_0^infty e^-7x/4 dx = frac 3 4 times frac 4 7 = frac 3 7 $$



                This is not a very elegant way but at least it is doable. Looking forward to someone to post a better solution.






                share|cite|improve this answer













                The lengths of the broken stick pieces $(Y_1, Y_2, Y_3, Y_4)$ are spacings of ordered uniform. The joint distribution is equivalent to
                $$ left(frac X_1 sum_i=1^4 X_i,
                frac X_2 sum_i=1^4 X_i,
                frac X_3 sum_i=1^4 X_i,
                frac X_4 sum_i=1^4 X_iright)$$
                where $X_i$ are iid exponential random variables. If we ordered the above spacings, the distribution of $(Y_(1), Y_(2), Y_(3), Y_(4))$ is equivalent to



                $$ left(frac X_1/4 sum_i=1^4 X_i,
                frac X_1/4 + X_2/3 sum_i=1^4 X_i,
                frac X_1/4 + X_2/3 + X_3/2 sum_i=1^4 X_i,
                frac X_1/4 + X_2/3 + X_3/2 + X_4 sum_i=1^4 X_iright)$$



                Any $3$ of the $4$ pieces cannot form a triangle if and only if
                $$ begincases
                Y_(1) + Y_(2) < Y_(3) \
                Y_(1) + Y_(2) < Y_(4) \
                Y_(1) + Y_(3) < Y_(4) \
                Y_(2) + Y_(3) < Y_(4)
                endcases$$
                Note that the second inequality is implied by the first, and the third inequality is implied by the fourth. So the probability of no triangle being formed is
                $$ beginalign
                &Pr Y_(1) + Y_(2) < Y_(3), Y_(2) + Y_(3) < Y_(4) \
                =& PrBiggfrac X_1 4 + frac X_1 4 + frac X_2 3 <
                frac X_1 4 + frac X_2 3 + frac X_3 2, \
                & frac X_1 4 + frac X_2 3 + frac X_1 4 + frac X_2 3
                + frac X_3 2 <
                frac X_1 4 + frac X_2 3 + frac X_3 2 + X_4Bigg \
                =& PrX_1 < 2X_3, 3X_1 + 4X_2 < 12X_4 \
                =& int_0^infty Pr2X_3 > xPr12X_4 - 4X_2 > 3xe^-xdx
                endalign$$



                Note that
                $$ beginalign
                &Pr12X_4 - 4X_2 > 3x \
                =& int_0^inftyPr12X_4 - 4u > 3xe^-udu \
                =& int_0^infty e^-(3x+4u)/12 e^-udu \
                =& e^-x/4int_0^infty e^-4u/3du \
                =& frac 3 4e^-x/4
                endalign
                $$



                So the integral become
                $$ int_0^infty e^-x/2frac 3 4 e^-x/4e^-xdx
                = frac 3 4 int_0^infty e^-7x/4 dx = frac 3 4 times frac 4 7 = frac 3 7 $$



                This is not a very elegant way but at least it is doable. Looking forward to someone to post a better solution.







                share|cite|improve this answer













                share|cite|improve this answer



                share|cite|improve this answer











                answered Jul 25 at 17:35









                BGM

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