A different approach to a common questionA unit stick is randomly broken into 3 pieces, it is given that these three pieces can make a triangle, what is the expected length of the medium-sized piece?This a question we are all familiar with and anyone who has seen it anywhere/solved it knows that the expected length of the medium piece is 5 18 , we reach this conclusion by using E ( L + M + S ) = 1, and we know we can calculate E(L) and E(S), so we just use E ( M ) = 1 − E ( L ) − E ( S ) to get our answer, is there any way we can solely calculate the E(M) without calculating the other two values?L: Length of the longest partS: Length of the smallest partM: Length of the medium part

Question

A different approach to a common questionA unit stick is randomly broken into 3 pieces, it is given that these three pieces can make a triangle, what is the expected length of the medium-sized piece?This a question we are all familiar with and anyone who has seen it anywhere/solved it knows that the expected length of the medium piece is       5    18  , we reach this conclusion by using   E  (  L  +  M  +  S  )  =  1, and we know we can calculate E(L) and E(S), so we just use   E  (  M  )  =  1  −  E  (  L  )  −  E  (  S  ) to get our answer, is there any way we can solely calculate the E(M) without calculating the other two values?L: Length of the longest partS: Length of the smallest partM: Length of the medium part

minotaurafe · Accepted Answer

Step 1An easy approach.Let&#039;s arrange the segments from smallest to largest. Let the three segments be x,   x  +  y and   x  +  y  +  z.Now sum of all segments   =  1     ⟹    3  x  +  2  y  +  z  =  1;with the conditions:   x  ⩾  0  ,  y  ⩾  0  ,  z  ⩾  0 and   and   x  ⩽  1      /    3  ,  y  ⩽  1      /    2  ,  z  ⩽  1.for normalisation, let n be a quantity   ⩽  1      /    6, then,      x    2    ⩽      1    6    ,      y    3    ⩽      1    6    ,      z    6    ⩽      1    6  x can be chosen randomly among a pool from [0,2n], y be chosen randomly from a pool [0,3n] and z from a pool [0,6n] where (x,y,z,n)   ∈ R.Step 2Therefore we can say:Expected value of   x  =  n, expected value of   y  =  1.5  n and expected value of   z  =  3  n.Expected length of middle segment =   x  +  y  =  2.5  nExpected total length =   3  x  +  2  y  +  z  =  9  nThe the expected length of middle segment is  (  x  +  y  )      /    (  3  x  +  2  y  +  z  )  =  2.5      /    9  =  5      /    18