A different approach to a common questionA unit stick is randomly broken into 3 pieces,...

Luz Stokes

Luz Stokes

Answered

2022-07-16

A different approach to a common question
A 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 part
S: Length of the smallest part
M: Length of the medium part

Answer & Explanation

minotaurafe

minotaurafe

Expert

2022-07-17Added 22 answers

Step 1
An easy approach.
Let'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 2
Therefore 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 n
Expected total length = 3 x + 2 y + z = 9 n
The the expected length of middle segment is
( x + y ) / ( 3 x + 2 y + z ) = 2.5 / 9 = 5 / 18

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