Luz Stokes

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 $\frac{5}{18}$, we reach this conclusion by using $E\left(L+M+S\right)=1$, and we know we can calculate E(L) and E(S), so we just use $E\left(M\right)=1-E\left(L\right)-E\left(S\right)$ 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

minotaurafe

Expert

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 $=1$ $\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}3x+2y+z=1$;
with the conditions: $x⩾0,y⩾0,z⩾0$ and .
for normalisation, let n be a quantity $⩽1/6$, then,
$\frac{x}{2}⩽\frac{1}{6},\frac{y}{3}⩽\frac{1}{6},\frac{z}{6}⩽\frac{1}{6}$
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) $\in$ R.
Step 2
Therefore we can say:
Expected value of $x=n$, expected value of $y=1.5n$ and expected value of $z=3n$.
Expected length of middle segment = $x+y=2.5n$
Expected total length = $3x+2y+z=9n$
The the expected length of middle segment is
$\left(x+y\right)/\left(3x+2y+z\right)=2.5/9=5/18$

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