This question is from 50 problems in probability.
Q. A glass rod is coloured red at one end and blue at other. Many such rods got broken into 3 parts. What is the average length of part colored blue
The author argues that the answer should be one third as all things being symmetric the average length of each part must be equal. This seems alright
I tried using calculus maybe incorrectly. I denote length of rod as $1$ unit then let us suppose length of blue rod is $x$. Then probability of other point of breaking must be $1-x$. Expected length will be
$$\int p(x) x\, dx = \int_0^1 (1-x)x \, dx = \frac{1}{6}$$
This is wrong. But what is the error. I think the error is that when writing $p(x)$ I forget to normalise it, that is $\int_{0}^{1}(1-x)N\, dx = 1 \implies N = 2$. So multiplying by this factor fixes the probability and we get $1/3$.
Is this correct
To me it is not obvious that the average length of each part is ${1\over3}$, even if the two cuts are assumed independently uniformly distributed. Note that there are two independent cuts at points $x$ and $y$ in the interval $[0,1]$. When the first cut is at $x$ you have no idea at which point $y$ the second cut would be. In particular there is no question of enforced $y=1-x$.
But we can compute it. Assume the rod to be $[0,1]$, and that it is cut at two independently uniformly placed points $x$, $y\in[0,1]$. We may assume that the end at $0$ is colored blue. The length of the blue part then is $$\min\{x,y\}=\left\{\eqalign{u\qquad&(v>u)\cr v\qquad&(v<u)\cr}\right.$$
Since $(x,y)$ is uniformly distributed in $[0,1]^2$ the expected length of the blue part then is $$E=\int_0^1\int_u^1 u\>dv\>du+\int_0^1\int_0^u v\>dv\>du={1\over3}\ .$$