Problem Statement: Given a stick of length 1, the stick is to be cut randomly into 3 pieces. What is the expected length of the shortest stick?
Attempt:
Let $S$ be the random variable for the shortest piece's length. Let $s$ be a particular value of $S$. To solve this problem, we can find the CDF of $S$, i.e., $F_S = P(S\leq s)$, differentiate this to get the pdf, and finally use this to get $E[S]$.
Let $x, y, 1-x-y$ be length of each piece. So $S =\min(x,y,1-x-y)$. In order for the event $S\leq s$ to occur, we must have $x \geq s,\ \ y\geq s,\ \ 1-x-y\geq s$ and that $\max(s) = \frac{1}{3}$.
I used these relations to obtain the CDF graphically. See image below. The shaded triangle represents the relative probability area bounded by the 3 conditions $x \geq s,\ \ y\geq s,\ \ 1-x-y\geq s$.
The horizontal and vertical lengths of this triangle are the same. To find the side length, I found $y_0=1-2s$ by solving $x_0=s$ and $1-x_0-y_0=s$. So we get the side length to be $1-2s – s = 1-3s$.
$$
F(S \leq s) = \frac{0.5(1-3s)^2}{0.5} = (1-3s)^2.
$$
The numerator in the middle expression is the area of the shaded triangle, and the denominator is the area of the outer triangle formed by the vertices (0,1), (0,0), (1,0). The outer triangle's area is the entire sample space.
The issue with the CDF I obtained is that it is decreasing, and $F_S(S = 0)$ = 1, which doesn't make any sense. It should be $F_S(S = \frac{1}{3}) = 1$.
The solution to the problem is $\frac{1}{9}$ and if I proceed with my incorrect CDF, I get the $E[S] = -\frac{1}{9}$. This implies that I have a sign switch somewhere, or I did this completely wrong and just happened to get the same magnitude.
Does anyone notice an error with my approach?
EDIT 1:
Including some more details of the outer triangle. If we let $x, y, 1-x-y$ be the lengths of the 3 separate pieces, then we know that $x \geq 0, \ \ y \geq 0, \ \ 1-x-y \geq 0$ (because length can't be negative). The outer triangle is bounded by these 3 constraints. All ways of breaking the stick are represented by points inside the triangle.
Best Answer
Your problem is that what you have obtained (the ratio of hatched triangle's area to big triangle area) is
$$P(X\color{red}{\ge s})=(1-3s)^2$$
Therefore :
$$F(s):=P(X<s)=1-P(X \geq s)=1-(1-3s)^2$$
Deriving it, you will get :
$$f(s)=6(1-3s) \ \ for \ 0 \leq s \leq \frac13$$
from which the mean is easily obtained.