# Stick of unit length is broken into three random pieces, what is the expected length of the longest piece?

In regards to this question: Average length of the longest segment

Can anybody explain why the cumulative distribution function is in the form that is given?

Take a one-foot stick,lay it down, and you simultaneously cut it at two points in order to get three pieces. It is equivalent to choose two points $X$ and $Y$ on the stick.

From left to right, the first piece $A$ will be of size the minimum between $X$ and $Y$ minus $0$.

The piece on the right will be of size $1$ minus the maximum between $X$ and $Y$, or $1-B$ where $B=max(X,Y)$

The one in the middle is of size $X-Y$ fs $X>Y$, or $Y-X$ if $X \leq Y$, in other words , $|X-Y|=B-A$

Finally, $C$ is the longest stick, that is $max(A,1-B,B-A)$

The cdf is then defined as follows $$F_C(a)=P(C \leq a)= P(max(A,1-B,B-A)\leq a)$$

If the maximum between $A,1-B$ and $B-A$ is smaller than $a$, therefore each of them are smaller than $a$, and reciprocally.

We can rewrite the cdf as

$$F_C(a)=P(C \leq a)= P(A\leq a,1-B\leq a,B-A\leq a)$$

$X$ and Y are totally symmetrical, you can write $$P(A\leq a,1-B\leq a,B-A\leq a)=P(A\leq a,1-B\leq a,B-A\leq a,X<Y)+P(A\leq a,1-B\leq a,B-A\leq a,X \geq Y)$$

Thus, $$P(A\leq a,1-B\leq a,B-A\leq a)=2P(X\leq a,1-Y\leq a,Y-X\leq a,X<Y)$$

A bit of geometry will give you the result : draw a square $[0,1]*[0,1]$, the vertical line $\{x=a\}$, the horizontal line $\{y=1-a\}$, the straight lines $y=a+x$ and $y=x$. Locate the intersection of the surfaces defined by $P(A\leq a,1-B\leq a,B-A\leq a)$ and you will get the results.