Splitting Line Segments and Finding Expected Value Consider a line segment which has a length of $2n-3$. It is split into $n$ segments at random. It is guaranteed that $n\ge 3$ and $n\in \mathbb{Z}$. These smaller lines are then used as the sides of a convex $n$-polygon. However, the lines are arranged in a way that the area is maximum. Additionally, there are ${2n-4}\choose {n-1}$ total ways to split the original line and occur with same probability. 
How can we find the expected value of the area?
If we work out the case for $n=3$, there will have to be $2$ splits. All three smaller lines have to be of the same size if we look at the above conditions. So now, we have a triangle whose sides are all $1$ and so the area is $\frac{1}{4}\sqrt{3}$. 
$n=3$ seems to be a simple case. What about for larger $n$? Can we derive a formula for general $n$, or is each case independent of one another?
 A: Finding a nice general formula for all $n$ is, I'm afraid, very difficult if not impossible. Just for fun, I found the areas for $n=4$ and $n=5$.
Case $n=4$ is simple, because the only way to split $5$ into four positive integers is $(1,1,1,2)$. Maximum area polygon in this case is a trapezoid, having bases of length $3$ and $1$. Its area is easy to compute and turns out to be $Area_{1112}=3\sqrt3/4$.
In case $n=5$ there are two possible ways to split $7$ into five positive integers: $(1,1,1,1,3)$ and $(1,1,1,2,2)$. Maximum area polygons look like in picture below (only one of the two possible polygons is drawn for the the second case, but that doesn't matter because their area is the same).

Their areas can be found exactly, but for the polygon on the left this requires solving a third degree equation (I haven't been able to find a simpler way):
$$
Area_{11113}=\frac{1}{4 \sqrt{48-\sqrt[3]{2} \left(81-3
   \sqrt{633}\right)^{2/3}-\sqrt[3]{2} \left(3
   \left(27+\sqrt{633}\right)\right)^{2/3}}}
\left(4 \sqrt[6]{2} \sqrt[3]{81-3 \sqrt{633}}+4 \sqrt[6]{2} \sqrt[3]{3\left(27+\sqrt{633}\right)}-9 \sqrt{-40+\sqrt[3]{2} \left(81-3
   \sqrt{633}\right)^{2/3}+\sqrt[3]{2} \left(3
   \left(27+\sqrt{633}\right)\right)^{2/3}}\right)
\approx2.34301;
$$
$$
Area_{11122}=\frac{1}{4} \left(5+\sqrt{2}\right) \sqrt{1+2 \sqrt{2}}
\approx 3.13757.
$$
To compute the expected value of the area, just notice that, out of $15$ possible choices, $5$ times one gets $Area_{11113}$ and $10$ times $Area_{11122}$:
$$
<Area>={5Area_{11113}+10Area_{11122}\over15}\approx 2.87272.
$$
