Find expected value of coloured segment Suppose picking $n$ random points from $[0,1]$ interval. For each point we colour segment of least length to it's neighboring point. If length to both neighboring points is the same then we colour to the left neighbour. Find expeceted value of length coloured sements. 
 A: Add an $n+1$-th point that doesn't affect the colouring. The expected value of the coloured length is the probability that this extra point lies in a coloured segment.
By symmetry, the extra point is the left-most or right-most with probability $\frac2{n+1}$, in which case it's not coloured, and it's the second from the left or right with probability $\frac2{n+1}$, and then it's coloured.
That leaves the cases in which the extra point has second-degree neighbours on either side. Conditional on their position, the extra point and its immediate neighbours are uniformly distributed between them. The probability that either of the extra point's immediate neighbours is the nearest neighbour of the other (remember that the extra point doesn't affect the colouring) is
\begin{align}
&3!\left(\int_0^\frac13\mathrm dx\int_x^\frac{x+1}2\mathrm dy(y-x)+\int_\frac13^\frac12\mathrm dx\int_x^{2x}\mathrm dy(y-x)+\int_\frac12^1\mathrm dx\int_x^1\mathrm dy(y-x)\right)\\
={}&6\left(\frac{19}{649}+\frac{19}{1296}+\frac1{48}\right)\\
={}&\frac7{18}\;.
\end{align}
Thus the expected value of the coloured length is
$$
\frac2{n+1}\cdot0+\frac2{n+1}\cdot1+\left(1-\frac4{n+1}\right)\cdot\frac7{18}=\frac7{18}+\frac49\frac1{n+1}\;.
$$
This is valid for $n\ge3$; for lower $n$ the same approach yields the values that Hagen gave in a comment, $0$ for $n=0$ and $n=1$ and $\frac13$ for $n=2$.
Here's code that checks this result numerically.
