The lengths of the segments are uniformly distributed over the unit $3$-simplex (see Simulating uniformly on $S^1=\{x \in \mathbb{R}^n \mid \|x\|_1=1\}$). Thus, the desired probability is the fraction of the part of the unit $3$-simplex with $x_1\le x_2\le x_3\le x_4$ for which $x_2+x_3\ge\frac12$.
The unit simplex has volume $1/6$, so the restriction to one particular permutation of the coordinates leads to a volume of $1/6\cdot1/4!=1/144$. We can check this to make sure the integration bounds are set up correctly:
$\int_0^{1/4}\int_{x_1}^{(1-x_1)/3}\int_{x_2}^{(1-x_1-x_2)/2}\mathrm dx_3\mathrm dx_2\mathrm dx_1=\frac1{144},$
as computed here.
The integration bounds under the condition $x_2+x_3\ge\frac12$ are a bit trickier, since there are two possibilities, depending on $x_2$. For $x_2\ge\frac14$, all values of $x_3$ with $x_3>x_2$ fulfill the condition, whereas for $x_2<\frac14$ there's a new lower bound $\frac12-x_2$ for $x_3$. Thus we split the integral into two parts at $x_2=\frac14$, with different lower bounds for $x_3$:
$\int_0^{1/4}\int_{x_1}^{1/4}\int_{1/2-x_2}^{(1-x_1-x_2)/2}\mathrm dx_3\mathrm dx_2\mathrm dx_1=\frac1{768},$
$\int_0^{1/4}\int_{1/4}^{(1-x_1)/3}\int_{x_2}^{(1-x_1-x_2)/2}\mathrm dx_3\mathrm dx_2\mathrm dx_1=\frac1{2304},$
as computed here and here, respectively. So the desired probability is indeed
$\frac{\frac1{768}+\frac1{2304}}{\frac1{144}}=144\cdot\frac{4}{2304}=\frac{144}{576}=\frac14\;.$