On uniform randomness of the weight of the remaining edges of a graph after deleting some of them
Let us consider the simplest case, that of a path of length two edges, which are drawn uniformly and independently from $[0,1]$. Let us suppose that the first edge is sub-dominant, which happens with probability 1/2. The probability that the weight of second edge is at most $t \in [0,1]$ is
$$
\Pr[x \leq y \leq t \mid x \leq y] = \Pr[x,y \leq t] = t^2.
$$
Here $x$ is the weight of the first edge, and $y$ is the weight of the second edge. We see that even before doubling, the distribution of $y$ is non-uniform.

Next, let us examine the case of a star with three edges, again drawn uniformly and independently from $[0,1]$. Again suppose that the first edge is sub-dominant, which happens with probability 1/3. The probability that the weight of the second edge is at most $t \in [0,1]$ while that of the third edge is at most $u \in [0,1]$ is
$$
\Pr[y \leq t, z \leq u \mid x \leq y,z] =
3 \int_0^{\min(t,u)} (t-x)(u-x) \, dx = \\
3tum - \frac{3}{2}(t+u)m^2 + m^3.
$$
Here $x,y,z$ are the weights of the first, second, and third edges, respectively, and $m = \min(t,u)$. Substituting $u = 1$, we obtain
$$
\Pr[y \leq t \mid x \leq y,z] = \frac{3t^2-t^3}{2}.
$$
Calculation shows that for generic $t,u$,
$$
\Pr[y \leq t, z \leq u \mid x \leq y,z] \neq
\Pr[y \leq t \mid x \leq y,z] \Pr[z \leq u \mid x \leq y,z],
$$
that is, the remaining weights are no longer independent.