I'm following the book Networks by Mark Newman. He considers an extension to the BA model where edges are removed uniformly at random. He computes the probability that a particular node $i$ loses an edge when a single edge is removed from the network to be
$$ p_i = \frac{2k_i}{\sum_j k_j}, $$
where $k_i$ is the degree of node $i$ and he says the factor two coming from the two ends of the edge.
I understand why the probability is proportional to the node degree, because the number of edges attached to a node is the degree (in the case there are no self-loops).
What I don't understand is why there is a factor $2$. It seems to me this probability is not properly normalized, as $\sum_i p_i = 2$.
Best Answer
That is because the events "Node $i$ loses an edge" are not disjoint - whenever an edge is removed, two nodes will lose an edge, so it makes sense that the probabilities will sum to 2.
Regarding the factor of $2$ in the $p_i$ formula, there are $\frac{\sum_{j} {k_j}}{2}$ edges, and there are $k_i$ edges with an end at node $i$, so the probability that such an edge will be chosen is $$\frac{k_i}{\frac{\sum_j {k_i}}{2}} = \frac{2k_i}{\sum_j {k_j}}$$