I have seen similar arguments at other places but couldn't convince myself so far about it. I am reading some literature related to graph theory but let me post the analogous problem which avoids the graph theory terminology:
There are $M$ boxes and $m$ balls which we want to randomly uniformly distribute among those boxes.
Now author essentially says that the expected number of balls in each box $i$ is $p_{i} = \frac{m}{M}$. However, then he says that if $m << M$, this can also be considered as the probability of filling that box with one of the $m$ balls.
I don't understand this. I tried deriving expression for the expected value but that itself involves the probability that a ball goes into the box. Can somebody please clarify this? Thanks
The expected value of a random variable $X$ whose values are natural numbers is $$ 0\cdot P(X=0) + 1\cdot P(X=1) + 2\cdot P(X=2) + 3\cdot P(X=3) + \cdots $$
In your case, if $m\ll M$, then the probability that there's more than one ball in the box will be vanishingly small, so the terms with $P(X=2)$ and so forth can be ignored. So what is left is $$ 0\cdot P(X=0) + 1\cdot P(X=1)$$ which is of course just $P(X=1)$.