For the expectation, linearity of expectation should help a lot. There are ${n\choose 2}$ pairs of chords, and if each chord is drawn i.i.d. then each pair has some probability of intersection of $p$, so the answer is $p{n\choose 2}$.

In the case where both endpoints of each chord are drawn uniformly at random, I believe $p=1/3$. Let's call the two chords AB and CD and imagine we first randomly place A; this divides the circle into a line segment, say, clockwise. Now the chords intersect if the ordering on this line segment is C, B, D or D, B, C, but do not intersect for any of the other four orderings. All six orderings are equally likely, so there's a $1/3$ chance that two chords will intersect.

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