If I have a graph with 10 nodes and probability of an edge being 0.6, what is the expected number of 'common friends' of two nodes A and B? e.g. by 1 common friend I mean that for Node A and B (just both of them), only one of the remaining nodes connect to both A and B.
I already have a correct solution which is 1*P(1)+2*P(2)+3*P(3)+...+8*P(8) = 2.xx (where xx is hidden because of homework reasons) (pls inform me if i should reveal the numbers)
But... there is a much simpler solution. Someone I know has the answer and the clue is it's composed of a mere 2-3 terms, the last one ending with a number (9 i think). Can anyone give me a hint on how to solve it?
P.S. please inform me if there is anything I need to clarify or add. thanks!
The simple way to do it is to remember that expectation is additive. That is, suppose the nodes are $A, B, C, D, E$. Then the expected total number of common friends is the equal to the sum of the expected chance that $C$ is a common friend, plus the expected chance that $D$ is a common friend, plus the expected chance that $E$ is a common friend.
So with $10$ nodes, you have $8$ candidates for a common friend. Each candidate has a $(.6) \cdot (.6)$ chance of being a common friend of $A$ and $B$ (why?) To find the total expected number of common friends of $A$ and $B$, just add up the values for each of the $8$ candidates.