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Consider a bipartition $\{A_1, A_2, ..., A_n\}$ and $\{B_1, B_2, ..., B_m\}$. No edges exists between $(A_i, A_j)$ or $(B_i, B_j)$, for any $i$ and $j$.

The probability that the edge $(A_i, B_j)$ exists is $p$. Node $A_i$ is said to be a 2-hop neighbor of $A_1$ if there exists a connected path $A_1 \rightarrow B_j \rightarrow A_i$. (Also, we don't consider $A_1$ as a 2-hop neighbor of itself.)

What is the probability that there exists at least one 2-hop neighbor of $A_1$, and that at the same time, for every 2-hop neighbor of $A_1$, the 2-hop path is unique?

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Let $X_1,X_2,\dots X_n$ be the sets of neighbours of $A_1,A_2,\dots,A_n$. We need to calculate the probability each of these events happens: $|X_1\cap X_j|=1$.

If we condition this to the size of $X_1$ being equal to $k$ then the events are independent.

The probability that $|X_1\cap X_2|=1$ is clearly $kp(1-p)^{k-1}$.

The probability that $X_1$ has size $k$ is $\binom{m}{k}p^k(1-p)^{m-k}$

So the answer is $p\sum\limits_{k=0}^m k\binom{m}{k}p^k(1-p)^{m-k}(1-p)^{k-1}=p(1-p)^{m-1}\sum\limits_{k=0}^mk\binom{m}{k}p^k$.

Notice that $f(x)=(x+1)^m=\sum\limits_{k=0}^m\binom{m}{k}x^k$ so $pf'(p)=\sum\limits_{k=0}^m k\binom{m}{k}p^k$.

Hence what we want is $p^2(1-p)^{m-1}f'(p)$

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    Does $X_i$ contain 2-hop neighbors or direct neighbors of $A_i$? – 2017-02-02
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    direct neighbours ${}{}{}{}$ – 2017-02-02
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    *The probability that $|{X_1 \cap X_2}| = 1$ is clearly $kp(1-p)^m$* -- can you explain a bit? – 2017-02-02
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    that is actually false, it should be $kp(1-p)^{k-1}$, my bad. – 2017-02-02