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For every $v$, we have $|N(v)|\approx np$ (I assume $p$ is not very small so the fluctuations are small enough to ignore). You want two of these such that $N(x)\cap N(y)={v}$, or in other words, if you now forget about $v$ itself, you want these two sets to be disjoint.

For any specific $x$ and $y$ the probability of this is roughly $(1-p^2)^n \approx e^{-np^2}$. If we take $p=\sqrt{\alpha \log n / n}$ then we get $n^{-\alpha}$. There are $np/2$ disjoint pairs of vertices in $N(v)$ and the choices are independent (well, almost, I'll leave it for you to sort it out). Hence, the probability of not seeing such a pair is bounded by $$(1-n^{-\alpha})^{np/2}\approx e^{-n^{1-\alpha}p/2}=e^{-n^{\frac12-\alpha}\sqrt{\alpha \log n}}$$.

Taking $\alpha<\frac12$ we get that the failure probability is less than polynomially small, and then union bound over all $v$ is enough.

If I had to bet, I'd say the right $\alpha$ is 1, but I don't have to bet.

What is the motivation for the question, if I may ask?

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For every $v$, we have $|N(v)|\approx np$ (I assume $p$ is not very small so the fluctuations are small enough to ignore). You want two of these such that $N(x)\cap N(y)={v}$, or in other words, if you now forget about $v$ itself, you want these two sets to be disjoint.

For any specific $x$ and $y$ the probability of this is roughly $(1-p^2)^n \approx e^{-np^2}$. If we take $p=\sqrt{\alpha \log n / n}$ then we get $n^{-\alpha}$. There are $np/2$ disjoint pairs of vertices in $N(v)$ and the choices are independent (well, almost, I'll leave it for you sort it out). Hence, the probability of not seeing such a pair is bounded by $$(1-n^{-\alpha})^{np/2}\approx e^{-n^{1-\alpha}p/2}=e^{-n^{\frac12-\alpha}\sqrt{\alpha \log n}}$$.

Taking $\alpha<\frac12$ we get that the failure probability is less than polynomially small, and then union bound over all $v$ is enough.

If I had to bet, I'd say the right $\alpha$ is 1, but I don't have to bet.

What is the motivation for the question, if I may ask?