Threshold probability of the 3-uniform hypertriangle I am stuck on a random hypergraph problem (I am encountering random hypergraphs for the very first time).
Let $G_{3}(n, p)$ be a binomial 3-uniform hypergraph. Find a threshold probability for containing a 3-uniform hypertriangle.
What I have tried: I saw a similar problem for the existence of triangles in $G(n, d/n)$. I am trying the same technique here for the hypergraphs. So according to my calculations I am getting $E[x]=\binom{n}{6}p^{3}$, where $x$ is the number of hypergraphs. Then for the variance part, I get here 5 cases: when the number of common edges are :
(i) up to 2, in which case its contribution is $E^{2}[x]$,
(ii) 3, the contribution is $\binom{n}{9}p^{6}$,
(iii) 4, the contribution is $\binom{n}{8}p^{5}$,
(iv) 5, the contribution is $\binom{n}{7}p^{4}$ and
(v) 6, the contribution is $\binom{n}{6}p^{3}$.
After this, I do not know how to proceed and somehow, I feel that all this is completely wrong and something entirely different needs to be done for hypergraphs. Any suggestions would be helpful. Thanks in advance.
$\textbf{Edit}$:
A 3-uniform hypergraph is a pair $(V, E)$, where $V$ is a set of its vertices, and $E\subset\binom{V}{3}$ is a set of its edges.
And a 3-uniform hypertriangle is a 3-uniform hypergraph with $|V | = 6$
with three edges such that every pair of edges share 1 vertex, and all three edges
do not have a common vertex.
$G_{3}(n, p)$ is a binomial 3-uniform hypergraph such that $V = \{1, . . . , n\}$ and every hyperedge is chosen independently with probability $p$.
And I got my ideas from a similar calculation for graphs on page 6 of this link : https://www.cs.cmu.edu/~avrim/598/chap4only.pdf
 A: The same technique works. If we let $\mathbf X$ be the number of hypertriangles in $G_3(n,p)$, then $\mathbb E[\mathbf X] = O(n^6 p^3)$. It is not $\binom n6 p^3$ because not all $6$ vertices play the same role: if we were being careful, it would be $\frac{6!}{3!} \binom n6 p^3$. But even without being that careful, we see that for $p \ll \frac1{n^2}$, $\mathbb E[\mathbf X] \to 0$, and with high probability there are no hypertriangles.
To show that with high probability, hypertriangles are present when $p \gg \frac1{n^2}$, we must first to compute $\text{Var}[\mathbf X]$, which requires computing $\mathbb E[\mathbf X^2]$: the expected number of ordered pairs of hypertriangles in $G_3(n,p)$. It will be convenient to consider $p$ in some small range, say $\frac1{n^2} \ll p \ll \frac1{n^{1.99}}$. For larger $p$, the claim follows by monotonicity.
However, I'm not sure your division into cases for computing $\mathbb E[\mathbf X^2]$ makes sense. A hypertriangle has only $3$ edges, so how can two hypertriangles share $6$ edges?
Instead, consider:

*

*We have pairs of hypertriangles that share no edges. An upper bound on the number of pairs like this is $\mathbb E[\mathbf X]^2$. Here, we must be slightly careful, because a bound of $O(n^{12}p^6)$ is not precise enough. However, there is some $f(n) = O(n^6)$ which tells us the number of potential hypertriangles. We have $\mathbb E[\mathbf X] = f(n)p^3$. Meanwhile, $f(n)^2$ is an upper bound on the number of potential pairs that share no edges - it is the total number of potential pairs. Each of the ones that share no edges has probability $p^6$ of appearing, so $f(n)^2 p^6 = \mathbb E[\mathbf X]^2$ is our upper bound.


*We have pairs of hypertriangles that share one edge. An upper bound on the number of pairs like this is $O(n^9 p^5)$.


*We have pairs of hypertriangles that share two edges. An upper bound on the number of pairs like this is $O(n^7 p^4)$.


*We have pairs of hypertriangles that share all three edges. There are exactly $\mathbb E[\mathbf X]$ of these.
Remember how I decided to look at $p$ such that $\frac1{n^2} \ll p \ll \frac1{n^{1.99}}$? That's so that the second and third cases are $o(1)$ terms. More precisely, we see that this holds as long as $p \ll \frac1{n^{1.8}} = n^{-9/5}$.
Anyway, putting together all three cases, we get $\mathbb E[\mathbf X^2] \le \mathbb E[\mathbf X]^2 + \mathbb E[\mathbf X] + o(1)$, so $\text{Var}[\mathbf X] \le \mathbb E[\mathbf X] + o(1)$, so $\frac{\text{Var}[\mathbf X]}{\mathbb E[\mathbf X]^2} \le \frac{\mathbb E[\mathbf X] + o(1)}{\mathbb E[\mathbf X]^2}$. For $p \gg \frac1{n^2}$, $\mathbb E[\mathbf X] \to \infty$, so $\frac{\text{Var}[\mathbf X]}{\mathbb E[\mathbf X]^2}  \to 0$.
By a standard application of Chebyshev's inequality, $\mathbf X > 0$ with high probability (and actually, $\mathbf X = (1+o(1))\mathbb E[\mathbf X]$ with high probability). So we get the threshold we wanted.
