An inequality in Erdős–Rényi random graph model Let's consider Erdős–Rényi random graph model. This means that the probability assigned to a graph $G_{n,p}$ with $n$ vertexes and $m$ edges is equal to
$$P(G_{n, p}) = p^{m}(1-p)^{N-m},$$
where $N = {n\choose 2}$.
I am to prove that
$$P(G_{n, p_1} \in \mathcal{P}) \le P(G_{n, p_2} \in \mathcal{P}) \tag{1}$$
for $0 \le p_1 \le p_2 \le 1$ and any graph property $\mathcal{P}$.
I'm a bit confused because let's assume that $n=4, m=2, p_1=0.1, p_2=0.9$. Then $P(G_{n, p_1})=  0.006561$ and $P(G_{n, p_2}) = 0.000081$. What do I not understand? I would appreciate to get a sketch of a proof.
 A: First, a note on $P(G_{n,p})=p^m(1-p)^{N-m}$ : This is the probability that your random graph return a specific graph $H$ on $m$ edges. So intuitively if $H$ has very few edges ($m\ll N$), then it's more probable to get this specific graph if you chose a small $p$.
Now, the statement
$$P(G_{n, p_1} \in \mathcal{P}) \le P(G_{n, p_2} \in \mathcal{P}) $$
does not hold for  any graph property $\mathcal{P}$ and any $0 \le p_1 \le p_2 \le 1$
For instance, let $\mathcal{P}$ be the property of being disconnected, or having diameter at least 2, and take the extreme case $p_2=1$, then
$$P(G_{n, p_2} \in \mathcal{P})=0$$
while for any $0<p_1<1$, $P(G_{n, p_1} \in \mathcal{P})>0$
In general note that by defining $\mathcal{Q}$ as the negation of $\mathcal{P}$,
$$P(G_{n, p} \in \mathcal{P}) +P(G_{n, p} \in \mathcal{Q}) = 1$$
Therefore if
$$P(G_{n, p_1} \in \mathcal{P}) < P(G_{n, p_2} \in \mathcal{P}) $$
Then
$$1-P(G_{n, p_1} \in \mathcal{Q}) < 1-P(G_{n, p_2} \in \mathcal{Q}) $$
$$P(G_{n, p_1} \in \mathcal{Q}) > P(G_{n, p_2} \in \mathcal{Q}) $$
