Consider the random graph $G(n,\frac{1}{n})$. I'm trying to estimate the size of the maximum matching in $G$.
If we look at one vertex, the expected value of its degree is $\frac{n-1}{n}$ so it seems like with high prob it should be 1.
So if I can show that with high probability half of the vertices has degree $1$, then with high probability the size of maximum match in $G$ would be of size $\frac{n}{4}$, but I couldn't prove it, and I'm looking for a hint on how to show that or something similar to that claim.
First, just because the expected value of a fixed vertex is $\frac{n-1}{n}$ does not mean w.h.p. its degree is 1. In fact, $$ \text{Pr(deg(}v\text{)=1)} = {n-1 \choose 1} \frac{1}{n} \left(1-\frac{1}{n}\right)^{n-2} \approx 1 \cdot e^{-1}. $$ Now if $X$ is the random number of vertices with degree exactly 1, then $E[X] \approx e^{-1} n$.
Let $Y$ be the number of isolated edges. Note that there is clearly a matching that is at least $Y$. Now $$ E[Y] = {n \choose 2} \frac{1}{n} \left(1-\frac{1}{n}\right)^{2(n-2)} \approx \frac{n}{2e^2}. $$ Now at that remains is to show that $Y$ is concentrated around its mean.