With probability 1/2, a random binary tree consists only of the root node.
Otherwise, it consists of two branches of height 1 with independent random
binary subtrees hanging from each of them.
Let $H$ denote the height of the random binary tree and set
$h(n):=\mathbb{P}(H>n)$, the probability that the tree's height exceeds $n$.
The tree's height exceeds $n+1$ if it has more than the root node, and
at least one of the subtrees has height exceeding $n$. It follows that
$$h(0)={1\over2},\qquad h(n+1)={1\over 2}-{1\over 2}[1-h(n)]^2,\quad n\geq 0.\tag1$$
Now, $h(n)$ is decreasing and by equation (1) converges to some root of $s={1\over 2}-{1\over 2}(1-s)^2$.
In other words, $h(n)\to 0$ so the tree has finite length: $\mathbb{P}(H<\infty)=1$.
To investigate $\mathbb{E}(H)$, let's rewrite equation (1) as
$${1\over h(n+1)}- {1\over h(n)}={1\over 2}+{h(n)\over2(2-h(n))}.\tag2$$
Adding these increments and dividing by $N$ gives
$${1\over Nh(N)}- {1\over Nh(0)}={1\over 2}+{1\over N}\sum_{n=0}^{N-1} {h(n)\over2(2-h(n))}.\tag3$$
Letting $N\to\infty$ in (3) shows that $Nh(N)\to 2$.
In particular, $h(N)>1/N$ for large $N$ and hence
$$\mathbb{E}(H)=\sum_{N=0}^\infty \mathbb{P}(H>N)=\infty.$$