Let's consider the case where we have a single noodle. Clearly, we can only connect the noodle to itself, so the expected number of noodles is simply 1
$$
\text{E}[f(1)] = 1.
$$
In the case of two noodles, if we take a random noodle end then there are 3 other ends available to connect the noodle to. One of these ends will give us a single noodle loop, whereas the other two ends involve us connecting our selected noodle to the other remaining noodle. So,
$$
\text{E}[f(2)] = \frac{2}{3} \text{E}[f(1)] + \frac{1}{3} \text{E}[1 + f(1)].
$$
Rearranging gives
$$
\text{E}[f(2)] = \frac{1}{3} + \text{E}[f(1)].
$$
Continuing, in the case of three noodles we have a $\frac{1}{5}$ probability of connecting our chosen noodle to itself giving an extra loop, and a $\frac{4}{5}$ chance of connecting the noodle to another noodle, reducing the problem to the case of two noodles.
$$
\text{E}[f(3)] = \frac{1}{5} + \text{E}[f(2)]
$$
We can infer from these results that generally,
$$
\text{E}[f(n)] = \frac{1}{2n-1} + \text{E}[f(n-1)].
$$
We can represent this relation as a sum
$$
\text{E}[f(n)] = \sum_{k=1}^n \frac{1}{2k-1}.
$$
We evaluate this sum using a computer to get the result
$$
E[f(100)] = 3.28.
$$