Connecting noodles probability question

Let's do it in a more general case. Let the expected number of loops from $n$ noodles be $E(n)$.

Obviously, $E(1)=1$.

For $n>1$, if you pick up two ends, the possibility for those two ends belongs to the same noodle will be $\frac{1}{2n-1}$. Then you have one loop now and there are $n-1$ noodles to keep on going. Otherwise you get no loop and still have $n-1$ noodles to go(The two you pick before are now connected as one). Hence $E(n)=E(n-1)+\frac{1}{2n-1}$

Inductively you may expect that $E(n) = 1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{2n-1}$.

The accepted answer is correct, but I think should be more carefully detailed.

$E(1) = 1$

Pick up two ends. They either belong to the Same Noodle ($S$) or they don't ($\bar{S}$). The $S$ event can be used to "compute expectations by conditioning" in the same spirit of this question. So:

$$E(n) = E(n | S)p(S) + E(n | \bar{S}) p(\bar{S})$$

Now $p(S) = \frac{1}{2n-1}$ and $p(\bar{S}) = 1-\frac{1}{2n-1}$. Now the trick is that $E(n|S) = E(n-1)+1$ (i.e. if you picked the two ends from the same noodle you've added one loop to the bowl, leaving $n-1$ noodles) and $E(n|\bar{S}) = E(n-1)$ (i.e if you've picked two ends from different noodles you didn't increase the number of loops).

$$E(n) = E(n | S)p(S) + E(n| \bar{S}) p(\bar{S}) = \\ = \left[E(n-1)+1\right]\frac{1}{2n-1} + E(n-1)\left[1-\frac{1}{2n-1}\right] = \\ = E(n-1) + \frac{1}{2n-1}$$