The answer is equal to \(\sum_{i=1}^{10} \frac{1}{2i-1} \simeq 2.133\).
Let \(f(n)\) be the answer to the problem in which we have \(n\) ropes. Now, take one end of one of the ropes and randomly connect it to one of the other \(2n-1\) ends. If both ends are from the same rope, we end up having one loop and \(n-1\) disconnected ropes, otherwise we end up with \(n-2\) disconnected ropes and two rops that are connected from one end. Note that when we connect two ropes, the combination is technically equivalent to a single rope and therefore in both cases the average number of loops that are made for the rest of the ropes is \(f(n-1)\). Since the former case happens with probability \(1/(2n-1)\) and the likelihood of the latter case is \((2n-2)/(2n-1)\) , we have \[f(n) = \frac{1}{2n-1} + f(n-1) = \sum_{i=1}^{n} \frac{1}{2i-1}.\]