Tying noodles into loops

Idea. Look at one tie at a time. With $k$ strands left, a tie closes a loop exactly when the two chosen ends belong to the same strand, which has probability $\frac{1}{2k-1}$. Either way the number of strands drops by one. Add the probabilities.

Step 1: one tie. With $k$ strands there are $2k$ free ends. Whatever the first chosen end, the second is one of the other $2k-1$, and exactly one of them is the other end of the same strand. So the tie makes a loop with probability $\frac{1}{2k-1}$. In all cases one strand fewer remains.

Step 2: add up. The ties happen with $k=100,99,\dots ,1$ strands, so $$\begin{array}{rl}\mathbb{E}[\text{loops}]& =\sum _{k=1}^{100}\frac{1}{2k-1}\\ & =1+\frac{1}{3}+\frac{1}{5}+\cdots +\frac{1}{199}.\end{array}$$

Step 3: estimate. The sum of reciprocals of odd numbers up to $2n-1$ is $H_{2n}-\frac{1}{2}{H}_n$, where $H_m=1+\frac{1}{2}+\cdots +\frac{1}{m}$. With $H_m\approx \ln m+\gamma$: $$\begin{array}{rl}{H}_{200}-\frac{1}{2}{H}_{100}& \approx \ln 200+\gamma -\frac{1}{2}(\ln 100+\gamma )\\ & =\ln 2+\ln 10+\frac{\gamma }{2}\approx \mathrm{3.284.}\end{array}$$ The exact sum is $3.28434\dots$: the approximation errors cancel almost completely.

The surprise. 100 noodles make only about 3.3 loops on average: most ties lengthen a strand rather than close it.