Suppose you start with $n$ ropes. You pick two free ends and tie them together:
If you happened to pick two ends of the same rope, you've added one additional loop (which you can set aside, since you'll never pick it now on), and have $n-1$ ropes
If you happened to pick ends of different ropes, you've added no loop, and effectively replaced the two ropes with a longer rope, so you have $n-1$ ropes in this case too.
Of the $\binom{2n}{2}$ ways of choosing two ends, $n$ of them result in the first case, so the first case has probability $\frac{n}{2n(2n-1)/2} = 1/(2n-1)$. So the expected number of loops you add in the first step, when you start with $n$ ropes, is $$\left(\frac{1}{2n-1}\right)1 + \left(1-\frac{1}{2n-1}\right)0 = \frac{1}{2n-1}.$$
After this, you start over with $n-1$ ropes. Since what happens in the first step and later are independent (and expectation is linear anyway), the expected number of loops is $$ \frac{1}{2n-1} + \frac{1}{2n-3} + \dots + \frac{1}{3} + 1 = H_{2n} - \frac{H_n}{2}$$
In particular, for $n=100$, the answer is roughly $3.28$, which come to think of it seems surprisingly small for the number of loops!
Ok, I think you are right now.
Identify procedure by labeling each string end from 1 to 200. Let's say (1,2) identifies string 1, (3,4) identifies string 2, etc, (2k-1,2k) identifies string k.
We can think of this procedure as generating a permutation over 200 elements, and pairing numbers.
Specifically, generate a random ordering of the numbers from 1 to 200,
and processing the order from left to right, pair the numbers together,
indicating which ends are tied together.
To figure out the probability that string 1 is going to end up as a self-loop, we need the probability that in the permutation, (1,2) is paired
together.
So, count the number of ways (1,2) appears together in the following manner:
First, 1 can appear in one of 200 slots. Given the location of 1, 2 has to be in the right position (only 1 possibility) for the configuration to indicate that the ends are tied together. After this, the other slots do not matter, and there are 198! ways to generate those. Then the answer is 200*198! / 200! = 1/199.
Alternatively, we can follow the process of generating a random permutation, where we first place 1 in one of 200 slots, and then place 2 in one of the remaining 199 slots, and the rest of the process can be ignored. The probability that 2 lands in the right place to indicate ends 1 and 2 are tied together is 1/199.
The same reasoning works for any other string $(2k-1,2k)$.
Best Answer
I'll explain the thought process more explicitly. We can model random choice of tying with the following algorithm that ties knots one at a time:
Label the spaghetti strings $S_1, \dots, S_{50}$.
Begin with one end of $S_1$. Tie it to a random end of the $99$ remaining. If it makes a knot (there is only one end that does), then start again with any of the remaining spaghetti strings, for simplicity we can just choose $S_2$. If it doesn't make a knot, let's say we tie it to $S_i$. Now choose which end to tie the remaining end of $S_i$.
Notice that no matter whether or not we made a knot on the first tying, we have $97$ ends remaining and still only $1$ of them causes a knot. Hence, we can justify simply adding up the expectations from each step and the answer will end up just being $1/99 + 1/97 + \dots + 1$.