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 will assume for this answer that there are $N$ entirely unattached pieces of rope in the bucket, and that a loop may consist of any number of pieces of rope attached in a closed chain.
We can write a recurrence for this expected value, as follows. Suppose the expected number of loops for $N-1$ pieces of rope is denoted $L(N-1)$. Consider the bucket of $N$ pieces of rope; there are thus $2N$ rope ends.
Pick an end of rope. Of the remaining $2N-1$ ends of rope, only one end creates a loop—the other end of the same piece of rope; there are then $N-1$ untied pieces of rope. The rest of the time, two separate pieces of rope are tied together, and there are effectively $N-1$ untied pieces of rope. The recurrence is therefore
$$ L(N) = \frac{1}{2N-1}+L(N-1) $$
Clearly, $L(1) = 1$, so
$$ L(N) = \sum_{k=1}^N \frac{1}{2k-1} = H_{2N}-\frac{H_N}{2} $$
where $H_k$ is the $k$th harmonic number.
ETA: Since $H_k \doteq \gamma+\ln k$ for large-ish $k$, where $\gamma \doteq 0.57722$ is the Euler-Mascheroni constant, we have
$$ L(N) \doteq \ln 2N - \frac{\ln N}{2} = \ln 2\sqrt{N} $$