This problem is already been discussed in several posts.
"You have $n=100$ noodles in your soup bowl. Being blindfolded, you are told to take two ends of some noodles (each end of any noodle has the same probability of being chosen) in your bowl and connect them. You continue until there are no free ends. The number of loops formed by the noodles this way is stochastic. Calculate the expected number of circles."
I know the standard solution for this problem, but I don't understand why my reasoning is not correct.
Let's reduce the difficulty of the problem and take $n=3$.
Let's compute how many ways we can generate three loops. In this case, we need to close each noodle with itself. Let's indicate with $(i,i+1)$ the ends of the $i-$th noodle. There are always two different ways to close each noodle, by picking first $i$ and then $i+1$, or vice versa. Therefore, we have $$\#(3loops)=2^3 \times 3 = 24 $$ possible cases to end up with three loops.
Let's compute how many ways we can generate two loops. There are $\left[2\times\binom{2n}{2}-n\right]$ combinations of two-ends that do not belong to the same noodle. In other words, there are $\left[2\times\binom{2n}{2}-n\right]$ possible ways of gluing two different noodles. The rest of the noodles are untouched, can we can close each of them in two different ways. Therefore, there are
$$
\#(2loops)=\left[2\times\binom{2n}{2}-n\right] 2^{n-1} (n-1) = 192
$$
possible ways of generating two loops, for $n=3$.
I can repeat the same argument now, for the case with one loop, getting
$$
\#(1loop)=\left[2\times\binom{2n}{2}-n\right] \left[2\times\binom{2(n-1)}{2}-(n-1)\right] 2^{n-2}(n-2) = 384
$$
Therefore,
Thefore, I was associating the following probabilities
$$p(1loop) =\frac{\#(1loop)}{\#(1loop)+\#(2loops)+\#(3loops)} = \frac{1}{25}\\
p(2loops) =\frac{\#(2loops)}{\#(1loop)+\#(2loops)+\#(3loops)}= \frac{8}{25}\\
p(3loops) =\frac{\#(3loops)}{\#(1loop)+\#(2loops)+\#(3loops)}= \frac{16}{25}$$
and computing the mean value as
$$
E[\text{number of loops}] = p(1loop) + 2 p(2loops) + 3 p(3loops) = \frac{7}{5}\,.
$$
The result is not correct, since it should be $E[\text{number of loops}] = 1 + 1/3 + … + 1/(2n-1) = 23/15$ for $n=3$, but I cannot understand where is the flaw in the argument.
Best Answer
You seem to be aiming for $(2n)!$ possible choices of the ends in order, so $720$ when $n=3$.
To get three loops you need to choose one of the six ends, then its other end to make a loop, then one of the other four ends, then its other end to make the second loop, then one of the last two ends, and then its other end to make the third loop, so $$6\times 1\times 4 \times 1 \times 2 \times 1 = 48$$ ways rather than your $24$. The probability of this is $\frac{48}{720}=\frac1{15}$.
To get two loops there are two possibilities: (a) you need to choose one of the six ends, then its other end to make a loop, then one of the other four ends, then an end on a different noodle to make a longer noodle, then one of the last two ends, and then its other end to make the second loop; or (b) you need to choose one of the six ends, then an end on a different noodle to make a longer noodle, then one of the other four ends, then its other end to make the first loop, then one of the last two ends, and then its other end to make the second loop, so $$6\times 1 \times 4 \times 2\times 2 \times 1 +6\times 4 \times 4 \times 1 \times 2 \times 1 = 288$$ ways rather than your $192$. The probability of this is $\frac{288}{720}=\frac25$.
To get one loop you need to choose one of the six ends, then an end on a different noodle to make a longer noodle, then one of the other four ends, then an end on a different noodle to make a longer noodle, then one of the last two ends, and then its other end to make the first loop, so $$6\times 4 \times 4 \times 2\times 2 \times 1 = 384$$ ways as you found. The probability of this is $\frac{384}{720}=\frac8{15}$.
So the expectation number of loops is $3 \times \frac1{15}+2\times \frac25 +1 \times \frac8{15} = \frac{23}{15}$ which is indeed equal to $1+\frac13+\frac15$.