At a banquet, there are $n$ people who shake hands according to the following process: In each round, two idle hands are randomly selected and shaken (these two hands are no longer idle). After $n$ rounds, there will be no idle hands left, and the $n$ people will form several cycles. For example, when $n=3$, the following situation may occur: the left and right hands of the first person are held together, the left hand of the second person and the right hand of the third person are held together, and the right hand of the second person and the left hand of the third person are held together. In this case, three people form two cycles.
My idea is to use the linearity of expectation to solve, & consider the probability of a certain $k$ individual forming a circle, which is the expectation of the corresponding indicator, my idea is that the likelihood of a $k$ individual forming a circle is $\frac{k!2^k \frac{(2n-2k)!}{2^{n-k}}}{\frac{(2n)!}{2^n}}$, but proceeding this way gives me a very complicated equation, and I'd like to know if it makes sense for me to calculate it this way, and ask if there is a more elegant way.
