Suppose each person is a vertex and every hand is an edge, and we can also assume that every right hand holds someone else's left hand. If the right hand is a directed edge outwards of the vertex (left hand inwards), we can see that this problem is equal to the expected number of cycles in a permutation, which is easier to google.

As Noam527 correctly pointed out, this is equivalent to having random permutation. If you focus on cycle containing one you can easily see that for every k=1,..., n there is $$$\frac{1}{n}$$$ probability it has length $$$k$$$. From that you can easily derive that expected number of cycle is $$$H_n = \frac{1}{1} + \frac{1}{2} + \ldots + \frac{1}{n}$$$.

In such team-building exercises, there is normally a check to make sure there are no cycles of size 2 or 3 before starting (usually to increase the difficulty). How would we tackle the problem from here?