Correctly labelling people in a lineup I've been watching video's like this* where someone is asked to label a group of people. All people in a line have a label that is distinct from all the others in the line. The labeller receives $n$ signs with labels and gives all the people in the lineup one of those signs. For example: in the video below, there are five people in line. One from each of the following countries: Japan, Korea, Thailand, Vietnam and China. The labeller tries to guess where the individuals in the lineup come from.  
In an attempt to understand how good the labellers are at labelling, I wanted to compute the expected number of correctly labeled individuals. Intuitively I would say 1, and in fact I was able to informally confirm this value with a computational approach. Let's rename the labels by choosing the following labels: $\{1,2,\ldots,n\}$. Without loss of generality I can claim $(1,2,\ldots,n)$ is the correct labelling of the lineup. Then I check the number of correctly labelled individuals by comparing $(1,2,\ldots,n)$ with all permutations of $\{1,2,\ldots,n\}$. I found that the expected number of assignments is indeed 1, regardless of $n$. However, although this value makes some sense, I'm not very happy with this computational approach and I'd rather see a more rigorous approach. 
Therefore my question is: how can I formally derive (and also intuitively understand) the expected number of correct assignments by the labeller?
*https://www.youtube.com/watch?v=POyEK5kb1OU&index=3&list=PLJic7bfGlo3qJcIXUJteaUm_3-3tgQSXw
 A: The linearity of expectation justifies this.  Among all permutations, the chance that the first person is correctly identified is $\frac 1n$.  By symmetry, the chance each person is identified correctly is also $\frac 1n$.  The expected number correct is therefore $n \cdot \frac 1n=1.$
A: Permutations by cycles with fixed points marked have species equation
$$\mathfrak{P}(\mathcal{U}\mathfrak{C}_{=1}(\mathcal{Z})
+ \mathfrak{C}_{=2}(\mathcal{Z})
+ \mathfrak{C}_{=3}(\mathcal{Z})
+ \mathfrak{C}_{=4}(\mathcal{Z})
+ \cdots)$$
and hence the mixed EGF of permutations under this scenario is
$$G(z, u) = \exp\left(uz - z + \sum_{q\ge 1} \frac{z^q}{q}\right)
= \exp(uz-z) \frac{1}{1-z}.$$
Differentiate to get
$$\frac{\partial}{\partial u} G(z, u)
= z \exp(uz) \exp(-z) \frac{1}{1-z}.$$
Set $u=1$ to get
$$\left. \frac{\partial}{\partial u} G(z, u) \right|_{u=1}
= \frac{z}{1-z}.$$
These   two  operations   will  turn   a  term   $u^q  \frac{z^n}{n!}$
representing a  permutation of  $n$ elements  having $q$  fixed points
into $q\frac{z^n}{n!},$  and the  latter generating function  now sums
the count of all fixed points, attaching them to $\frac{z^n}{n!}.$
Hence the expected number of fixed points is
$$[z^n] \frac{z}{1-z} = 1$$
confirming the  observation by OP.  We did  not multiply by  $n!$ when
extracting  the coefficient  because we  need to  divide by  the total
number of permutations, which is $n!$, to get the expectation.
