Correctly labelling people in a lineup

Question

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

Answer 1

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 $n \cdot \frac 1n=1.$

Answer 2

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.