Problem source

A team of $m$ players, numbered from $1$ to $m$, 
puts on a set of a $m$ shirts, similarly numbered from $1$ to $m$. The players change in a hurry, so that the shirts are assigned to them randomly, 
one to each player.
Let $C_i$ be the random variable that takes the value $1$ if player $i$ is wearing shirt $i$, and 0 otherwise. Show that $\mathrm{E}\left(C_1\right)={1 \over m}$ 
and find $\var \left(C_1\right)$ and $\mathrm{Cov}\left(C_1 \, , \; C_2 \right) \,$. Let $\, N = C_1 + C_2 + \cdots + C_m \,$ be the random variable whose value is the number of players who are wearing the correct shirt. 
Show that $\mathrm{E}\left(N\right)= \var \left(N\right) = 1 \,$. Explain why a Normal approximation to $N$ is not likely to be appropriate for any $m$, but that a Poisson approximation might be reasonable. In the case $m = 4$, find, by listing equally likely possibilities or otherwise, the probability that no player is wearing the correct shirt and verify that an appropriate Poisson approximation to $N$ gives this probability with a relative error  of about $2\%$. [Use $\e \approx 2\frac{72}{100} \,$.]

Solution source

There are $m!$ different ways of assigning the shirts, and in $(m-1)!$ of them player $1$ gets their own shirt, ie $\mathbb{E}(C_1) = \mathbb{P}(\text{player }1\text{ gets own shirt}) = \frac{(m-1)!}{m!} = \frac{1}{m}$. $\var(C_1) = \mathbb{E}(C_1^2) - [\mathbb{E}(C_1)]^2 = \frac{1}{m} - \frac{1}{m^2} = \frac{m-1}{m^2}$.

If we have two players, there are $(m-2)!$ ways they both get their own shirts, therefore $\textrm{Cov}(C_1,C_2) = \mathbb{E}(C_1C_2) - \mathbb{E}(C_1)\mathbb{E}(C_2) = \frac{(m-2)!}{m!} - \frac{1}{m^2} = \frac{1}{m(m-1)} - \frac{1}{m^2} = \frac{m-m+1}{m^2(m-1)} = \frac{1}{m^2(m-1)}$.

\begin{align*}
\mathbb{E}(N) &= \mathbb{E}(C_1 + C_2 + \cdots + C_m) \\
&= \mathbb{E}(C_1) +  \mathbb{E}(C_2)  + \cdots +  \mathbb{E}(C_m)  \\
&= \frac{1}{m} + \frac{1}{m} +\cdots+ \frac1m \\
&= 1 \\
\\
\var(N) &= \sum_{r=1}^m \var(C_r) + 2\sum_{r=1}^{m-1} \sum_{s=2}^{m} \textrm{Cov}(C_r,C_s) \\
&= m \frac{m-1}{m^2} + 2 \frac{m(m-1)}{2}\frac{1}{m^2(m-1)} \\
&=\frac{m-1}{m} + \frac{1}{m} \\
&= 1
\end{align*} 

If we were to take a normal approximation, we would want to take $N(1,1)$, but this would say things like $-1$ is as likely as $3$ shirts being correct, which is clearly a bad model.

A Poisson is much more likely to be a sensible model as they have the same mean and variance as the parameter, and if $m$ is large, the covariance between shirts is going to be very small, so it will appear similar to random events occurring.

We can have

\begin{align*}
BADC \\
BCDA \\
BDAC \\
CADB \\
CDAB\\
CDBA \\
DABC\\
DCAB \\
DCBA
\end{align*}

Ie $\frac{9}{24}$ ways to have no player wearing their own shirt with $4$ players.

$Po(1)$ would say this probability is $e^{-1}$, giving a relative error of:

\begin{align*}
\frac{e^{-1}-\frac{9}{24}}{\frac9{24}} &\approx \frac{\frac{100}{272} - \frac{9}{24}}{\frac9{24}} \\
&= -\frac{1}{51} \\
&\approx -2\%
\end{align*}