An abstract animal lives in pairs and trios.

In a forest, a pair and a trio live. Every day, a new animal arrives in the forest and randomly selects one of the forest's inhabitants. If the selected animal belongs to a trio, the group splits into two pairs; if it belongs to a pair, they transform into a trio. What is the probability that the $n$-th arriving animal joins a pair?
Suppose that before the arrival of the $n$-th animal, $K_{n}$ animals in the forest live in pairs. Since there are now a total of $(n+4)$ animals living in the forest, the $n$-th animal joins a pair with a probability of $K_{n} /(n+4)$. In this case, before the arrival of the $(n+1)$-th animal, only $\left(K_{n}-2\right)$ animals will live in pairs, and if the $n$-th animal joins a group of three, then $\left(K_{n}+4\right)$. Therefore, if $K_{n}$ is given, the conditional expected value of $K_{n+1}$ is

$$
k_{n+1}=\left(K_{n}-2\right) \frac{K_{n}}{n+4}+\left(K_{n}+4\right)\left(1-\frac{K_{n}}{n+4}\right)
$$

Divide both sides by the new number of animals:

$$
\frac{k_{n+1}}{n+5}=\frac{n-2}{n+5} \frac{K_{n}}{n+4}+\frac{4}{n+5}
$$

On the left side, we have the conditional probability that the $(n+1)$-th animal joins a pair, and on the right side, $\frac{K_{n}}{n+4}$ represents the same for the $n$-th animal. If we take the expected value of both sides, we get a recursion for the desired probability:

$$
p_{n+1}=\frac{n-2}{n+5} p_{n}+\frac{4}{n+5}
$$

Based on this, starting from the probability $p_{1}=\frac{2}{5}$, we get the values $p_{2}=\frac{3}{5}, p_{3}=\frac{4}{7}, p_{4}=\frac{4}{7}$, and every subsequent term will be $\frac{4}{7}$, since if $p_{n}=\frac{4}{7}$, then

$$
p_{n+1}=\frac{n-2}{n+5} \cdot \frac{4}{7}+\frac{4}{n+5}=\frac{4}{n+5}\left(\frac{n-2}{7}+1\right)=\frac{4}{7}
$$