What is the probability...? There are 5 people in a party and there will be exchanging gifts.
Now, they conducted a lottery writing their names.
What is the probability that they will get the paper with their names?
Initial Analysis:
Suppose we fix the order of draw, say $P_1 , P_2, ..., P_5$ letting $P_1$ be the first person to draw.
If the question would be what is the probability of $P_1$ to get the paper with his name, then the answer would be $20$%. 
What about $P_2$?
The probability depends on the outcome of $P_1$.
If $P_1$ got the paper except of $P_2$'s, then the probability is $25$%.
If $P_1$ got $P_2$'s name, then the probability will be $0$%. 
I can't go any further.
It seems complicated to me or I am just complicating the problem.
 A: The probability that the first person gets his own name is, as you say, $1/5$.
The same is true for each of the others:
It doesn't matter where a person comes in the order of drawing.
There are $5!=120$ possible orders in which the $5$ slips can be drawn. I claim that exactly one-fifth of those orders have $P_2$’s slip in the second position.
To see this, consider the $4!=24$ possible orders for the slips for $P_1,P_3,P_4$, and $P_5$.
You can insert $P_2$’s slip into one of these orders in $5$ possible places. For instance, starting with $$P_3,P_1,P_5,P_4$$ you can get any of these:
$$\begin{align*}
&\color{red}{P_2},P_3,P_1,P_5,P_4\\
&P_3,\color{red}{P_2},P_1,P_5,P_4\\
&P_3,P_1,\color{red}{P_2},P_5,P_4\\
&P_3,P_1,P_5,\color{red}{P_2},P_4\\
&P_3,P_1,P_5,P_4,\color{red}{P_2}
\end{align*}$$
Thus, each of the $24$ possible orders for the slips for $P_1,P_3,P_4$, and $P_5$ gives you one order of all five slips in which $P_2$’s slip is first, one in which it’s second, one in which it’s third, one in which it’s fourth, and one in which it’s last.
Consequently, $P_2$’s slip is equally likely to appear in any one of the five possible positions: for any position, there are $24$ orders in which it appears in that position.
In particular, it appears in the second position $24$ times out of $120$, or one time in five.
Similarly, $P_3$’s slip is equally likely to appear in each position, so the probability that it’s in the third position (so that $P_3$ gets it) is $1/5$, and $P_4$ and $P_5$ also have probability $1/5$ of drawing their own slips.
Added: The probability that at least one person gets his own slip is $$1-\mathbb{P}(\text{no one gets his own slip})\;.$$ Calculating the probability that no one gets his own slip is the classic problem of derangements.
It turns out that $44$ of the $120$ possible orders of drawing are derangements, i.e., orders in which no one gets his own slip, so the probability that you want is $\frac{120-44}{120}=\frac{76}{120}=\frac{19}{30}=0.63$
As the number of participants increases, the probability that at least one of them gets his own slip approaches $1-\frac1e\approx 0.63212\;.$
As you can see, the approximation is quite good already when $n=5$.
