Is this lot drawing fair? Sorry for a stupid question, but it is bugging me a lot. 
Let's say there are $30$ classmates in my class and one of us has to clean the classroom. No one wants to do that. So we decided to draw a lot - thirty pieces of paper in a hat, one of which is with "X" on it. The one who draws "X" has to do the cleaning. Each one starts to draw... 
Is this kind of lot drawing fair or not fair? It looks to me like the first one's chances to get an "X" are equal to $1/29$, while the second one's chances would be equal either to  $1/28$ (in case the first one didn't draw an "X") or zero $0/29 = 0$ (in case the first one drew an "X"). However, neither $1/28$, nor $0/29$ is equal to $1/29$.     

The second person's chance is still $1/30$. Indeed, Person 2 draws an X exactly when Person 1 does NOT draw an X, and so:
\begin{align*}
\Pr[P_2 = X]
&= \Pr[P_1 \neq X] \cdot \Pr[P_2 = X \mid P_1 \neq X] \\
&= \frac{29}{30} \cdot \frac{1}{29} \\
&= \frac{1}{30}
\end{align*}
Likewise, Person 3 draws an X exactly when the previous two people do NOT draw an X:
\begin{align*}
\Pr[P_3 = X]
&= \Pr[P_1 \neq X] \cdot \Pr[P_2 \neq X \mid P_1 \neq X] \cdot \Pr[P_3 = X \mid P_2 \neq X] \\
&= \frac{29}{30} \cdot \frac{28}{29} \cdot \frac{1}{28} \\
&= \frac{1}{30}
\end{align*}
This continues all the way up to Person 29 and Person 30:
$$
\Pr[P_{29} = X] = \frac{29}{30} \cdot \frac{28}{29} \cdot \frac{27}{28} \cdot \ldots \cdot \frac{3}{4} \cdot \frac{2}{3} \cdot \frac{1}{2} \\
\Pr[P_{30} = X] = \frac{29}{30} \cdot \frac{28}{29} \cdot \frac{27}{28} \cdot \ldots \cdot \frac{3}{4} \cdot \frac{2}{3} \cdot \frac{1}{2} \cdot \frac{1}{1} = \frac{1}{30}
$$
