In an urn, there are 9 slips numbered from 1 to 9. We draw one slip, note the number, and put the slip back into the urn. We repeat this process 8 more times, writing the drawn single-digit numbers from left to right next to each other. What is the probability that in the resulting 9-digit number, at least one digit will be in its correct position? (Correct position: counting from left to right, the $n$-th digit is in the $n$-th position).
Let's calculate the opposite probability, i.e., that no digit lands in its own place. Each draw has a $\frac{8}{9}$ probability of not drawing the correct slip. According to the multiplication rule, the probability of drawing an incorrect slip 9 times is $\left(\frac{8}{9}\right)^{9}$, and thus the sought probability that at least one digit lands in its own place is

$$
v=1-\left(\frac{8}{9}\right)^{9}
$$

Using logarithm tables,

$$
\left(\frac{8}{9}\right)^{9}=0.3468
$$

and thus

$$
v \approx 0.653
$$

Péter Theisz (Kaposvár, Táncsics g. IV. o. t.)