In an urn, there are 9 slips numbered from 1 to 9. We draw all 9 slips one by one and place them side by side from left to right. What is the probability that at least one slip does not end up in its proper place? (Proper place: $n$ - from left to right - is the $n$-th position).
Solution: The number of possible cases is equal to the number of permutations of 9 elements, that is, 9!. Except for the single group 123456789, every permutation group is favorable, so the number of favorable cases is 9! - 1, and thus the sought probability is

$$
v=\frac{9!-1}{9!}=\frac{362879}{362880} \approx 1
$$

The sought probability is therefore very close to one, and in everyday life, we would say that it is "almost certain."

Szendrei István (Kunszentmiklós, Damjanich g. II.o.t.)