I am trying to understand why this solution to a specific case of the hat matching problem is incorrect.
The problem goes like this: $n$ professors each place their hat for a total $n$ hats into a bag. The professors each randomly select a hat. What is the probability that none of the professors receive their own hat?
Can we not interpret each draw by a professor as a single independent event? As in, the probability we have for the first professor drawing a hat other than his hat is $\frac{n-1}{n}$. Continuing on with successive independent events where there is one less hat and one less professor leads to the probability that no professor draws their own hat: $P(\text{no match}) = \frac{n-1}{n} * \frac{n-2}{n-1} * \frac{n-3}{n-2} * \dots * \frac{1}{2} * 1$. The reasoning for the last item being that there is only one hat left for the last professor to draw.
Can someone explain where my logic is flawed?