I have four people (n) and every time each is getting a random three digit combination (000-999). This means there are 1000 possible combinations (k).
I want to know how probable it is for two or more people to get the same three digits in idk let's say 10,000 trys.
I found this but I calculated it with n=4 and k=1000 and it doesn't seem to be the right formula for my use case. How would I calculate something like that?
The probability is $$ p=1-\left[\frac{(k-1)(k-2)(k-3)}{k^3}\right]^N, $$ where $N$ is the number of trials.