Probability of last 3 digits matching in 9 digit student ID Here is the scenario:
There are a number of students (n), each with a random 9-digit integer student ID number (including zeros). At what n-value would there be a 0.5 probability at least two students having the same "last 3 digits" (identical integers in identical order)?
I am not sure how to set this up.  
 A: As MikeP points out in the comments, this problem is an instance of the famous Birthday problem.  There are $10^3 = 1000$ different combinations of the last-three digits that can occur in this case, and for simplicity we will assume these are equiprobable.  Let $\mathcal{S}_n$ be the event that at least two of the $n$ students share these last three digits on their cards.  Then we have:$^\dagger$
$$\mathbb{P}(\mathcal{S}_n) = 1 - \mathbb{P}(\bar{\mathcal{S}}_n) = 1 - \frac{(1000)_n}{1000^n}.$$
(Note that if $n > 1000$ then $(1000)_n = 0$ so $\mathbb{P}(\mathcal{S}_n) = 1$ as we would expect.  In this case there are more students than number-combinations, so at least two student must share their number-combination.)  It is quite simple to program a vectorised function in R to calculate values of these probabilities for a series of values of $n$.  This is accomplished by the following code:
#Create function to calculate probability for any n
PROBS <- function(n) { T <- rep(1,n);
                       for (i in 1:(n-1)) { T[i+1] <- T[i]*((1000-i)/1000); }
                       1-T; }

We can plot these probabilities for different values of $n$ to get a sense of how this probability increases when we have more students.
library(ggplot2);
DATA   <- data.frame(n = 1:100, P = PROBS(100));
FIGURE <- ggplot(data = DATA, aes(x = n, y = P)) +
          geom_line(size = 1) +
          theme(plot.title    = element_text(hjust = 0.5, size = 14, face = 'bold')) +
          ggtitle('Probability of sharing digits on student card') +
          xlab('Number of students') +
          ylab('Probability');
FIGURE;



$^\dagger$ The values $(1000)_n = \prod_{i=1}^n (1000-i+1)$ are the falling factorial.
