Say, for the sake of arguement, I made of a random list of size n, consisting of s number of unique elements. What is the probability of consecutive elements, when I pick r at a time?
For example, I have a random list of size 1400, with elements consisting of integers 1-42. When I picked 40 items at a time, what is their probability that they're consecutive integers?
Let's first consider the probability for one particular sequence of r values. Each picked element has to fall somewhere in the sequence and cannot be repeated. So if we consider the picks one at a time, the first pick has probability $\frac{r}{s}$. The next element has to be something in the sequence of length r that hasn't been picked yet, so the probability is $\frac{r-1}{s}$, then $\frac{r-2}{s}, \frac{r-3}{s}$, ... This leads to the general form of $\prod_0^{r-1}\frac{r-i}{s} = \frac{r!}{s^r}$.
But there isn't just one sequence of length r. There are $s-r+1$ possible sequences. If we multiply that by the probability for one individual sequence, we've covered the full probability space of all possible consecutive sequences. So the general form answer becomes $(s-r+1)\cdot\frac{r!}{s^r}$
Using your example, P(consecutive) = $(42-40+1)\cdot\frac{40!}{42^{40}} = 2.876\times10^{-17}$.
Note that the list size (n) does not appear anywhere in the answer because this solution assumes that the distribution of possible values is uniform for each element in the list, which is independent of the size of the list.