challenge: Closing hash, linear probing, probe length 1
This problem was formulated and solved in 1962 by Donald Knuth; apparently, it was his first exercise in algorithmic analysis. It's often referred to as the "parking problem".

Knuth's analysis is beautifully simple. (Note: Knuth, and I think most computer scientists, used $n$ as the size of the hash table, and $k$ as the number of items inserted. So do I, even though the question is expressed differently.)

Let's start with the set $A$ of hash values of $k$ items, consisting of all sequences $\langle a_1, \ldots, a_k \rangle$, where $0 \le a_i &lt; n$. Clearly there are $n^k$ possible sequences.

Now, suppose we have a bug-free hash table of size $n+1$, and look the set $B$ of all possible hash sequences for $k$ items,  $\langle b_1, \ldots, b_k \rangle$ with $0 \le b_i &lt; n+1$. There are $(n+1)^k$ such sequences.

Of those sequences, we select the subset $B'$ for which the last table entry ($n$) is empty at the end of the insertion. Since once a slot is occupied, it remains occupied at the end, we know that for every sequence in $B'$, every $b_i &lt; n$. Therefore, every sequence in $B'$ is also in $A$.

Moreover, we know that in the course of the insertion, no probe passes through position $n$. In other words, every sequence in this subset of $B$ is a sequence in $A$ in which there is no wrap-around, which are precisely the sequences in $A$ which do not trigger the bug. 

So it will only remains to count these sequences. But that's trivial, because all of the slots in the (bug-free) hash table are identical. So the probability that a particular slot will be unoccupied after inserting $k$ items is $1 - {k \over {n+1}}$, with the result that there are $(n+1-k)(n+1)^{k-1}$ such sequences.

Thus, the probability that a random sequence of $k$ insertions into the buggy hash table will trigger the bug is $1 - {(n+1-k)(n+1)^{k-1}\over n^k}$.

By the way, the output of the program you paste-binned is not correct.  For example, it lists $4, 4, 4$ twice in the list for three insertions into a table of size 5.