A fair die is rolled $N > 6$ times. What is the probability the last 6 rolls were exactly $1,2,3,4,5,6$? You rolls a fair die $N > 6$ times and you want to rolls the sequence $1,2,3,4,5,6$ in this order.
What is the probability that the last 6 rolls were (in consecutive order) $1,2,3,4,5,6$, (So, on the Nth roll you get a 6, on the $(N-1)$th roll you get a 5, etc.), AND you did not roll this sequence any times before the last 6 rolls.
If I state another way, what is the probability that you roll the sequence 1,2,3,4,5,6 starting from the $(N-5)$th roll and you did not roll this sequence starting from any roll before the $(N-5)$th roll?
This is not a homework problems, just something I'm thinking about.

For seven rolls the last six being $1,2,3,4,5,6$ rules out having $1,2,3,4,5,6$ before that, so the chance you get the first $1,2,3,4,5,6$ starting on roll $2$ is $1/6^6$.  For any number of rolls below $12$ the odds of ending with the first $1,2,3,4,5,6$ are $1/6^6$.  It is only with $12$ or more rolls that you can get $1,2,3,4,5,6$ before the end, so for $N\ge 12$ the probability is $P($no $1,2,3,4,5,6$ in the first $N-6$ rolls)$1/6^6$  Roughly speaking the chance of not getting $1,2,3,4,5,6$ in $M$ rolls is $(1-1/6^6)^{(M-5)}$ (this assumes independence, which is not quite true), so an approximate answer for the chance of getting $1,2,3,4,5,6$ for the first time at the end of $N$ rolls is $$\begin {cases} 0&N\lt 6\\1/6^6& 6\le N \lt 12 \\(1-1/6^6)^{N-11}/6^6&N \ge 12 \end {cases}$$
