TITLE: A fair die is rolled $N > 6$ times. What is the probability the last 6 rolls were exactly $1,2,3,4,5,6$?
QUESTION [7 upvotes]: 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.

REPLY [2 votes]: For $n\ge 6, 0\le r\le 6$, let $p_r(n)$ be the probability that $n$ rolls end in $1,2,\ldots,r$ and $1,2,3,4,5,6$ never occurs  in the sequence. Then $p(n)=p_0(n)+\ldots+p_6(n)$ denotes the probability that $1,2,3,4,5,6$ never occurs within $n$ rolls.
Clearly $p_r(6)=1$ for $r<6$ and $p_6(n)=0$ for all $n\ge 6$. Otherwise we have the following recursions:
$$ p_r(n+1)= \begin{cases}0&\text{if }r=6\\\frac16p_{r-1}(n)&\text{if }2\le r\le 5\\
\frac16(p_0(n)+\ldots+p_6(n))&\text{if }r=1\\
\frac56(p_0(n)+p_6(n))+\frac23(p_1(n)+\ldots+p_5(n))&\text{if }r=0\end{cases}$$
This allows some manual computations for small $n$. For the general solution this means that we have to find the eigenvalues and eigenvectors of the matrix
$$\frac16\begin{pmatrix}5&4&4&4&4&4\\
1&1&1&1&1&1\\
0&1&0&0&0&0\\
0&0&1&0&0&0\\
0&0&0&1&0&0\\
0&0&0&0&1&0\\
\end{pmatrix} $$
to find an explicit formula for the $p_r(n)$ and finally for $p(n)$ of the form
$$\tag1p(n)=c_1\lambda_1^n+\ldots+c_6\lambda_6^n.$$ 
The answer to your question is then finally
$$\tag2P= 6^{-6}\cdot p(N-6).$$
If my eigenvalue computations are right, we have
$$\begin{align} \lambda_1&\approx\hphantom{-}0.999978564232\\
\lambda_2&\approx \hphantom{-}0.119473055129\\
\lambda_{3,4}&\approx \hphantom{-}0.033606052210\pm 0.112188066230\,i\\
\lambda_{5,6}&\approx-0.093331861891\pm 0.066102276039\,i\end{align}$$
and I leave the determination of the $c_\nu$ as an exercise.
For large $N$, the $c_1\lambda_1^n$ term in $(1)$ becomes dominant so that $(2)$ becomes
$$ \tag{2'}P\approx \alpha\cdot \lambda_1^N $$
with $\alpha = 6^{-6}c_1\lambda_1^{-6}$.

Remark: In the light of Ross Millikan's answer, note that $1-6^{-6}\approx 0.9999785665$. At first glance, the difference  to $\lambda_1$ might be  an artefact of my numeric calculation (PARI/GP with standard precision). But redoing my calculation with higher precision suggests that the difference ($\approx 2.297\cdot 10^{-9}$) is real.