I throw a die $N$ times and the results are observed to be a monotonic sequence. What is probability that all 6 numbers occur in the sequence?
I'm having trouble with this. There are two cases: when the first number is 1, and when the first number is 6. By symmetry, we can just consider one of them and double the answer at the end. I've looked at individual cases of $N$, and have that
For $ N = 6 $, the probability is $ \left(\frac{1}{6}\right)^2 \frac{1}{5!} $.
For $ N = 7 $, the probability is $ \left(\frac{1}{6}\right)^2 \frac{1}{5!}\left(\frac{1}{6} + \frac{1}{5} + \frac{1}{4} + \frac{1}{3} + \frac{1}{2} + 1\right) $.
I'm not sure if the above are correct. When it comes to $ N = 8 $, there are many more cases to consider. I'm worried I may be approaching this the wrong way.
I've also thought about calculating the probability that a number doesn't occur in the sequence, but that doesn't look to be any easier.
Any hints/corrections would be greatly appreciated. Thanks