TITLE: Throw a die $N$ times, observe results are a monotonic sequence. What is probability that all 6 numbers occur in the sequence?
QUESTION [5 upvotes]: 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

REPLY [2 votes]: I have a slightly different answer to the above, comments are very welcome :)
The number of monotonic sequences we can observe when we throw a dice $N$ times is 2$ N+5\choose5$-$6\choose1$ since the six sequences which consist of the same number repeatedly are counted as both increasing and decreasing (i.e. we have counted them twice so need to subtract 6 to take account of this).
The number of increasing sequences involving all six numbers is $N-1\choose5$ (as has already been explained). Similarly the number of decreasing sequences involving all six numbers is also $N-1\choose5$. 
Therefore I believe that the probability of all seeing all six numbers given a monotonic sequence is
2$ N+5\choose5$-$6\choose1$ divided by 2$N-1\choose5$.
This is only slightly different to the above answers but if anyone has any comments as to whether you agree or disagree with my logic or if you require further explanation I'd be interested to hear from you.