Call a consecutive difference" the absolute value of the difference between two consecutive rolls of a $6$ sided die. For example, the sequence of rolls $143511$ has the corresponding sequence of consecutive differences $31240$. What is the expected number of rolls until all $6$ consecutive differences have appeared?
I can see that there are $6$ possible consecutive differences: $0, 1,…,5$ and they appear among any pair of dice with probability $6/36, 10/36, 8/36, 6/36, 4/36, 2/36$ respectively. However, they could come in all sorts of orders.
I am happy with answers that are close approximations. By simulation the answer is around 25.8.
Best Answer
This can be solved using standard Markov chain methods – let $E(S, r)$ denote the expected rolls needed if the differences in $S$ have been seen and the current roll is $r$ – but an observation is needed to keep the computation time reasonable: the set of seen differences never shrinks, so we can compute expectations six at a time (per each possible $S$) working backwards from $E(\{1,2,3,4,5,6\},r)=0$.
The final answer is $1+(E(\varnothing,1)+E(\varnothing,2)+E(\varnothing,3))/3$ since $E(S,r)=E(S,7-r)$ by symmetry considerations, or $$\frac{672875275767847611958914137}{26031560987606728347794000}=25.848441\cdots$$