Throw 40 dice and arrange them in a row - check my proof 
Assume you throw $40$ dice and arrange them in a row so that you got a $40$-tuple. Then you throw antoher die. The result of this die indicates the number of the starting die of the $40$-tuple. E.g. you throw a $5$ then you start at the $5$-th die of the $40$-tuple. Then you traverse the $40$-tuple in steps determined by the die you are standing on. Let's say the $40$-tuple looks like this:
$$ (4, 4, 5, 1, 6, 6, 4, 2, 1, 1, 4, 6, ....., 2)$$
If you throw a $3$ you start at the die which shows a $5$, then you go 5 steps further to the die that shows a $2$ and from there you go $2$ steps further and so on ...
If you cannot complete the steps because you would exceed the $40$-tuple then you remain on the actual die and remove all the subseqeunt dice. Let's say you are on the die which shows the $6$
$$ (3,.., 2, 1, 4, \color{red}6, 4, 5, 5, 2, 1)$$
You cannot go $6$ steps further so you stay there and remove all subsequent dice. You get:
$$ (3,.., 2, 1, 4,\color{red}6)$$
So in this case you finally would get a $35$-tuple beacuse you have removed 5 dice.


The question is:
If I throw a die a second time to traverse the remaining tuple again how can I estimate the probability of exactly landing on the last die of the remaining tuple?
Note that if I once arrive a die which was already visited during the first run I will never leave the path and will automatically end on the last die.
My approach:
Let be $A$ the event of landing excatly on the last die and $B= A^c$ the complement. If I find an upper boundary $U$ for $P(B)$ I can then estimate $P(A)$ by $P(A)\geq 1-U$. So I need to figure out when $P(B)$ is highest.
If I stand somewhere in the tuple and I am about to take my next steps then the maximum probability of not finishing on a die which I have already visited is $\frac{5}{6}$. So $P(B)$ must be something like $\left(\frac{5}{6}\right)^k$. Where $k$ refers to each time I have finished my steps. The smaller $k$ is the greater becomes $\left(\frac{5}{6}\right)^k$. So $\left(\frac{5}{6}\right)^k$ is highest if I always end my steps on a die showing a $6$ beacuse then I traverse the tuple fastest. So after ending my steps six times on a $6$ I will at least have traversed the tuple. But If I had removed 5 dice after the first run I am even able to traverse the tuple after ending my steps five times on a $6$. So, $P(B)\leq\left(\frac{5}{6}\right)^5$. Hence, $P(A)\geq 1-\left(\frac{5}{6}\right)^5$.
What do you think, is this correct?
I have absoultey no idea how to tackle this problem in a rigorous way. May be you have different approaches. Any comments or suggestions are welcome!
 A: In case you dont know, you are describing the dice-based version of a card trick called Kruskal count.  Instead of $40$ dice you deal out a regular $52$ card deck, and instead of dice pips you use the numeric value of the cards (J/Q/K each counts as 5 in that article).
Anyway, I agree your bound is valid.  To lose (event $B$) you certainly have to miss "visited" dice at least $T \ge 5$ times, and every miss has probability $\le 5/6$.  So $P(B) \le (5/6)^T \le (5/6)^5$.  No real need to "assume" you land on a $6$ every round in the 2nd sequence.
A much better estimate is to use expected values, as discussed in the Kruskal Count article - it has an estimate based on expected total number of rounds and expected "density" of "visited" cards, estimated at ${1 \over 5.38}$.  While not a bound, it is a much better estimate.  In the case of cards the article estimates that overall win prob $\approx 0.826$.
Following the same techniques, and since the average dice roll is $3.5$:


*

*The average number of rounds $\approx 40 / 3.5 \approx 11.4 \approx 11$

*The "density" of "visited" dice $= 1/3.5$

*So the prob to lose, estimated as the prob of not landing on a visited dice all those $11$ times, $\approx (1 - {1 \over 3.5})^{11} \approx 2.47\%$ which is very close to the simulated result.
