What is the probability that the accumulated sum is L at some point?

Question

Several days ago I was asked the following:

Question 1: Repeatedly roll a fair 6 sided dice. What is the probability that at some point, the accumulated sum of the rolls is 10000?

At first, I dismissed the problem as a simple exercise, but I still don't know the answer. Obviously, the number 10000 is not really relevent, so a more precise question is the following:

Question 2: Let $E_L$ be the event that at some point, the accumulated sum of the rolls is $L$. What is the value $\lim_{L \to \infty} \mathbb{P}(E_L)$?

I can solve the following related but easier question:

Question 3: Repeatedly flip a fair coin with faces 1 and 2. Let $F_L$ be the event that at some point, the accumulated sum of the flips is $L$. What is the value of $ \lim_{L \to \infty} \mathbb{P}(F_L)$?

In this case, the probability that at some point your accumulated total is $L$ is given by $$ \mathbb{P}(F_L) = \sum_{n=0}^{\infty} \frac{{L-n} \choose {L - 2n}}{2^{L-n}}$$ and evaluating this sum (using a computer) for large values of $L$ gives 2/3. I suspect a similar approach will work for the dice case, but writing down the correct sum if harder. This raises the following:

final question: Is there a more conceptual reason why $ \lim_{L \to \infty} \mathbb{P}(F_L) = 2/3$?

Answer 1

Since each number is hit at most once, the probability $p$ of hitting any particular number and the expected step size $E$ must satisfy $pE=1$. In the coin example, the expected step size is $\frac32$, so the probability is $\frac23$. In the die example, the expected step size is $\frac72$, so the probability is $\frac27$.

Answer 2

here is another solution following lulu's comment. The event $E_L^{\rm c}$ is the union of:

$E_{L-1}$ and then roll a 2,3,4,5 or 6

$E_{L-2}$ and then roll a 3,4,5 or 6

$E_{L-3}$ and then roll a 4,5 or 6

$E_{L-4}$ and then roll a 5 or 6

$E_{L-5}$ and then roll a 6

Moreover, these events are pairwise disjoint. Therefore the probability we care about satisfies the equation $$ 1-p = \frac{5}{6}p + \frac{4}{6} p + \frac{3}{6} p + \frac{2}{6} p + \frac{1}{6} p$$ solving gives $p = 6/21 = 2/7$.