Intuitively, why does $\lim_{n \to \infty} \frac16 (p_{n - 1} + p_{n - 2} ... + p_{n - 6}) = 2/7$? Blitzstein, Introduction to Probability (2019 2 edn), Chapter 2, Exercise 48, p 94.



*A fair die is rolled repeatedly, and a running total is kept (which is, at each time, the total of all the rolls up until that time). Let $p_n$ be the probability that the running total is ever exactly n (assume the die will always be rolled enough times so that the running total will eventually exceed n, but it may or may not ever equal n).

(c) Give an intuitive explanation for the fact that $p_n \rightarrow  1/3.5 = 2/7 \quad as \quad n \rightarrow \infty$.

From p 17 in the publicly downloadable PDF of curbed solutions.

(c)An intuitive explanation is as follows. The average number thrown by the die is (total of dots)/6, which is 21/6 = 7/2, so that every throw adds on an average of 7/2. We can therefore expect to land on 2 out of every 7 numbers, and the probability of landing on any particular number is 2/7.

That's the line I don't get it, why we can transfer
 A: I explain this informal argument slightly differently. Let $p(i)$ be the probability that the running total equals $i$ at some point. 


*

*Note that $k = p(1) + \dots + p(n)$ equals the expected number of values between $1$ and $n$ that the running sum takes (by the linearity of expectation). Specifically, let ${\cal E}_i$ be the event that the running sum takes value $i$ and $e_i$ be the indicator variable of the event ${\cal E}_i$ (that is, $e_i =1$ if ${\cal E}_i$ happens; $e_i=0$, otherwise). Then the expected number of values the running sum takes equals
$$\mathbb{E}[e_1 + \dots + e_n] = \mathbb{E}[e_1] + \dots + \mathbb{E}[e_n] = \mathrm{Pr}[{\cal E}_1] + \dots + \mathrm{Pr}[{\cal E}_n] = p(1) + \dots p(n).$$

*Since the average increment is $7/2$, we have that $k \approx \frac{2}{7} n$.

*On the other hand, intuitively all values $p(i)$ are approximately equal (this is not true for small values of $i$; but it is true for large values of $i$).


Thus $\frac{2}{7} n \approx p(1) + \dots p(n) \approx n p(n)$. Therefore, $p(n) \approx 2/7$.
A: Here's a way to think of it. Suppose you throw the die 1000 times. What sum will you get? Well, of course, it could be anything from 1000 to 6000, but you expect it to be pretty near 3500, right? So, you expect to get 1000 numbers on the way to the sum 3500. Well, getting 1000 numbers out of 3500 is getting, on average, 2 numbers out of every 7. Landing on 1000 numbers on the way to 3500 is landing on two-sevenths of the numbers. So the probability of landing on any particular number is 2/7.  
A: For the sake of simplicity, consider that you may start with any small random number. Then each step you increase by '1'. What is the probability that you will reach a specific big number? Intuitively, your answer should be 1.
Now, what if your increment is '$2$'? Again, the probability should be $(0.5=\frac{1}{2})$ to reach a specific large number. 
For our problem, it's found that on average we increase the total by (7/2=3.5). This should give intuitive impression that our probability to reach a large number is $(\frac{1}{3.5}=\frac{2}{7})$.
