# Intuitive thinking of fair die rolling P(n) = 1 + … +n = rolling number

I am solving one question like this

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 pn 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).

the question is
Give an intuitive explanation for the fact that p(n) <- 1/3.5 = 2/7 as n = infinity.

the explanation is below,

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

-

I explain this informal argument slightly differently. Let $p(i)$ be the probability that the running total equals $i$ at some point.

1. 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 $$\mathbb{E}[e_1 + \dots + e_n] = \mathbb{E}[e_1] + \dots + \mathbb{E}[e_n] = \mathrm{Pr}[{\cal E}_1] + \dots + \mathrm{Pr}[{\cal E}_n] = p(1) + \dots p(n).$$
2. Since the average increment is $7/2$, we have that $k \approx \frac{2}{7} n$.
3. 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, $p(n) \approx 2/7$.

-
That's quite interesting, but i don't get the idea of k=p(1)+⋯+p(n), since i think P(.) is probability, it should multiply the corresponding value to become expectation, is that right? – user1489975 Jul 15 '12 at 5:26
I think i need more example for this, since now it seems i fell into mess and still don't get the idea... – user1489975 Jul 15 '12 at 5:27
I added an explanation why k = p(1) + ... + p(n). – Yury Jul 15 '12 at 5:52
the indicator variable is very nice, thanks for your explanation! Can u also help me identify the 2nd point? – user1489975 Jul 15 '12 at 6:04
After each step, the sum increases by 7/2 in expectation. Thus after r steps it increases by 7r/2 in expectation. Moreover, the sum is “very close” to 7r/2 with high probability (by the Law of Large Numbers or the Central Limit Theorem). Therefore, the sum reaches n after approximately 2n/7 steps. – Yury Jul 15 '12 at 6:17

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.

-
"So, you expect to get 1000 numbers on the way to the sum 3500." I got the idea that i expect there are 1000 numbers (each of them could be 1 to 6) and sum them up to be 3500. Well, getting 1000 numbers out of 3500 is getting, on average, 2 numbers out of every 7. – user1489975 Jul 15 '12 at 14:35
but i still can't get the next sentence that 1000 numbers out of 3500 .... – user1489975 Jul 15 '12 at 14:45
Let me rephrase it: you expect to land on 1000 numbers on the way to 3500. So on average you land on two-sevenths of the numbers; two numbers out of every seven. Is that better? – Gerry Myerson Jul 16 '12 at 10:47
Sorry that i kept asking the same idea, when we expect to land on 1000 numbers on the way to 3500, does that mean i expect throw 1000 times dice and get the sum 3500? – user1489975 Jul 17 '12 at 13:14
If you throw the die 1000 times, you expect to get a sum pretty near 3500. If you throw it 10,000 times, you expect a sum near 35,000. If you throw it 1,000,000 times, you expect a sum near 3,500,000. Wait. let's try this: say you throw it 600 times. You expect to get about 100 1s, 100 2s, ..., 100 6s, and those numbers add up to $100+200+\cdots+600=2100$, and 600 is to 2100 as 2 is to 7. If you roll the die $Q$ times, you expect a sum near $(7/2)Q$; so, to get the sum $n$, you expect to roll the die about $(2/7)n$ times (and thus land on about $(2/7)n$ numbers). – Gerry Myerson Jul 17 '12 at 13:23

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 $(\frac{1}{3.5}=\frac{2}{7})$.

-