MHB Probability of Rolling Sum of Die-Rolling

[magneto1]

I ran into this problem, and would like to see if there is something more elegant.

Suppose we have a sequence $a_1, a_2, \dotsc, a_n, \dotsc$ where $a_k$
is the (running) sum of rolling a standard 6-side die $k$ times.

E.g. What's the chance of saying the number $2$ appears in this sequence? There are two possibilities, rolling a $2$ on the first roll, so $a_1 = 2$, or rolling two $1$'s, so $a_1 = 1, a_2 = 2$. Therefore, the probability is $\frac 16 + \frac 1{36} = \frac 7{36}$.

The question now is what is more likely to occur in this sequence? The number $6$ or the number $1006$?

There are formula to determine the probability of getting a sum of $t$ after $n$ rolls of a $6$-sided die:
\[
p = \frac 1{6^n} \sum_{k=0}^n (-1)^k \binom{n}{k} \binom{t - 6t - 1}{n-1}.
\]
We can further reduce the upper limit of the summation from $n$, and utilize code/Mathematica to evaluate this. If the code is carried out, I find that it is more likely you will see a $6$ than seeing a $2$ or $1006$.

Is there a "cleaner" way to show this without doing the actual arithemtics?

 

[Opalg]

I ran into this problem, and would like to see if there is something more elegant.

Suppose we have a sequence $a_1, a_2, \dotsc, a_n, \dotsc$ where $a_k$
is the (running) sum of rolling a standard 6-side die $k$ times.

E.g. What's the chance of saying the number $2$ appears in this sequence? There are two possibilities, rolling a $2$ on the first roll, so $a_1 = 2$, or rolling two $1$'s, so $a_1 = 1, a_2 = 2$. Therefore, the probability is $\frac 16 + \frac 1{36} = \frac 7{36}$.

The question now is what is more likely to occur in this sequence? The number $6$ or the number $1006$?

There are formula to determine the probability of getting a sum of $t$ after $n$ rolls of a $6$-sided die:
\[
p = \frac 1{6^n} \sum_{k=0}^n (-1)^k \binom{n}{k} \binom{t - 6t - 1}{n-1}.
\]
We can further reduce the upper limit of the summation from $n$, and utilize code/Mathematica to evaluate this. If the code is carried out, I find that it is more likely you will see a $6$ than seeing a $2$ or $1006$.

Is there a "cleaner" way to show this without doing the actual arithemtics?

Let $p(n)$ denote the probability that the number $n$ occurs in the sequence $\{a_k\}.$ I doubt whether there is a "clean" way to find $p(n).$ The mean score for a roll of the die is $\frac72$. So the average gap between consecutive terms in the sequence $\{a_k\}$ is $\frac72$. It follows that the asymptotic probability of a number occurring in the sequence is $\frac27 \approx 0.2857$. In other words, $p(n) \to \frac27$ as $n\to\infty.$ For small values of $n$, you can calculate $p(n)$ fairly easily. In fact, for $1\leqslant n\leqslant6$ the formula is $p(n) = \dfrac{7^{n-1}}{6^n}.$ When $n=7$ it is no longer possible to reach that score with one roll of the die, and the probability drops quite substantially (which I found quite surprising and counter-intuitive). After that, it gets rapidly more complicated to calculate $p(n).$ For $1\leqslant n\leqslant12$ the values of $p(n)$ (rounded to four decimal places) are $$\begin{array}{r|cccccccccccccc} n&1&2&3&4&5& 6&7&8&9&10 &11&12&\ldots & \infty \\ \hline p(n) & .1667 & .1944 & .2269 & .2647 & .3088 & .3602 & .2536 & .2681 & .2804 & .2893 & .2934 & .2908 && .2857 \end{array}$$ It looks as though $p(n)$ is going to converge quite rapidly to its asymptotic value $\frac27$, so I guess that $p(1006)$ will be very close to $.2857$, and will certainly be less than $p(6).$