A symmetrical die is thrown independently 7 times, what is the probability that the sum of the faces is 14?

Question

A 3 dice version has been asked and answered.

However it is much simpler to count when you only throw the dice 3 times.

I am interested in a solution using probability generating functions (PGF). (Would be nice to see a counting one too).

I got that the PGF of one throw is: $$\frac16\left(s+s^2+s^3+s^4+s^5+s^6\right)=\frac16s\left(1+s+s^2+s^3+s^4+s^5\right).$$ Hence the PGF of 7 independent throws is: $$\frac1{6^7}s^7\left(1+s+s^2+s^3+s^4+s^5\right)^7.$$ However I do not know where to take it from here.

Answer 1

Let $$f(s) = \frac{1}{6^7}s^7{(1+s+s^2+s^3+s^4+s^5)}^7 $$ We want to find the coefficient of $s^{14}$ when $f(s)$ is expanded. One way to do this is to notice that $1 + s + s^2 + \dots + s^5$ is the sum of a geometric series, for which we have a formula. $$\begin{align} f(s) &= \frac{1}{6^7} s^7 \left( \frac{1-s^6}{1-s} \right) ^7 \\ &= \frac{1}{6^7} s^7 \; (1-s^6)^7 \;(1-s)^{-7} \\ &= \frac{1}{6^7} s^7 \cdot \sum_{i=0}^7 (-1)^i \binom{7}{i} s^{6i} \cdot \sum_{j=0}^{\infty} \binom{7+j-1}{j} s^j \end{align}$$ by use of the Binomial Theorem.

We have a coefficient of $s^{14}$ in the product whenever $6i + j = 7$ with $0 \le i \le 7$ and $0 \le j$ in the last equation above. There are only two possibilities: $(i,j) = (0,7)$ and $(i,j) = (1,1)$.

Can you take it from there?

Answer 2

We could use the probability generating function here... However, let me say there is a much simpler way to solve this problem using what are called compositions. In fact, we only need to find the 7-compositions of 14 using the restricted numbers $1,2,3,4,5,6$. This happens to have a closed form solution. What this is asking, is that in exactly 7 sums, how can we make the number 14? One way is $(1,1,1,1,1,3,6)$. Another is $(1,1,1,6,1,1,3)$. Another is $(3,2,1,2,3,2,1)$. You will notice there are exactly $7$ elements and they add up to $14$. But for each of these they can be rearranged to with exactly the same numbers in a different order. https://en.wikipedia.org/wiki/Composition_(combinatorics)

So that is what a composition does. The only thing remaining is the probability, and well if we take 7 rolls, for each way we can get 14 in 7 goes, it is just $(\frac 16)^7$. Or rather there are $6^7$ outcomes when we roll the dice in 7 rolls.

You will find a closed-form solution here:

https://www.fq.math.ca/Scanned/14-5/abramson.pdf

Use equation (E) with $n=14$, $k=7$ and $w=6$.

$$\sum_{j=0}^{k}(-1)^j{k \choose j}{n-6j-1 \choose k-1}$$

$$=(-1)^0{7 \choose 0}{14-1 \choose 7-1}+(-1)^1{7 \choose 1}{14-6-1 \choose 7-1}+(-1)^2{7 \choose 2}{14-12-1 \choose 7-1}$$

$$={13 \choose 6}-7{7 \choose 6}=1667$$

So probability of getting 14 in 7 rolls, $P(T=14|R=7)=\frac{1667}{6^7}= 0.005954933$

Does that sound right?

Answer 3

Here is another simple solution using only basic knowledge in combinatorics:

We know that the number of possible outcomes of the seven throws of the die is $6^7$ . Now we need to find all outcomes for which the sum of the points, after the seven throws, is 14. It is obvious that each throw would give us at least 1 point. It remains to count the number of ways the other 7 points could be distributed to the throws. For this objective we will use slashes where each slash will separate points which would be distributed to one of the throws.

For example, (··/·//···//·/). Two points will be given to the first throw which will result in an outcome of three points from the first throw, 1 point will be given to the second throw, no points will be given to the third, since there are two consecutive slashes in the row, and etc.

Hence the number of such possible distributions is $\frac{13!}{6!7!}$ . Note that some of these distributions add more than 5 points to one of the throws, so we will have to exclude them from the upper result. Their number is

$ \frac{8!}{6!} - \frac{7!}{6!}=49$.

We subtract $\frac{7!}{6!}$ since we have counted twice the number of outcomes where the 7 points are distributed into the same throw. Thus the probability that the total number of points is 14 is

$P(A=14)= \frac{1}{6^7} . (\frac{13!}{6!.7!} - 49)$