A dice is rolled $7$ times. What is the probability the sum of the results is $14$?

Question

In a certain exercise I am asked to find the probability of the event in which a dice is rolled seven times and the sum of the results is $14$, given the fact that the dize is regular (this is, the probability of each of the six possible results is the same: $\frac{1}{6}$.

I think I know the approach: let $X_i$ denote the random discrete variable that assigns each roll to the result, this is, $X_i = \{1 \dots 6\}$. Then, I consider the variable $Z = X_1 + \cdots X_7$, and I want to find $P(Z = 14)$. Note that each of the $X_i$ are uniform: each result has possibility $\frac{1}{6}$.

Is the sum of $n$ uniform discrete variables also a uniform one? Is there any trick to find this number (just a hint)? I would be grateful if somebody could give me his/her thoughts about the exercise, but not the full solution, just a hint (at least for the moment).

Answer 1

Define the following generating function: $f(x)=(x+x^2+x^3+...+x^6)^7$. The powers of $x$ correspond to the face value of the die and the power of $7$ indicates the fact that it is being rolled $7$ times. What we want is $[x^{14}]f(x)$; that is, the coefficient of the $x^{14}$ term.

Begin by factoring out $x$ from the expression. We are left with $f(x)=x^7(1+x+x^2+...+x^5)^7$. The right factor can be rewritten as $(\frac{1-x^6}{1-x})^7$, so $f(x)=x^7(\frac{1-x^6}{1-x})^7$.

If $g(x)$ is a function of $x$, it is a property of generating functions that $[x^n]=x^mg(x)=[x^{n-m}]g(x)$, so our problem is reduced to finding $[x^7](1-x^6)^7(1-x)^{-7}$.

Beginning with this, we see that there are $\binom{7}{1}=7$ ways of choosing a $x^6$ term from the first factor, which means we need an $x$ term from the second. There are $\binom{-7}{1}=\binom{7+1-1}{1}=\binom{7}{1}=7$ ways of doing so. Therefore there are $7\cdot7=49$ ways of making $x^7$ if $x^6$ is taken from the first factor. We must then remember to multiply this by $-1$ — one comes from $-x^6$, another from $-x$, and the last comes from the fact that $\binom{-n}{k}=\binom{n+k-1}{k}(-1)^k$, with $k=1$.

The next case would be choosing $7$ $x$ terms from the second factor. There are $\binom{-7}{7}=\binom{7+7-1}{7}(-1)^7=-\binom{13}{7}$ ways of doing so. Note that there is no need to multiply by $-1$ this time as the $(-1)^7$ multiplied by $(-x)^7$ equals just $x^7$.

So in total, the coefficient of $x^7$ in our modified generating function is $\binom{13}{7}-7\binom{7}{1}=1667$. This forms the numerator of our probability. The denominator consists of all possible outcomes, which is $6^7=279936$. The overall probability of getting a sum of $14$ is therefore equal to $\frac{1667}{279936}\approx 0.0059$

Answer 2

The total number of possible outcomes is $6^7 = 279936$.

If $x_i$, $1 \le i \le 7$ represents one less than the number on the $i^{\text{th}}$ die, for favorable outcomes we have

$$x_1 + x_2 + \cdots + x_7 = 7, \;\;\;0 \le x_i \le 5$$

The number of whole number solutions to this equation is ${{7 + 7 - 1} \choose {7-1}} = {{13 \choose 6}} = 1716$.

But in $7$ of those solutions one variable would have the value $7$. And in $7 \times 6 = 42$ of them, one variable will have $6$ (with one other having $1$) as its value. These are invalid solutions because the $x_i$ can't exceed $5$.

So, the number of valid solutions is $1716 - (42+7) = 1667$. Hence, the required probability would be

$$\frac{1667}{279936} \approx 0.006 $$

Answer 3

We want the coefficient of $x^{14}$ in $(x+x^2+x^3+x^4+x^5+x^6)^7$

= coeff. of $x^7$ in $(1+x+x^2+x^3+x^4+x^5)^7$

= coeff. of $x^7$ in $(1-x^6)^7\frac{1}{(1-x)^7}\quad$ (sum of G.P.)

Expanding the first term, and the second (using Taylor series expansion at $x=0$), we have

$(1- 7x^6 + ...)(1 + \binom71x + \binom82x^2 +...+\binom{13}7x^7 + ..)$

= $1\cdot\binom{13}7x^7 - 7x^6\cdot\binom71x$

and Pr $ = \dfrac{\binom{13}7 - 7\cdot7}{6^7} =\dfrac{1667}{6^7}$