A die is thrown $k$ times. How to determine the sum equal to $k$?

Question

A die is thrown $k$ times. Compute the probability of these resulting sums:

• $k$
• $k+1$

In this problem I assume that the die is $6$-sided and that the answer should be in terms of $k$.

Formula: $p(x)^n = x^n (1 - x^6) (1-x)^n$

Is my approach correct?

Answer 1

If you want to use probability generating functions, then a single die corresponds to $\frac16(x+x^2+x^3+x^4+x^5+x^6)$ and so for $k$ dice: $$\left(\frac{x+x^2+x^3+x^4+x^5+x^6}{6}\right)^k=\left(\frac{x(1-x^6)}{6(1-x)}\right)^k$$

If you expand this you get $$\tfrac{1}{6^k}x^k +\tfrac{k}{6^k}x^{k+1}+\tfrac{k(k+1)}{2\times 6^k}x^{k+2}+\cdots + \tfrac{k(k+1)}{2\times 6^k}x^{6k-2}+\tfrac{k}{6^k}x^{6k-1}+\tfrac{1}{6^k}x^{6k}$$

so from the coefficients of the first two terms

• the probability of getting $k$ is $\frac{1}{6^k}$
• the probability of getting $k+1$ is $\frac{k}{6^k}$

Answer 2

The only way to get $k$ as an answer is if all rolls are $1$, so the probability is $\frac{1}{6^k}$.

To get $k+1$ all rolls must be $1$ except for one roll which is a $2$; there are therefore $k$ different ways to do this and the probability is $\frac{k}{6^k}$.