Probability mass function for the number of dice throws to obtain a second six

Question

You throw a fair six-sided die until you get two sixes. Let $X$ be the number of throws you need to obtain the second six. Write down a formula for $P(X = k)$.

What is the thinking behind the answer?

Answer 1

Say we have gotten one six in $k-1$ attempts. Based on the binomial distribution, the probability of this happening equals:

$${k-1 \choose 1} \bigg(\frac{1}{6}\bigg)^1 \bigg(\frac{5}{6}\bigg)^{k-2}$$

Then, the probability of hitting the second six on the $k^{th}$ turn equals $\frac{1}{6}$. Thus, we get:

$$P[X=k] = {k-1 \choose 1} \bigg(\frac{1}{6}\bigg)^2 \bigg(\frac{5}{6}\bigg)^{k-2} = (k-1)\frac{5^{k-2}}{6^{k}}$$

Answer 2

Suppose that the $n$th throw is the second one to return a six. You will have one six in the $n-1$ previous throws, so:

$$P(X=n)=\binom{1}{1}\dfrac16 \times \binom {n-1}{1}\dfrac{1}{6} \times\left(\dfrac{5}{6}\right)^{n-1-1}\\=\binom{n-1}{1}\left(\frac16\right)^2\left(\frac56\right)^{n-2}$$