TITLE: How many times to roll a die before getting two consecutive sixes?
QUESTION [35 upvotes]: Basically, on average, how many times do you have to roll a fair six-sided die before getting two consecutive sixes?

REPLY [4 votes]: Adding up to Andre's answer. The framework proposed in the solution can be generalized to obtain a formula of expected time as of $t = 0$ for $n$ consecutive throws of any number in the dice as follows. Let $a_1$ = expected additional waiting time if we have not just tossed the number of interest (say, six), $a_{n}$ = expected additional waiting time given we have tossed $n-1$ sixes at $t = n-1$. Then based on Andre's framework, we can solve the following system of equations recursively to obtain a formula of the expected time to obtain $n$ consecutive throws of any number in the dice:
$$a_1 = 1 + \frac{5}{6}a_1 + \frac{1}{6}a_2$$
$$a_2 = 1 + \frac{5}{6}a_1 + \frac{1}{6}a_3$$
$$...$$
$$a_{n-1} = 1 + \frac{5}{6}a_1 + \frac{1}{6}a_n$$
$$a_{n} = 1 + \frac{5}{6}a_1$$
Solving recursively, we can obtain an function for the expected waiting time in terms of the number of consecutive throws we would like to observe:
$$a_1 = \sum^{n-1}_{i=0} (\frac{1}{6})^i +\frac{5}{6}a_1 \sum^{n-1}_{i=0} (\frac{1}{6})^i + (\frac{1}{6})^na_n$$
Using the properties of geometric series and substituting $a_n$, we obtain the following expression for $a_1$
$$a_1 = (6^{n+1} - 6)\frac{6}{5} + 6$$
The result indicated by Andre is obtained when we set $n = 1$, i.e. the expected waiting time for one consecutive throw. Additionally, we see that for zero consecutive throws, we are expected to throw 6 times to obtain a six, which is in line with what we know about fair dices.