consecutive dice throw problem Consider a fair die with 6 faces - the probability of each number appearing on any throw is equal to 1/6.
What is the minimum number of times should a player throw the die before his/her probability of
getting two consecutive sixes is greater than 1/2?
 A: using bernoilles trials , if an event is repeated $n$ times and if probability  of success is $p$ and $q$ the probability of failure . then the event happening $r$ times in $n$ trials is
$$P(r) = ^{n}C_{r}p^{r}q^{n-r}$$
here $r =2$.
as you are requiring consecutive terms $p= \frac {1}{6}*\frac{1}{6} =\frac{1}{36}$ and $q =\frac{35}{36}$. find $n$
A: This is too long for a comment, but isn't a full solution.
Let $a_n$ be the number of sequences of n throws which do not contain two consecutive $6$s and in which the last throw is not a 6, and $b_n$ be the number of sequences without a double six in which the last throw is a $6$.
A sequence of $n+1$ throws without a double six in which the last throw is a $6$ is obtained by having the first $n$ throws not end in a six, and then throwing a six - thus $$b_{n+1}=a_n$$
A sequence of $n+1$ throws without a double six and which don't end in a six can be obtained from any sequence of length $n$ without a double six by throwing $1,2,3,4,5$ hence$$a_{n+1}=5(a_n+b_n)$$
Substituting $b_n=a_{n-1}$ we obtain $$a_{n+1}=5a_n+5a_{n-1}$$
Solving the recurrence gives $a_n$ and hence $b_n$, from which $a_n+b_n$ can be determined.
The number of sequences of length $n$ is $6^n$.
