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?

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 $n$
