Rolling six, six-sided dice. I throw six, six-sided dice together repeatedly until I get three or more sixes in a single throw. What is the probability that this takes me no more than twelve throws of the six dice........I am struggling to see what to do, I presume I must use a cumulative binomial probability table. I might have missed something obvious but really looking for some help, thank you.
 A: Find the probability that in a given throw of the six dice, you achieve your goal. This can be done with the binomial distribution by adding the probability of 3 sixes, 4 sixes, 5 sixes, and 6 sixes.
Now find the probability of this failing 12 times in a row. Call this $p$.
Then your answer is $1-p$.
A: Let $P$ denote the probability you're after.  Then
$$P=1-q^{12}$$
where $q$ is the probability that you do not get $3$ or more sixes on a single roll of the $6$ dice.  To not get $3$ or more sixes is to get either exactly $2$ sixes, exactly $1$ six, or none at all.  Thus
$$q={6\choose2}\left(1\over6\right)^2\left(5\over6\right)^4+{6\choose1}\left(1\over6\right)^1\left(5\over6\right)^5+{6\choose0}\left(1\over6\right)^0\left(5\over6\right)^6$$
Putting it all together is a matter of doing some arithmetic.
A: Each roll of the dice is a Bernoulli trial: you either roll three or more sixes or you don’t. This will have some probability $p$ of success, which you can indeed work out using the c.d.f. of a binomial distribution. The number of failures before a success, however, is distributed geometrically.
